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1lgxgzye9 | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $${{x + 2} \over 1} = {y \over { - 2}} = {{z - 5} \over 2}$$ and $${{x - 4} \over 1} = {{y - 1} \over 2} = {{z + 3} \over 0}$$ is :</p> | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "9"}] | ["D"] | null | Given, the lines are
<br/><br/>$$
\begin{aligned}
\frac{x+2}{1} & =\frac{y}{-2}=\frac{z-5}{2} ~~~~..........(i)\\\\
\text { and } \frac{x-4}{1} & =\frac{y-1}{2}=\frac{z+3}{0} ~~~~..........(ii)
\end{aligned}
$$
<br/><br/>Formula for shortest distance between two skew-lines,
<br/><br/>$$
\begin{aligned}
S D & =\left|\fr... | mcq | jee-main-2023-online-10th-april-morning-shift |
1lgzydtfy | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $$\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$$ and $$\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$$ is :</p> | [{"identifier": "A", "content": "$$3 \\sqrt{6}$$"}, {"identifier": "B", "content": "$$6 \\sqrt{2}$$"}, {"identifier": "C", "content": "$$6 \\sqrt{3}$$"}, {"identifier": "D", "content": "$$2 \\sqrt{6}$$"}] | ["A"] | null | The given lines are
<br/><br/>$$\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$$ and $$\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$$
<br/><br/>$$
\begin{aligned}
& \text { So, } \vec{b}_1=4 \hat{i}+5 \hat{j}+3 \hat{k} \\\\
& \vec{b}_2=3 \hat{i}+4 \hat{j}+2 \hat{k} \\\\
& \vec{a}_1=4 \hat{i}-2 \hat{j}-3 \hat{k} \\\\
&\vec{a}_2... | mcq | jee-main-2023-online-8th-april-morning-shift |
1lh2197tv | maths | 3d-geometry | lines-in-space | <p>One vertex of a rectangular parallelopiped is at the origin $$\mathrm{O}$$ and the lengths of its edges along $$x, y$$ and $$z$$ axes are $$3,4$$ and $$5$$ units respectively. Let $$\mathrm{P}$$ be the vertex $$(3,4,5)$$. Then the shortest distance between the diagonal OP and an edge parallel to $$\mathrm{z}$$ axis,... | [{"identifier": "A", "content": "$$\\frac{12}{\\sqrt{5}}$$"}, {"identifier": "B", "content": "$$12 \\sqrt{5}$$"}, {"identifier": "C", "content": "$$\\frac{12}{5}$$"}, {"identifier": "D", "content": "$$\\frac{12}{5 \\sqrt{5}}$$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnq8po6g/d220c54e-806e-4b56-ac43-a5312272c160/0b7f4280-6aad-11ee-822f-a9d01571f2aa/file-6y3zli1lnq8po6h.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnq8po6g/d220c54e-806e-4b56-ac43-a5312272c160/0b7f4280-6aad-11ee-82... | mcq | jee-main-2023-online-6th-april-morning-shift |
1lh2z2249 | maths | 3d-geometry | lines-in-space | <p>If the lines $$\frac{x-1}{2}=\frac{2-y}{-3}=\frac{z-3}{\alpha}$$ and $$\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta}$$ intersect, then the magnitude of the minimum value of $$8 \alpha \beta$$ is _____________.</p> | [] | null | 18 | Given, lines
<br/><br/>$$
\begin{aligned}
& \frac{x-1}{2} =\frac{2-y}{-3}=\frac{z-3}{\alpha} \\\\
&\Rightarrow \frac{x-1}{2} =\frac{y-2}{3}=\frac{z-3}{\alpha}=\lambda .........(i)
\end{aligned}
$$
<br/><br/>Any point on the line (i)
<br/><br/>$$
x=2 \lambda+1, y=3 \lambda+2, z=\alpha \lambda+3
$$
<br/><br/>and line $\... | integer | jee-main-2023-online-6th-april-evening-shift |
lsamoe4e | maths | 3d-geometry | lines-in-space | Let $\mathrm{P}$ and $\mathrm{Q}$ be the points on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ which are at a distance of 6 units from the point $\mathrm{R}(1,2,3)$. If the centroid of the triangle PQR is $(\alpha, \beta, \gamma)$, then $\alpha^2+\beta^2+\gamma^2$ is : | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "26"}, {"identifier": "D", "content": "36"}] | ["A"] | null | Any point on line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ <br/><br/>can be taken as $(8 \lambda-3,2 \lambda+4,2 \lambda-1)$
<br/><br/> If at a distance of 6 units from $R(1,2,3)$
<br/><br/>$$
\begin{aligned}
& \Rightarrow(8 \lambda-3-1)^2+(2 \lambda+4-2)^2+(2 \lambda-1-3)^2=36 \\\\
& \left.\Rightarrow \lambda^2-\la... | mcq | jee-main-2024-online-1st-february-evening-shift |
lsamrrrt | maths | 3d-geometry | lines-in-space | If the mirror image of the point $P(3,4,9)$ in the line
<br/><br/>$\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$, then 14 $(\alpha+\beta+\gamma)$ is : | [{"identifier": "A", "content": "102"}, {"identifier": "B", "content": "138"}, {"identifier": "C", "content": "132"}, {"identifier": "D", "content": "108"}] | ["D"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsohd1pb/c1eafb9f-de3f-4c37-82e0-908506938702/622ee6f0-ccb2-11ee-aa98-13f456b8f7af/file-6y3zli1lsohd1pc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsohd1pb/c1eafb9f-de3f-4c37-82e0-908506938702/622ee6f0-ccb2-11ee-aa... | mcq | jee-main-2024-online-1st-february-evening-shift |
lsaptbc8 | maths | 3d-geometry | lines-in-space | If the shortest distance between the lines <br/><br/>$\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is 1 , then the sum of all possible values of $\lambda$ is : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$2 \\sqrt{3}$"}, {"identifier": "C", "content": "$3 \\sqrt{3}$"}, {"identifier": "D", "content": "$-2 \\sqrt{3}$"}] | ["B"] | null | <p>Given the two lines:</p>
<p>$$ L_1: \frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1} $$</p>
<p>$$ L_2: \frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1} $$</p>
<p>We observe that these lines are not parallel as their directional vectors are not proportional. The directional vector for $L_1$ is $(-2,1,1)$ ... | mcq | jee-main-2024-online-1st-february-morning-shift |
lsaqawqo | maths | 3d-geometry | lines-in-space | Let the line of the shortest distance between the lines
<br/><br/>$$
\begin{aligned}
& \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \text { and } \\\\
& \mathrm{L}_2: \overrightarrow{\mathrm{r}}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{... | [] | null | 21 | $\begin{array}{ll}L_1 \equiv \vec{r}=(1,2,3)+\lambda(1,-1,1) & \left(\vec{r}=\vec{a}_1+\lambda \vec{b}_1\right) \\\\ L_2 \equiv \vec{r}=(4,5,6)+\mu(1,1,-1) & \left(\vec{r}=\vec{a}_2+\lambda \vec{b}_2\right)\end{array}$
<br/><br/>$P \equiv(\lambda+1,-\lambda+2, \lambda+3)$
<br/><br/>$Q \equiv(\mu+4, \mu+5,-\mu+6)$
<br/>... | integer | jee-main-2024-online-1st-february-morning-shift |
lsbklvv1 | maths | 3d-geometry | lines-in-space | The distance, of the point $(7,-2,11)$ from the line <br/><br/>$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is : | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "18"}, {"identifier": "C", "content": "21"}, {"identifier": "D", "content": "14"}] | ["D"] | null | <p>$$\mathrm{B}=(2 \lambda+7,-3 \lambda-2,6 \lambda+11)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1c0zyx/3ef0ec08-6ff8-4ca7-97e0-1c5c0fde9c3c/b5ee7090-d3c3-11ee-a50b-bb659a2e1d74/file-1lt1c0zyy.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1c0zyx/3ef0ec08-6f... | mcq | jee-main-2024-online-27th-january-morning-shift |
lsbkxlh7 | maths | 3d-geometry | lines-in-space | If the shortest distance between the lines <br/><br/>$\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is : | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "8"}] | ["D"] | null | <p>$$\begin{aligned}
& \frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3} \\
& \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}
\end{aligned}$$</p>
<p>the shortest distance between the lines</p>
<p>$$=\left|\frac{(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \cdot\left(\overrightarrow{\mathrm{d}_1} \times \overrighta... | mcq | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lscn35vz | maths | 3d-geometry | lines-in-space | <p>Let the image of the point $$(1,0,7)$$ in the line $$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$$ be the point $$(\alpha, \beta, \gamma)$$. Then which one of the following points lies on the line passing through $$(\alpha, \beta, \gamma)$$ and making angles $$\frac{2 \pi}{3}$$ and $$\frac{3 \pi}{4}$$ with $$y$$-axis an... | [{"identifier": "A", "content": "$$(1,-2,1+\\sqrt{2})$$\n"}, {"identifier": "B", "content": "$$(3,-4,3+2 \\sqrt{2})$$\n"}, {"identifier": "C", "content": "$$(3,4,3-2 \\sqrt{2})$$\n"}, {"identifier": "D", "content": "$$(1,2,1-\\sqrt{2})$$"}] | ["C"] | null | <p>$$\mathrm{L}_1=\frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}=\lambda$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1toimr/45988a05-1bb2-47f1-9316-43cb63fa6ccd/bf271320-d408-11ee-b9d5-0585032231f0/file-1lt1toimt.png?format=png" data-orsrc="https://app-content.cdn.exa... | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lscogowm | maths | 3d-geometry | lines-in-space | <p>The lines $$\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$$ and $$\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$$ intersect at the point $$P$$. If the distance of $$\mathrm{P}$$ from the line $$\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$$ is $$l$$, then $$14 l^2$$ is equal to __________.</p> | [] | null | 108 | <p>$$\begin{aligned}
& \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-7}{8}=\lambda \\
& \frac{\mathrm{x}+3}{4}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}+2}{1}=\mathrm{k} \\
& \Rightarrow \lambda+2=4 \mathrm{k}-3 \\
& -\lambda=3 \mathrm{k}-2 \\
& \Rightarrow \mathrm{k}=1, \lambda=-1 \\
... | integer | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsd3h061 | maths | 3d-geometry | lines-in-space | <p>Let $$(\alpha, \beta, \gamma)$$ be the mirror image of the point $$(2,3,5)$$ in the line $$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$. Then, $$2 \alpha+3 \beta+4 \gamma$$ is equal to</p> | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "31"}, {"identifier": "C", "content": "33"}, {"identifier": "D", "content": "34"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjwr9eg/7ae9597b-3dbe-4745-92d2-f1eee5f94c62/ab25ed80-ca2e-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwr9eh.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsjwr9eg/7ae9597b-3dbe-4745-92d2-f1eee5f94c62/ab25ed80-ca2e-11ee... | mcq | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lsd4flem | maths | 3d-geometry | lines-in-space | <p>The shortest distance, between lines $$L_1$$ and $$L_2$$, where $$L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$$ and $$L_2$$ is the line, passing through the points $$\mathrm{A}(-4,4,3), \mathrm{B}(-1,6,3)$$ and perpendicular to the line $$\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$$, is</p> | [{"identifier": "A", "content": "$$\\frac{141}{\\sqrt{221}}$$\n"}, {"identifier": "B", "content": "$$\\frac{24}{\\sqrt{117}}$$\n"}, {"identifier": "C", "content": "$$\\frac{42}{\\sqrt{117}}$$\n"}, {"identifier": "D", "content": "$$\\frac{121}{\\sqrt{221}}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \mathrm{L}_2=\frac{\mathrm{x}+4}{3}=\frac{\mathrm{y}-4}{2}=\frac{\mathrm{z}-3}{0} \\
& \therefore \mathrm{S} . \mathrm{D}=\frac{\left|\begin{array}{ccc}
\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\
2 & -3 & 2 \\
3 & 2 & 0
\end{array}\right|}{\left|\overrigh... | mcq | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lsd5a4ux | maths | 3d-geometry | lines-in-space | <p>A line passes through $$A(4,-6,-2)$$ and $$B(16,-2,4)$$. The point $$P(a, b, c)$$, where $$a, b, c$$ are non-negative integers, on the line $$A B$$ lies at a distance of 21 units, from the point $$A$$. The distance between the points $$P(a, b, c)$$ and $$Q(4,-12,3)$$ is equal to __________.</p> | [] | null | 22 | <p>$$\begin{aligned}
& \frac{x-4}{12}=\frac{x+6}{4}=\frac{z+2}{6} \\
& \frac{x-4}{\frac{6}{7}}=\frac{y+6}{\frac{2}{7}}=\frac{z+2}{\frac{3}{7}}=21 \\
& \left(21 \times \frac{6}{7}+4, \frac{2}{7} \times 21-6, \frac{3}{7} \times 21-2\right) \\
& =(22,0,7)=(a, b, c) \\
& \therefore \sqrt{324+144+16}=22
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lse5wwlu | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{Q}$$ and $$\mathrm{R}$$ be the feet of perpendiculars from the point $$\mathrm{P}(a, a, a)$$ on the lines $$x=y, z=1$$ and $$x=-y, z=-1$$ respectively. If $$\angle \mathrm{QPR}$$ is a right angle, then $$12 a^2$$ is equal to _________.</p> | [] | null | 12 | <p>$$\begin{aligned}
& \frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r \rightarrow Q(r, r, 1) \\
& \frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k \rightarrow R(k,-k,-1) \\
& \overline{P Q}=(a-r) \hat{i}+(a-r) \hat{j}+(a-1) \hat{k} \\
& a=r+a-r=0 \\
& 2 a=2 r \rightarrow a=r \\
& \overline{P R}=(a-k) i+(a+k) \hat{j}+(a+1) \hat{k} \\
... | integer | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lsf08uic | maths | 3d-geometry | lines-in-space | <p>Let $$P Q R$$ be a triangle with $$R(-1,4,2)$$. Suppose $$M(2,1,2)$$ is the mid point of $$\mathrm{PQ}$$. The distance of the centroid of $$\triangle \mathrm{PQR}$$ from the point of intersection of the lines $$\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$$ and $$\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$$ is</p> | [{"identifier": "A", "content": "69"}, {"identifier": "B", "content": "$$\\sqrt{99}$$"}, {"identifier": "C", "content": "$$\\sqrt{69}$$"}, {"identifier": "D", "content": "9"}] | ["C"] | null | <p>Centroid $$G$$ divides MR in $$1: 2$$</p>
<p>$$\mathrm{G}(1,2,2)$$</p>
<p>Point of intersection $$A$$ of given lines is $$(2,-6,0)$$</p>
<p>$$\mathrm{AG}=\sqrt{69}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsf0lnlm | maths | 3d-geometry | lines-in-space | <p>A line with direction ratios $$2,1,2$$ meets the lines $$x=y+2=z$$ and $$x+2=2 y=2 z$$ respectively at the points $$\mathrm{P}$$ and $$\mathrm{Q}$$. If the length of the perpendicular from the point $$(1,2,12)$$ to the line $$\mathrm{PQ}$$ is $$l$$, then $$l^2$$ is __________.</p> | [] | null | 65 | <p>Let $$\mathrm{P}(\mathrm{t}, \mathrm{t}-2, \mathrm{t})$$ and $$\mathrm{Q}(2 \mathrm{~s}-2, \mathrm{~s}, \mathrm{~s})$$</p>
<p>D.R's of PQ are 2, 1, 2</p>
<p>$$\begin{aligned}
& \frac{2 \mathrm{~s}-2-\mathrm{t}}{2}=\frac{\mathrm{s}-\mathrm{t}+2}{1}=\frac{\mathrm{s}-\mathrm{t}}{2} \\
& \Rightarrow \mathrm{t}=6... | integer | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsfkztkt | maths | 3d-geometry | lines-in-space | <p>Let O be the origin, and M and $$\mathrm{N}$$ be the points on the lines $$\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$$ and $$\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$$ respectively such that $$\mathrm{MN}$$ is the shortest distance between the given lines. Then $$\overrightarrow{O M} \cdot \overrightarrow{O N}$$ ... | [] | null | 9 | <p>$$\mathrm{L}_1: \frac{\mathrm{x}-5}{4}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{3}=\lambda \quad \operatorname{drs}(4,1,3)=\mathrm{b}_1$$</p>
<p>$$\begin{aligned}
& \mathrm{M}(4 \lambda+5, \lambda+4,3 \lambda+5) \\
& \mathrm{L}_2: \frac{\mathrm{x}+8}{12}=\frac{\mathrm{y}+2}{5}=\frac{\mathrm{z}+11}{9}=\mu \\
& \mat... | integer | jee-main-2024-online-29th-january-evening-shift |
1lsg4acyi | maths | 3d-geometry | lines-in-space | <p>Let $$L_1: \vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in \mathbb{R}$$,</p>
<p>$$L_2: \vec{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k}), \mu \in \mathbb{R} \text {, and } L_3: \vec{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k}), \delta \in \mathbb{R}$$</p>
<p>be three lines... | [{"identifier": "A", "content": "$$(1,7,-4)$$\n"}, {"identifier": "B", "content": "$$(1,-7,4)$$\n"}, {"identifier": "C", "content": "$$(-1,7,4)$$\n"}, {"identifier": "D", "content": "$$(-, 1-7,4)$$"}] | ["C"] | null | <p>$$\mathrm{L}_1 \perp \mathrm{L}_2 \quad \mathrm{~L}_3 \perp \mathrm{L}_1, \mathrm{~L}_2$$</p>
<p>$$\begin{aligned}
& 3-1+2 \mathrm{P}=0 \\
& \mathrm{P}=-1 \\
& \left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
1 & -1 & 2 \\
3 & 1 & -1
\end{array}\right|=-\hat{\mathrm{i}}+7 \hat{\math... | mcq | jee-main-2024-online-30th-january-evening-shift |
1lsg550xe | maths | 3d-geometry | lines-in-space | <p>Let a line passing through the point $$(-1,2,3)$$ intersect the lines $$L_1: \frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2}$$ at $$M(\alpha, \beta, \gamma)$$ and $$L_2: \frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}$$ at $$N(a, b, c)$$. Then, the value of $$\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}$$ equals __________.</p> | [] | null | 196 | <p>$$\begin{aligned}
& \mathrm{M}(3 \lambda+1,2 \lambda+2,-2 \lambda-1) \quad \therefore \alpha+\beta+\gamma=3 \lambda+2 \\
& \mathrm{~N}(-3 \mu-2,-2 \mu+2,4 \mu+1) \quad \therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=-\mu+1
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3... | integer | jee-main-2024-online-30th-january-evening-shift |
1lsg95okx | maths | 3d-geometry | lines-in-space | <p>Let $$(\alpha, \beta, \gamma)$$ be the foot of perpendicular from the point $$(1,2,3)$$ on the line $$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$$. Then $$19(\alpha+\beta+\gamma)$$ is equal to :</p> | [{"identifier": "A", "content": "99"}, {"identifier": "B", "content": "102"}, {"identifier": "C", "content": "101"}, {"identifier": "D", "content": "100"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqnidnf/3617c118-3469-40cf-b021-2c95fa43dbe8/fe8826b0-cde3-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnidng.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqnidnf/3617c118-3469-40cf-b021-2c95fa43dbe8/fe8826b0-cde3-11ee... | mcq | jee-main-2024-online-30th-january-morning-shift |
1lsgclxib | maths | 3d-geometry | lines-in-space | <p>If $$\mathrm{d}_1$$ is the shortest distance between the lines $$x+1=2 y=-12 z, x=y+2=6 z-6$$ and $$\mathrm{d}_2$$ is the shortest distance between the lines $$\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$$, then the value of $$\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}... | [] | null | 16 | <p>$$\mathrm{L}_1: \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}}{1 / 2}=\frac{\mathrm{z}}{-1 / 12}, \mathrm{~L}_2: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}+2}{1}=\frac{\mathrm{z}-1}{\frac{1}{6}}$$</p>
<p>$$\mathrm{d}_1=$$ shortest distance between $$\mathrm{L}_1 ~\& \mathrm{~L}_2$$</p>
<p>$$\begin{aligned}
& =\left|\frac{\left... | integer | jee-main-2024-online-30th-january-morning-shift |
luxwcs1z | maths | 3d-geometry | lines-in-space | <p>Consider the line $$\mathrm{L}$$ passing through the points $$(1,2,3)$$ and $$(2,3,5)$$. The distance of the point $$\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$$ from the line $$\mathrm{L}$$ along the line $$\frac{3 x-11}{2}=\frac{3 y-11}{1}=\frac{3 z-19}{2}$$ is equal to</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "4"}] | ["B"] | null | <p>$$L: \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{2}=\mu$$</p>
<p>Measured along $$L_2: \frac{x-\frac{11}{3}}{\frac{2}{3}}=\frac{y-\frac{11}{3}}{\frac{1}{3}}=\frac{z-\frac{19}{3}}{\frac{2}{3}}=\lambda$$</p>
<p>Any point on $$L_1:(\mu+1, \mu+2,2 \mu+3)$$</p>
<p>Any point on $$L_2\left(\frac{2}{3} \lambda+\frac{11}{3}, \fra... | mcq | jee-main-2024-online-9th-april-evening-shift |
luxwden8 | maths | 3d-geometry | lines-in-space | <p>The square of the distance of the image of the point $$(6,1,5)$$ in the line $$\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4}$$, from the origin is __________.</p> | [] | null | 62 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw1my772/a7f8567c-57ab-46bd-a498-b733b7e930a6/12a4b0e0-0f53-11ef-91cd-f19f7dc20f18/file-1lw1my773.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw1my772/a7f8567c-57ab-46bd-a498-b733b7e930a6/12a4b0e0-0f53-11ef-91cd-f19f7dc20f18... | integer | jee-main-2024-online-9th-april-evening-shift |
luy6z5hh | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $$\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$$ and $$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$$ is:</p> | [{"identifier": "A", "content": "$$\\frac{185}{\\sqrt{563}}$$\n"}, {"identifier": "B", "content": "$$\\frac{187}{\\sqrt{563}}$$\n"}, {"identifier": "C", "content": "$$\\frac{178}{\\sqrt{563}}$$\n"}, {"identifier": "D", "content": "$$\\frac{179}{\\sqrt{563}}$$"}] | ["B"] | null | <p>Given lines are</p>
<p>$$\frac{x-3}{4}=\frac{y-(-7)}{-11}=\frac{z-1}{5}$$ and</p>
<p>$$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z-(-2)}{1}$$</p>
<p>Shortest distance between two lines,</p>
<p>$$\begin{aligned}
& d=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\left... | mcq | jee-main-2024-online-9th-april-morning-shift |
luy6z4qf | maths | 3d-geometry | lines-in-space | <p>Let the line $$\mathrm{L}$$ intersect the lines $$x-2=-y=z-1,2(x+1)=2(y-1)=z+1$$ and be parallel to the line $$\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$$. Then which of the following points lies on $$\mathrm{L}$$ ?</p> | [{"identifier": "A", "content": "$$\\left(-\\frac{1}{3}, 1,-1\\right)$$\n"}, {"identifier": "B", "content": "$$\\left(-\\frac{1}{3},-1,1\\right)$$\n"}, {"identifier": "C", "content": "$$\\left(-\\frac{1}{3},-1,-1\\right)$$\n"}, {"identifier": "D", "content": "$$\\left(-\\frac{1}{3}, 1,1\\right)$$"}] | ["A"] | null | <p>$$\begin{aligned}
& L_1: \frac{x-2}{1}=\frac{y}{-1}=\frac{z-1}{1}=\lambda \\
& L_2: \frac{x+1}{(1 / 2)}=\frac{y-1}{(1 / 2)}=\frac{z+1}{1}=\mu
\end{aligned}$$</p>
<p>Any point of $$L_1$$ and $$L_2$$ will be $$(\lambda+2,-\lambda, \lambda+1)$$ and
$$\left(\frac{\mu}{2}-1, \frac{\mu}{2}+1, \mu-1\right)$$</p>
<p>Now Dr ... | mcq | jee-main-2024-online-9th-april-morning-shift |
lv0vxcns | maths | 3d-geometry | lines-in-space | <p>Let the point, on the line passing through the points $$P(1,-2,3)$$ and $$Q(5,-4,7)$$, farther from the origin and at a distance of 9 units from the point $$P$$, be $$(\alpha, \beta, \gamma)$$. Then $$\alpha^2+\beta^2+\gamma^2$$ is equal to :</p> | [{"identifier": "A", "content": "150"}, {"identifier": "B", "content": "155"}, {"identifier": "C", "content": "160"}, {"identifier": "D", "content": "165"}] | ["B"] | null | <p>Line through $$P Q$$</p>
<p>$$\frac{x-1}{4}=\frac{y+2}{-2}=\frac{z-3}{4}$$</p>
<p>Any point on $$P Q$$. be $$R(4 \lambda+1,-2 \lambda-2,4 \lambda+3)$$</p>
</p>$$P R=9$$ unit</p>
<p>$$(P R)^2=81$$</p>
<p>$$(4 \lambda+1-1)^2+(-2 \lambda-2+2)^2+(4 \lambda+3-3)^2=81$$</p>
<p>$$16 \lambda^2+4 \lambda^2+16 \lambda^2=81$$<... | mcq | jee-main-2024-online-4th-april-morning-shift |
lv2er3nv | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{P}$$ be the point of intersection of the lines $$\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$$ and $$\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$$. Then, the shortest distance of $$\mathrm{P}$$ from the line $$4 x=2 y=z$$ is </p> | [{"identifier": "A", "content": "$$\\frac{3 \\sqrt{14}}{7}$$\n"}, {"identifier": "B", "content": "$$\\frac{5 \\sqrt{14}}{7}$$\n"}, {"identifier": "C", "content": "$$\\frac{\\sqrt{14}}{7}$$\n"}, {"identifier": "D", "content": "$$\\frac{6 \\sqrt{14}}{7}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& L_1: \frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1} \\
& L_2: \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}
\end{aligned}$$</p>
<p>Point of intersection of $$L_1$$ and $$L_2$$ is $$(-1,1,-1)$$</p>
<p>Distance of point $$P$$ from $$L_3: 4 x=2 y=z$$</p>
<p>$$L_3: \frac{x}{\frac{1}{4}}=\frac{y}{\frac... | mcq | jee-main-2024-online-4th-april-evening-shift |
lv2er9tw | maths | 3d-geometry | lines-in-space | <p>Consider a line $$\mathrm{L}$$ passing through the points $$\mathrm{P}(1,2,1)$$ and $$\mathrm{Q}(2,1,-1)$$. If the mirror image of the point $$\mathrm{A}(2,2,2)$$ in the line $$\mathrm{L}$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha+\beta+6 \gamma$$ is equal to __________.</p> | [] | null | 6 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwhk80oi/a0341b0b-ecfa-449a-9d8f-e828635e1020/f2b47520-1814-11ef-a08c-4f00e8f9a4da/file-1lwhk80oj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwhk80oi/a0341b0b-ecfa-449a-9d8f-e828635e1020/f2b47520-1814-11ef-a08c-4f00e8f9a4da... | integer | jee-main-2024-online-4th-april-evening-shift |
lv3vef5l | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the lines $$\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$$ and $$\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$$ is $$\frac{13}{\sqrt{29}}$$, then a value of $$\lambda$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{13}{25}$$\n"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$-$$1\n"}, {"identifier": "D", "content": "$$-\\frac{13}{25}$$"}] | ["B"] | null | <p>$$\begin{aligned}
& \vec{l}_1=2 \hat{i}+3 \hat{j}+4 \hat{k} \\
& \vec{l}_2=4 \hat{i}+6 \hat{j}+8 \hat{k}
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4lkbz1/158cd7a8-1619-430b-bf76-8b2ddd399871/01d314d0-10f4-11ef-8553-fdfc6347789d/file-1lw4lkbz2.png?format=png" da... | mcq | jee-main-2024-online-8th-april-evening-shift |
lv3vefgs | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{P}(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{Q}(1,6,4)$$ in the line $$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$$. Then $$2 \alpha+\beta+\gamma$$ is equal to ________</p> | [] | null | 11 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4mrnop/6ea8698b-9387-47c8-8aca-29425e7a83aa/b6aa6990-10f8-11ef-b50f-395513bc7af2/file-1lw4mrnoq.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4mrnop/6ea8698b-9387-47c8-8aca-29425e7a83aa/b6aa6990-10f8-11ef-b50f-395513bc7af2... | integer | jee-main-2024-online-8th-april-evening-shift |
lv5gsxy6 | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the lines</p>
<p>$$\begin{array}{ll}
L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, & \lambda \in \mathbb{R} \\
L_2: \vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, & \mu \in \mathbb{R}
\end{array}$$</p>
<p>is $$\frac{m}{\sqrt{n}}$$, ... | [{"identifier": "A", "content": "384"}, {"identifier": "B", "content": "387"}, {"identifier": "C", "content": "390"}, {"identifier": "D", "content": "377"}] | ["B"] | null | <p>$$\begin{aligned}
& L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k} \\
& L_1=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(\hat{i}-3 \hat{j}+4 \hat{k}) \\
& L_2: \vec{r}=2 \hat{i}+3 \hat{j}+5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k}) \\
& \vec{a}_1=2 \hat{i}+\hat{j}+3 \hat{k} \\
& \vec{a}_2=2 \hat{... | mcq | jee-main-2024-online-8th-april-morning-shift |
lv7v3jvb | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{d}$$ be the distance of the point of intersection of the lines $$\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$$ and $$\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$$ from the point $$(7,8,9)$$. Then $$\mathrm{d}^2+6$$ is equal to :</p> | [{"identifier": "A", "content": "75"}, {"identifier": "B", "content": "78"}, {"identifier": "C", "content": "72"}, {"identifier": "D", "content": "69"}] | ["A"] | null | <p>$$\begin{aligned}
& P_1:(3 k-6,2 k, k-1) \\
& P_2(4 \alpha+7,3 \alpha+9,2 \alpha+4) \\
& P_1 \equiv P_2 \\
& 3 k-6=4 \alpha+7 \Rightarrow 3 k-4 \alpha=13 \\
& 2 k=3 \alpha+9 \Rightarrow 2 k-3 \alpha=9 \\
& \therefore k=3, \alpha=-1 \\
& \therefore P_1:(3,6,2)
\end{aligned}$$</p>
<p>Distance of $$(3,6,2)$$ and $$(7,8... | mcq | jee-main-2024-online-5th-april-morning-shift |
lv9s1zwj | maths | 3d-geometry | lines-in-space | <p>Let $$(\alpha, \beta, \gamma)$$ be the image of the point $$(8,5,7)$$ in the line $$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}$$. Then $$\alpha+\beta+\gamma$$ is equal to :</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "20"}, {"identifier": "C", "content": "18"}, {"identifier": "D", "content": "14"}] | ["D"] | null | <p>$$(x, y, z) \equiv(2 \lambda+1,3 \lambda+1,5 \lambda+2)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwenv6x5/39fb95d9-cd80-4c68-91b8-1f5c0166f573/d8931b90-167c-11ef-9070-f523f4c6bd4b/file-1lwenv6x6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwenv6x5/39fb95d9... | mcq | jee-main-2024-online-5th-april-evening-shift |
lv9s20kw | maths | 3d-geometry | lines-in-space | <p>Let the point $$(-1, \alpha, \beta)$$ lie on the line of the shortest distance between the lines $$\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$$ and $$\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$$. Then $$(\alpha-\beta)^2$$ is equal to _________.</p> | [] | null | 25 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lweuubn4/2752e8cb-54f7-4c91-8212-691518b0463e/20511700-1698-11ef-ad74-71e0a1829759/file-1lweuubn5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lweuubn4/2752e8cb-54f7-4c91-8212-691518b0463e/20511700-1698-11ef-ad74-71e0a1829759... | integer | jee-main-2024-online-5th-april-evening-shift |
lvb294m1 | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{P}(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{Q}(3,-3,1)$$ in the line $$\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}$$ and $$\mathrm{R}$$ be the point $$(2,5,-1)$$. If the area of the triangle $$\mathrm{PQR}$$ is $$\lambda$$ and $$\lambda^2=14 \mathrm{~K}$$, then $$\mathrm{K}$$ is equa... | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "81"}, {"identifier": "C", "content": "72"}, {"identifier": "D", "content": "36"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwad7urv/960c0402-e139-4dd9-89a4-53df9b260717/11b135b0-1420-11ef-90c1-6758c45d5635/file-1lwad7urw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwad7urv/960c0402-e139-4dd9-89a4-53df9b260717/11b135b0-1420-11ef-90c1-6758c45d5635... | mcq | jee-main-2024-online-6th-april-evening-shift |
lvb29516 | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the lines $$\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$$ and $$\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$$ is $$\frac{44}{\sqrt{30}}$$, then the largest possible value of $$|\lambda|$$ is equal to _________.</p> | [] | null | 43 | <p>$$\begin{aligned}
& L_1: \frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1} \\
& L_2: \frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& n_1 \times n_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -1 & 1 \\
-3 & 2 & 4
\end{array}\right| \\
& =-6 \hat{i}-15 \hat{j}+3 \h... | integer | jee-main-2024-online-6th-april-evening-shift |
lvc58e18 | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $$\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$$ and $$\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$$ is</p> | [{"identifier": "A", "content": "$$8 \\sqrt{3}$$\n"}, {"identifier": "B", "content": "$$6 \\sqrt{3}$$\n"}, {"identifier": "C", "content": "$$5 \\sqrt{3}$$\n"}, {"identifier": "D", "content": "$$4 \\sqrt{3}$$"}] | ["D"] | null | <p>Given two lines are represented as:</p>
<p>$ \frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5} $</p>
<p>and</p>
<p>$ \frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3} $</p>
<p>The formula for the shortest distance between two lines is:</p>
<p>$ d = \frac{|\left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \tim... | mcq | jee-main-2024-online-6th-april-morning-shift |
lvc58e9z | maths | 3d-geometry | lines-in-space | <p>Let $$P$$ be the point $$(10,-2,-1)$$ and $$Q$$ be the foot of the perpendicular drawn from the point $$R(1,7,6)$$ on the line passing through the points $$(2,-5,11)$$ and $$(-6,7,-5)$$. Then the length of the line segment $$P Q$$ is equal to _________.</p> | [] | null | 13 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwd5pdpk/204a0778-5148-47b3-a659-091fdb696421/09fd8280-15a9-11ef-99cc-63deb36dcc4a/file-1lwd5pdpl.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwd5pdpk/204a0778-5148-47b3-a659-091fdb696421/09fd8280-15a9-11ef-99cc-63deb36dcc4a... | integer | jee-main-2024-online-6th-april-morning-shift |
ZZ2mbSw3xnje5rne | maths | 3d-geometry | plane-in-space | The $$d.r.$$ of normal to the plane through $$(1, 0, 0), (0, 1, 0)$$ which makes an angle $$\pi /4$$ with plane $$x+y=3$$ are : | [{"identifier": "A", "content": "$$1,\\sqrt 2 ,1$$ "}, {"identifier": "B", "content": "$$1,1,\\sqrt 2 $$ "}, {"identifier": "C", "content": "$$1, 1, 2$$"}, {"identifier": "D", "content": "$$\\sqrt 2 ,1,1$$ "}] | ["B"] | null | Equation of plane through $$\left( {1,0,0} \right)$$ is
<br><br>$$a\left( {x - 1} \right) + by + cz = 0\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>$$(i)$$ passes through $$\left( {0,1,0} \right).$$
<br><br>$$ - a + b = 0 \Rightarrow b = a;$$
<br><br>Also, $$\cos {45^ \circ }$$
<br><br>$$ = {{a + a} \over {\sqrt ... | mcq | aieee-2002 |
tklE0iEL4MC3JSAh | maths | 3d-geometry | plane-in-space | The shortest distance from the plane $$12x+4y+3z=327$$ to the sphere <br/><br/>$${x^2} + {y^2} + {z^2} + 4x - 2y - 6z = 155$$ is | [{"identifier": "A", "content": "$$39$$ "}, {"identifier": "B", "content": "$$26$$ "}, {"identifier": "C", "content": "$$11{4 \\over {13}}$$ "}, {"identifier": "D", "content": "$$13$$"}] | ["D"] | null | Shortest distance $$=$$ perpendicular distance between the plane and sphere $$=$$ distance of plane from center of sphere $$-$$ radius
<br><br>$$ = \left| {{{ - 2 \times 12 + 4 \times 1 + 3 \times 3 - 327} \over {\sqrt {144 + 9 + 16} }}} \right| - \sqrt {4 + 1 + 9 + 155} $$
<br><br>$$ = 26 - 13 = 13$$ | mcq | aieee-2003 |
CFmX8122sUhvE9Dl | maths | 3d-geometry | plane-in-space | Two systems of rectangular axes have the same origin. If a plane cuts then at distances $$a,b,c$$ and $$a', b', c'$$ from the origin then | [{"identifier": "A", "content": "$${1 \\over {{a^2}}} + {1 \\over {{b^2}}} + {1 \\over {{c^2}}} - {1 \\over {a{'^2}}} - {1 \\over {b{'^2}}} - {1 \\over {c{'^2}}} = 0$$ "}, {"identifier": "B", "content": "$$\\,{1 \\over {{a^2}}} + {1 \\over {{b^2}}} + {1 \\over {{c^2}}} + {1 \\over {a{'^2}}} + {1 \\over {b{'^2}}} + {1 \... | ["A"] | null | Equation of planes be $${x \over a} + {y \over b} + {z \over c} = 1\,\,\& \,\,{x \over {a'}} + {y \over {b'}} + {z \over {c'}} = 1$$
<br><br>So the distance from (0, 0, 0) to both the plane is same.
<br><br>$$ \therefore $$ $$\left| {{{ - 1} \over {\sqrt {{1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}}} }... | mcq | aieee-2003 |
HYn9nj1UWdOirDBu | maths | 3d-geometry | plane-in-space | The radius of the circle in which the sphere
<br/><br>$${x^2} + {y^2} + {z^2} + 2x - 2y - 4z - 19 = 0$$ is cut by the plane
<br/><br>$$x+2y+2z+7=0$$ is </br></br> | [{"identifier": "A", "content": "$$4$$"}, {"identifier": "B", "content": "$$1$$"}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$$3$$"}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265346/exam_images/c0dqhctziwbekfugrolb.webp" loading="lazy" alt="AIEEE 2003 Mathematics - 3D Geometry Question 321 English Explanation">
<br><br>Center of sphere $$ = \left( { - 1,1,2} \right)$$
<br><br>Radius of sphere $$\sqrt {1 ... | mcq | aieee-2003 |
wXNJAE9OixV0o1Ul | maths | 3d-geometry | plane-in-space | The intersection of the spheres
<br/>$${x^2} + {y^2} + {z^2} + 7x - 2y - z = 13$$ and
<br/>$${x^2} + {y^2} + {z^2} - 3x + 3y + 4z = 8$$
<br/>is the same as the intersection of one of the sphere and the plane | [{"identifier": "A", "content": "$$2x-y-z=1$$ "}, {"identifier": "B", "content": "$$x-2y-z=1$$"}, {"identifier": "C", "content": "$$x-y-2z=1$$ "}, {"identifier": "D", "content": "$$x-y-z=1$$ "}] | ["A"] | null | The equation of spheres are
<br><br>$${S_1}:{x^2} + {y^2} + {z^2} + 7x - 2y - z - 13 = 0$$ and
<br><br>$${S_2}:{x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8 = 0$$
<br><br>Their plane of intersection is $${S_1} - {S_2} = 0$$
<br><br>$$ \Rightarrow 10x - 5y - 5z - 5 = 0$$
<br><br>$$ \Rightarrow 2x - y - z = 1$$ | mcq | aieee-2004 |
wUkvqWxVsZ5qPXxv | maths | 3d-geometry | plane-in-space | Distance between two parallel planes
<br/><br>$$\,2x + y + 2z = 8$$ and $$4x + 2y + 4z + 5 = 0$$ is :</br> | [{"identifier": "A", "content": "$${9 \\over 2}$$ "}, {"identifier": "B", "content": "$${5 \\over 2}$$"}, {"identifier": "C", "content": "$${7 \\over 2}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["C"] | null | The planes are $$2x + y + 2x - 8 = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>and $$4x + 2y + 4z + 5 = 0$$
<br><br>or $$2x + y + 2z + {5 \over 2} = 0\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>$$\therefore$$ Distance between $$\left( 1 \right)$$ and $$\,\left( 2 \right)$$
<br><br>$$ = \left| {{{{5 \over ... | mcq | aieee-2004 |
gsaP2wxbhrwI4MNS | maths | 3d-geometry | plane-in-space | The plane $$x+2y-z=4$$ cuts the sphere $${x^2} + {y^2} + {z^2} - x + z - 2 = 0$$ in a circle of radius | [{"identifier": "A", "content": "$$3$$ "}, {"identifier": "B", "content": "$$1$$ "}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$${\\sqrt 2 }$$"}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265266/exam_images/iljyyh25rrxltbqfibh1.webp" loading="lazy" alt="AIEEE 2005 Mathematics - 3D Geometry Question 311 English Explanation">
<br><br>Perpendicular distance of center $$\left( {{1 \over 2},0, - {1 \over 2}} \right)$$
<br... | mcq | aieee-2005 |
jSTkPlfBJ5I6fd65 | maths | 3d-geometry | plane-in-space | <b>Statement-1 :</b> The point $$A(3, 1, 6)$$ is the mirror image of the point $$B(1, 3, 4)$$ in the plane $$x-y+z=5.$$
<br/><br><b>Statement-2 :</b> The plane $$x-y+z=5$$ bisects the line segment joining $$A(3, 1, 6)$$ and $$B(1, 3, 4).$$ </br> | [{"identifier": "A", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is <b>not</b> a correct explanation for Statement - 1."}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is false."}, {"identifier": "C", "content": "Statement - 1 is false , Statement - 2 is true."}, {"id... | ["A"] | null | $$A\left( {3,1,6} \right);\,\,B = \left( {1,3,4} \right)$$
<br><br>Mid-point of $$AB = \left( {2,2,5} \right)$$ lies on the plane.
<br><br>and d.r's of $$AB=(2,-2,2)$$
<br><br>d.r's of normal to plane $$ = \left( {1, - 1,1} \right).$$
<br><br>Direction ratio of $$AB$$ and normal to the plane are proportional therefore... | mcq | aieee-2010 |
DuqJJduwK99t03Hr | maths | 3d-geometry | plane-in-space | A equation of a plane parallel to the plane $$x-2y+2z-5=0$$ and at a unit distance from the origin is : | [{"identifier": "A", "content": "$$x-2y+2z-3=0$$"}, {"identifier": "B", "content": "$$x-2y+2z+1=0$$"}, {"identifier": "C", "content": "$$x-2y+2z-1=0$$ "}, {"identifier": "D", "content": "$$x-2y+2z+5=0$$"}] | ["A"] | null | Given equation of a plane is
<br><br>$$x - 2y + 2z - 5 = 0$$
<br><br>So, Equation of parallel plane is given by
<br><br>$$x - 2y + 2z + d = 0$$
<br><br>Now, it is given that distance from origin to the parallel planes is $$1.$$
<br><br>$$\therefore$$ $$\left| {{d \over {\sqrt {{1^2} + {2^2} + {2^3}} }}} \right| = 1 ... | mcq | aieee-2012 |
1rZ38hZcB3hCW5WP | maths | 3d-geometry | plane-in-space | Distance between two parallel planes $$2x+y+2z=8$$ and $$4x+2y+4z+5=0$$ is : | [{"identifier": "A", "content": "$${3 \\over 2}$$ "}, {"identifier": "B", "content": "$${5 \\over 2}$$"}, {"identifier": "C", "content": "$${7 \\over 2}$$"}, {"identifier": "D", "content": "$${9 \\over 2}$$"}] | ["C"] | null | $$2x + y + 2z - 8 = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( {Plane\,\,1} \right)$$
<br><br>$$2x + y + 2z + {5 \over 2} = 0\,\,\,\,\,\,\,\,\,\,\,...\left( {Plane\,\,2} \right)$$
<br><br>Distance between Plane $$1$$ and $$2$$
<br><br>$$ = \left| {{{ - 8 - {5 \over 2}} \over {\sqrt {{2^2} + {1^2} + {2^2}} }}} \right|$$
<br><... | mcq | jee-main-2013-offline |
Mb2JUKkOpNuTwuvd2BgWO | maths | 3d-geometry | plane-in-space | The distance of the point (1, − 2, 4) from the plane passing through the point
(1, 2, 2) and perpendicular to the planes x − y + 2z = 3 and 2x − 2y + z + 12 = 0, is : | [{"identifier": "A", "content": "$$2\\sqrt 2 $$ "}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$$\\sqrt 2 $$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 2 }}$$ "}] | ["A"] | null | Equation of plane passing through point (1, 2, 2) is,
<br><br>a(x $$-$$ 1) + b(y $$-$$ 2) + c(z $$-$$ 2) = 0 . . .(1)
<br><br>This plane is perpendicular to the plane
<br><br>x $$-$$ y + 2z = 3 and 2x $$-$$ 2y + z + 12 = 0
<br><br>When two planes,
<... | mcq | jee-main-2016-online-9th-april-morning-slot |
HsclIbPpXGLLYwNuOmjqV | maths | 3d-geometry | plane-in-space | If x = a, y = b, z = c is a solution of the system of linear equations
<br/><br/>x + 8y + 7z = 0
<br/><br/>9x + 2y + 3z = 0
<br/><br/>x + y + z = 0
<br/><br/>such that the point (a, b, c) lies on the plane x + 2y + z = 6, then 2a + b + c equals : | [{"identifier": "A", "content": "$$-$$ 1"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["C"] | null | Given,
<br><br>x + 8y + 7z = 0
<br><br>9x + 2y + 3z = 0
<br><br>x + y + z = 0
<br><br>Solving those equations, we get
<br><br>x = $$\lambda $$, y = 6$$\lambda $$, z = -7$$\lambda $$
<br><br>This point lies on the plane x + 2y + z = 6
<br><br>$$ \therefore $$ $$\lambda $$ + 2(6$$\lambda $$) + (-7$$\lambda $$) = 0
<br>... | mcq | jee-main-2017-online-9th-april-morning-slot |
QOo7bAqiG06GOKThuvm05 | maths | 3d-geometry | plane-in-space | If a variable plane, at a distance of 3 units from the origin, intersects the coordinate
axes at A, B and C, then the locus of the centroid of $$\Delta $$ABC is :
| [{"identifier": "A", "content": "$${1 \\over {{x^2}}} + {1 \\over {{y^2}}} + {1 \\over {{z^2}}} = 1$$ "}, {"identifier": "B", "content": "$${1 \\over {{x^2}}} + {1 \\over {{y^2}}} + {1 \\over {{z^2}}} = 3$$ "}, {"identifier": "C", "content": "$${1 \\over {{x^2}}} + {1 \\over {{y^2}}} + {1 \\over {{z^2}}} = {1 \\over 9}... | ["A"] | null | Suppose centroid be (h, k, $$l$$)
<br><br>$$ \therefore $$ x $$-$$ intp $$=$$ 3h, y $$-$$ intp $$=$$ 3k, z $$-$$ intp $$=$$ 3$$l$$
<br><br>Equation $${x \over {3h}} + {y \over {3k}} + {z \over {3l}} = 1$$
<br><br>$$ \therefore $$ Distance from (0, 0, 0)
<br><br>$$\left| {{{ - 1} \over ... | mcq | jee-main-2017-online-9th-april-morning-slot |
By2W9q22rMEe3evacfhKc | maths | 3d-geometry | plane-in-space | The sum of the intercepts on the coordinate axes of the plane passing through the point ($$-$$2, $$-2,$$ 2) and containing the line joining the points (1, $$-$$1, 2) and (1, 1, 1) is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "$$-$$ 4"}, {"identifier": "C", "content": "$$-$$ 8"}, {"identifier": "D", "content": "12"}] | ["B"] | null | Equation of plane passing through three given points is :
<br><br>$$\left| {\matrix{
{x - {x_1}} & {y - {y_1}} & {z - {z_1}} \cr
{{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr
{{x_3} - {x_1}} & {{y_3} - {y_1}} & {{z_3} - {z_1}} \cr
} } \right| = 0$$
<br><br>$$ \Rightar... | mcq | jee-main-2018-online-16th-april-morning-slot |
EJOo70wLbGOQ1WD7CS7IU | maths | 3d-geometry | plane-in-space | A plane bisects the line segment joining the points (1, 2, 3) and ($$-$$ 3, 4, 5) at rigt angles. Then this plane also passes through the point : | [{"identifier": "A", "content": "($$-$$ 3, 2, 1)"}, {"identifier": "B", "content": "(3, 2, 1)"}, {"identifier": "C", "content": "($$-$$ 1, 2, 3)"}, {"identifier": "D", "content": "(1, 2, $$-$$ 3)"}] | ["A"] | null | Since the plane bisects the line joining the points (1, 2, 3) and ($$-$$3, 4, 5) then the plane passes through the midpoint of the line which is :
<br><br>$$\left( {{{1 - 3} \over 2},{{2 + 4} \over 2},{{5 + 3} \over 2}} \right)$$ $$ \equiv $$ $$\left( {{{ - 2} \over 2},{6 \over 2},{8 \over 2}} \right)$$ $$... | mcq | jee-main-2018-online-15th-april-evening-slot |
pFJKQ1n9aFLfryRPPa1Lz | maths | 3d-geometry | plane-in-space | A variable plane passes through a fixed point (3,2,1) and meets x, y and z axes at A, B and C respectively. A plane is drawn parallel to yz -plane through A, a second plane is drawn parallel zx-plane through B and a third plane is drawn parallel to xy-plane through C. Then the locus of the point of intersection of thes... | [{"identifier": "A", "content": "$${x \\over 3} + {y \\over 2} + {z \\over 1} = 1$$ "}, {"identifier": "B", "content": "x + y + z = 6"}, {"identifier": "C", "content": "$${1 \\over x} + {1 \\over y} + {1 \\over z} = {{11} \\over 6}$$"}, {"identifier": "D", "content": "$${3 \\over x} + {2 \\over y} + {1 \\over z} = 1$$"... | ["D"] | null | If a, b, c are the intercepts of the variable plane on the x,y,z axes respectively, then the equation of the plane is <br/><br/>$${x \over a} + {y \over b} + {z \over c} = 1$$
<br><br>And the point of intersection of the planes parallel to the xy, yz and zx planes is $$\left( {a,b,c} \right)$$.
<br><br>As the point (... | mcq | jee-main-2018-online-15th-april-morning-slot |
LfSBYkVJfNOMRk2S | maths | 3d-geometry | plane-in-space | If L<sub>1</sub> is the line of intersection of the planes 2x - 2y + 3z - 2 = 0, x - y + z + 1 = 0 and L<sub>2</sub> is the line of
intersection of the planes x + 2y - z - 3 = 0, 3x - y + 2z - 1 = 0, then the distance of the origin from the
plane, containing the lines L<sub>1</sub> and L<sub>2</sub>, is : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${1 \\over {4\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${1 \\over {3\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${1 \\over {2\\sqrt 2 }}$$"}] | ["C"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265244/exam_images/mtsizoslixc1qq9ocqig.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - 3D Geometry Question 287 English Explanation">
<br><br>L<sub>1</sub> is the line of intersection of plane 1 and plane 2.
<br><br... | mcq | jee-main-2018-offline |
h2kYrw078v0a3uPgCCURP | maths | 3d-geometry | plane-in-space | The perpendicular distance from the origin to the plane containing the two lines, <br/><br/>$${{x + 2} \over 3} = {{y - 2} \over 5} = {{z + 5} \over 7}$$ and <br/><br/>$${{x - 1} \over 1} = {{y - 4} \over 4} = {{z + 4} \over 7},$$ is : | [{"identifier": "A", "content": "$$6\\sqrt {11} $$"}, {"identifier": "B", "content": "$${{11} \\over {\\sqrt 6 }}$$"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "11$$\\sqrt 6 $$"}] | ["B"] | null | $$\left| {\matrix{
i & j & k \cr
3 & 5 & 7 \cr
1 & 4 & 7 \cr
} } \right|$$
<br><br>= $$\widehat i$$(35 $$-$$ 28) $$-$$ $$\widehat j$$(21.7) + $$\widehat k$$(12 $$-$$ 5)
<br><br>= 7$$\widehat i$$ $$-$$ 14$$\widehat j$$ + 7$$\widehat k$$
<br><br>= $$\widehat i$$ $$-$$ 2$$\widehat ... | mcq | jee-main-2019-online-12th-january-morning-slot |
ZNjWs09KjUBWLXz3C53rsa0w2w9jxadan6d | maths | 3d-geometry | plane-in-space | A plane which bisects the angle between the two given planes 2x – y + 2z – 4 = 0 and x + 2y + 2z – 2 = 0,
passes through the point : | [{"identifier": "A", "content": "(1, \u20134, 1)"}, {"identifier": "B", "content": "(1, 4, \u20131)"}, {"identifier": "C", "content": "(2, 4, 1)"}, {"identifier": "D", "content": "(2, \u20134, 1)"}] | ["D"] | null | Planes bisecting the given planes are<br><br>
$${{2x - y + 2z - 4} \over 3} = \pm {{x + 2y + 2z - 2} \over 3}$$<br><br>
$$ \Rightarrow $$ x - 3y = 2 or 3x + y + 4z = 6<br><br>
Out of the four given points in question's options only (2, -4, 1) lies on the plane 3x + y + 4z = 6 | mcq | jee-main-2019-online-12th-april-evening-slot |
pOnUJRdMYw4ui1JSKs3rsa0w2w9jx1zs3oi | maths | 3d-geometry | plane-in-space | If the plane 2x – y + 2z + 3 = 0 has the distances
$${1 \over 3}$$
and
$${2 \over 3}$$
units from the planes 4x – 2y + 4z + $$\lambda $$ = 0 and
2x – y + 2z + $$\mu $$ = 0, respectively, then the maximum value of $$\lambda $$ + $$\mu $$ is equal to : | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "15"}] | ["A"] | null | Distance formula<br><br>
(i) $${{\left| {\lambda - 6} \right|} \over {\sqrt {16 + 4 + 16} }} = \left| {{{\lambda - 6} \over 6}} \right| = {1 \over 3}$$<br><br>
$$ \Rightarrow $$ $$\left| {\lambda - 6} \right| = 2$$<br><br>
$$ \Rightarrow $$ $$\lambda = 8,4$$<br><br>
(ii) $${{\left| {\mu - 3} \right|} \over {\sqrt ... | mcq | jee-main-2019-online-10th-april-evening-slot |
ACCrs5Q2N2AyKcnuNg3rsa0w2w9jwxv0uci | maths | 3d-geometry | plane-in-space | If Q(0, –1, –3) is the image of the point P in the plane 3x – y + 4z = 2 and R is the point (3, –1, –2), then the
area (in sq. units) of $$\Delta $$PQR is : | [{"identifier": "A", "content": "$${{\\sqrt {65} } \\over 2}$$"}, {"identifier": "B", "content": "$$2\\sqrt {13} $$"}, {"identifier": "C", "content": "$${{\\sqrt {91} } \\over 2}$$"}, {"identifier": "D", "content": "$${{\\sqrt {91} } \\over 4}$$"}] | ["C"] | null | Image of Q in plane<br><br>
$${{\left( {x - 0} \right)} \over 3} = {{\left( {y + 1} \right)} \over { - 1}} = {{z + 3} \over { + 4}} = {{ - 2(1 - 12 - 2)} \over {9 + 1 + 16}} = 1$$<br><br>
x = 3, y = –2, z = 1<br><br>
P(3, –2, 1), Q(0, –1, –3), R(3, –1, –2)<br><br>
Now area of $$\Delta $$PQR is<br><br>
$${1 \over 2}\lef... | mcq | jee-main-2019-online-10th-april-morning-slot |
yodXktCxgD0IJ4xePL18hoxe66ijvwvwdsy | maths | 3d-geometry | plane-in-space | Let P be the plane, which contains the line of
intersection of the planes, x + y + z – 6 = 0 and
2x + 3y + z + 5 = 0 and it is perpendicular to the
xy-plane. Then the distance of the point (0, 0, 256)
from P is equal to : | [{"identifier": "A", "content": "205$$\\sqrt5$$"}, {"identifier": "B", "content": "63$$\\sqrt5$$"}, {"identifier": "C", "content": "11/$$\\sqrt5$$"}, {"identifier": "D", "content": "17/$$\\sqrt5$$"}] | ["C"] | null | P<sub>1</sub> : x + y + z – 6 = 0
<br><br>P<sub>2</sub> : 2x + 3y + z + 5 = 0
<br><br>Equation of plane which passes through the line of intersection of P<sub>1</sub> and P<sub>2</sub> is
<br><br>P<sub>1</sub> + $$\lambda $$P<sub>2</sub> = 0
<br><br>$$ \Rightarrow $$ (x + y + z – 6) + $$\lambda $$(2x + 3y + z + 5) = 0
... | mcq | jee-main-2019-online-9th-april-evening-slot |
oW3MTlGq7uoYSy1HduycQ | maths | 3d-geometry | plane-in-space | A plane passing through the points (0, –1, 0)
and (0, 0, 1) and making an angle $${\pi \over 4}$$ with the
plane y – z + 5 = 0, also passes through the
point | [{"identifier": "A", "content": "$$\\left( {\\sqrt 2 ,1,4} \\right)$$"}, {"identifier": "B", "content": "$$\\left(- {\\sqrt 2 ,1,4} \\right)$$"}, {"identifier": "C", "content": "$$\\left( -{\\sqrt 2 ,-1,-4} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {\\sqrt 2 ,-1,4} \\right)$$"}] | ["A"] | null | Let ax + by + cz = 1 be the equation of the plane
<br><br>it passed through point (0, –1, 0).
<br><br>$$ \therefore $$ -b = 1
<br><br>$$ \Rightarrow $$ b = -1
<br><br>Also it passes through point (0, 0, 1)
<br><br>$$ \therefore $$ c = 1
<br><br>So the plane is ax - y + z = 1.
<br><br>This plane an angle $${\pi \over 4... | mcq | jee-main-2019-online-9th-april-morning-slot |
1h4HQ5Q8R2RpdCkkTSLdT | maths | 3d-geometry | plane-in-space | The magnitude of the projection of the vector
$$\mathop {2i}\limits^ \wedge + \mathop {3j}\limits^ \wedge + \mathop k\limits^ \wedge $$ on the vector perpendicular to the plane
containing the vectors $$\mathop {i}\limits^ \wedge + \mathop {j}\limits^ \wedge + \mathop k\limits^ \wedge $$ and $$\mathop {i}\limi... | [{"identifier": "A", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "B", "content": "$$\\sqrt 6 $$"}, {"identifier": "C", "content": "$$\\sqrt {3 \\over 2} $$"}, {"identifier": "D", "content": "3$$\\sqrt 6 $$"}] | ["C"] | null | Let vector $$\overrightarrow p $$ is perpendicular to the both vectors $$\mathop {i}\limits^ \wedge + \mathop {j}\limits^ \wedge + \mathop k\limits^ \wedge $$ and $$\mathop {i}\limits^ \wedge + \mathop {2j}\limits^ \wedge + \mathop {3k}\limits^ \wedge $$.
<br><br>$$ \therefore $$ $$\overrightarrow p $$ = ($$\... | mcq | jee-main-2019-online-8th-april-morning-slot |
A3dJQnt8WpOkVMARcRG7E | maths | 3d-geometry | plane-in-space | Let S be the set of all real values of $$\lambda $$ such that a plane passing through the points (–$$\lambda $$<sup>2</sup>, 1, 1), (1, –$$\lambda $$<sup>2</sup>, 1) and (1, 1, – $$\lambda $$<sup>2</sup>) also passes through the point (–1, –1, 1). Then S is equal to : | [{"identifier": "A", "content": "{1, $$-$$1}"}, {"identifier": "B", "content": "{3, $$-$$ 3}"}, {"identifier": "C", "content": "$$\\left\\{ {\\sqrt 3 } \\right\\}$$"}, {"identifier": "D", "content": "$$\\left\\{ {\\sqrt 3 , - \\sqrt 3 } \\right\\}$$"}] | ["D"] | null | All four points are coplanar so
<br><br>$$\left| {\matrix{
{1 - {\lambda ^2}} & 2 & 0 \cr
2 & { - {\lambda ^2} + 1} & 0 \cr
2 & 2 & { - {\lambda ^2} - 1} \cr
} } \right| = 0$$
<br><br>($$\lambda $$<sup>2</sup> + 1)<sup>2</sup> (3 $$-$$ $$\lambda $$<sup>2</sup>) = 0
<br><br>$$\la... | mcq | jee-main-2019-online-12th-january-evening-slot |
vCpFMyMOEky4bJBVl2v9s | maths | 3d-geometry | plane-in-space | If the point (2, $$\alpha $$, $$\beta $$) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x – 5y = 15, then 2$$\alpha $$ – 3$$\beta $$ is equal to | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "17"}, {"identifier": "D", "content": "5"}] | ["B"] | null | Normal vector of plane
<br><br>$$ = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
2 & { - 5} & 0 \cr
4 & { - 4} & 5 \cr
} } \right|$$
<br><br>$$ = - 4\left( {5\widehat i + 2\widehat j - 3\widehat k} \right)$$
<br><br>equation of plane is
<br><br>5(x $$-$$ 7) ... | mcq | jee-main-2019-online-11th-january-evening-slot |
y1O9ASDQ4I3iqhyAwRaOV | maths | 3d-geometry | plane-in-space | The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an angle $${\pi \over 4}$$ with the plane y $$-$$ z + 5 = 0 are : | [{"identifier": "A", "content": "2, $$-$$1, 1"}, {"identifier": "B", "content": "$$2\\sqrt 3 ,1, - 1$$"}, {"identifier": "C", "content": "$$\\sqrt 2 ,1, - 1$$"}, {"identifier": "D", "content": "$$\\sqrt 2 , - \\sqrt 2 $$"}] | ["C"] | null | Let the equation of plane be
<br><br>a(x $$-$$ 0) + b(y + 1) + c(z $$-$$ 0) = 0
<br><br>It passes through (0, 0, 1) then
<br><br>b + c = 0 . . . . (1)
<br><br>Now cos $${\pi \over 4}$$ = $${{a\left( 0 \right) + b\left( 1 \right) + c\left( { - 1} \right)} \over {\sqrt 2 \sqrt {{a^2} + {b^2} + {... | mcq | jee-main-2019-online-11th-january-morning-slot |
FZp6IAeQMHnGO3LU7gCrN | maths | 3d-geometry | plane-in-space | The plane which bisects the line segment joining the points (–3, –3, 4) and (3, 7, 6) at right angles, passes through which one of the following points ? | [{"identifier": "A", "content": "(2, 1, 3)"}, {"identifier": "B", "content": "(4, $$-$$ 1, 2)"}, {"identifier": "C", "content": "(4, 1, $$-$$ 2)"}, {"identifier": "D", "content": "($$-$$ 2, 3, 5)"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264937/exam_images/uqar8tycbulapjs4smas.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Mathematics - 3D Geometry Question 261 English Explanation">
<br>p : 3... | mcq | jee-main-2019-online-10th-january-evening-slot |
X8MEKBrNXBZseySo39pJw | maths | 3d-geometry | plane-in-space | Let A be a point on the line $$\overrightarrow r = \left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k$$ and B(3, 2, 6) be a point in the space. Then the value of $$\mu $$ for which the vector $$\overrightarrow {AB} $$ is parallel to the plane x $$-$$ 4y ... | [{"identifier": "A", "content": "$${1 \\over 8}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over 4}$$"}, {"identifier": "D", "content": "$$-$$ $${1 \\over 4}$$"}] | ["C"] | null | Let point A is
<br><br>$$\left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k$$
<br><br>and point B is (3, 2, 6)
<br><br>then $$\overrightarrow {AB} = \left( {2 + 3\mu } \right)\widehat i + \left( {3 - \mu } \right)\widehat j + \left( {4 - 5\mu } \right)\wi... | mcq | jee-main-2019-online-10th-january-morning-slot |
xUxj1ys4ZXlUo5ebGRM5g | maths | 3d-geometry | plane-in-space | The plane passing through the point (4, –1, 2) and parallel to the lines $${{x + 2} \over 3} = {{y - 2} \over { - 1}} = {{z + 1} \over 2}$$ and $${{x - 2} \over 1} = {{y - 3} \over 2} = {{z - 4} \over 3}$$ also passes through the point - | [{"identifier": "A", "content": "(1, 1, $$-$$ 1)"}, {"identifier": "B", "content": "(1, 1, 1)"}, {"identifier": "C", "content": "($$-$$ 1, $$-$$ 1, $$-$$1)"}, {"identifier": "D", "content": "($$-$$ 1, $$-$$ 1, 1)"}] | ["B"] | null | Let $$\overrightarrow n $$ be the normal vector to the plane passing through (4, $$-$$1, 2) and parallel to the lines L<sub>1</sub> & L<sub>2</sub>
<br><br>then $$\overrightarrow n $$ = $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & { - 1} & 2 \cr
1 & 2 &... | mcq | jee-main-2019-online-10th-january-morning-slot |
8wJsTP566s6V7Wx2nqpZD | maths | 3d-geometry | plane-in-space | The equation of the plane containing the straight line $${x \over 2} = {y \over 3} = {z \over 4}$$ and perpendicular to the plane containing the straight lines $${x \over 3} = {y \over 4} = {z \over 2}$$ and $${x \over 4} = {y \over 2} = {z \over 3}$$ is : | [{"identifier": "A", "content": "x $$-$$ 2y + z = 0"}, {"identifier": "B", "content": "3x + 2y $$-$$ 3z = 0"}, {"identifier": "C", "content": "x + 2y $$-$$ 2z = 0"}, {"identifier": "D", "content": "5x + 2y $$-$$ 4z = 0"}] | ["A"] | null | Vector $$ \bot $$ to given plane = $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & 4 & 2 \cr
4 & 2 & 3 \cr
} } \right|$$
<br><br>= $$\widehat i\left( {12 - 4} \right) - \widehat j\left( {9 - 8} \right) + \widehat k\left( {6 - 16} \right)$$
<br><br>= $$8\wi... | mcq | jee-main-2019-online-9th-january-evening-slot |
OOzulklPdGI4Zf272LTtO | maths | 3d-geometry | plane-in-space | A tetrahedron has vertices P(1, 2, 1), Q(2, 1, 3), R(–1, 1, 2) and O(0, 0, 0). The angle between the faces OPQ and PQR is : | [{"identifier": "A", "content": "cos<sup>$$-$$1</sup>$$\\left( {{{17} \\over {31}}} \\right)$$"}, {"identifier": "B", "content": "cos<sup>$$-$$1</sup>$$\\left( {{{9} \\over {35}}} \\right)$$"}, {"identifier": "C", "content": "cos<sup>$$-$$1</sup>$$\\left( {{{19} \\over {35}}} \\right)$$"}, {"identifier": "D", "content"... | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265437/exam_images/kiidub3irc2kt9mrj7ft.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Mathematics - 3D Geometry Question 255 English Explanation">
<br>$$\ov... | mcq | jee-main-2019-online-12th-january-morning-slot |
iWc51afvvP3KvJRMFwjgy2xukf467nsf | maths | 3d-geometry | plane-in-space | The plane which bisects the line joining, the
points (4, –2, 3) and (2, 4, –1) at right angles
also passes through the point : | [{"identifier": "A", "content": "(4, 0, 1)"}, {"identifier": "B", "content": "(0, \u20131, 1)"}, {"identifier": "C", "content": "(0, 1, \u20131)"}, {"identifier": "D", "content": "(4, 0, \u20131)"}] | ["D"] | null | Direction ratios of normal to plane are < 2, –6, 4 >
<br><br> Also plane passes through (3, 1, 1)
<br><br>$$ \therefore $$ Equation of plane
<br><br/>2(x–3)–6(y–1)+4(z–1) = 0
<br><br>$$ \Rightarrow $$ x – 3y + 2z = 2
<br><br>By checking all options we can see this equation passes through (4, 0, –1) | mcq | jee-main-2020-online-3rd-september-evening-slot |
48HxDRJEsiZs9PgCGajgy2xukf8zwdb2 | maths | 3d-geometry | plane-in-space | If the equation of a plane P, passing through the intersection of the planes, <br/>x + 4y - z + 7 = 0
and 3x + y + 5z = 8 is ax + by + 6z = 15 for some a, b $$ \in $$ R, then the distance of the point
(3, 2, -1) from the plane P is........... | [] | null | 3 | Equation of plane P is<br><br>$$(x + 4y - z + 7) + \lambda (3x + y + 5z - 8) = 0$$<br><br>$$ \Rightarrow x(1 + 3\lambda ) + y(4 + \lambda ) + z( - 1 + 5\lambda ) + (7 - 8\lambda ) = 0$$<br><br>$${{1 + 3\lambda } \over a} = {{4 + \lambda } \over b} = {{5\lambda - 1} \over 6} = {{7 - 8\lambda } \over { - 15}}$$
<br><br>... | integer | jee-main-2020-online-4th-september-morning-slot |
4fJqBCT7tmvq1cvtwz7k9k2k5hk0mnw | maths | 3d-geometry | plane-in-space | The mirror image of the point (1, 2, 3) in a plane
is<br/><br> $$\left( { - {7 \over 3}, - {4 \over 3}, - {1 \over 3}} \right)$$. Which of the following
points lies on this plane ?</br> | [{"identifier": "A", "content": "(1, \u20131, 1)"}, {"identifier": "B", "content": "(\u20131, \u20131, \u20131)"}, {"identifier": "C", "content": "(\u20131, \u20131, 1)"}, {"identifier": "D", "content": "(1, 1, 1)"}] | ["A"] | null | Let A(1, 2, 3), B$$\left( { - {7 \over 3}, - {4 \over 3}, - {1 \over 3}} \right)$$
<br><br>$$ \therefore $$ Midpoint of AB = M = $$\left( {{{{{ - 7} \over 3} + 1} \over 2},{{{{ - 4} \over 3} + 2} \over 2},{{{{ - 1} \over 3} + 3} \over 2}} \right)$$
<br><br>= $$\left( {{{ - 2} \over 3},{1 \over 3},{4 \over 3}} \right)$... | mcq | jee-main-2020-online-8th-january-evening-slot |
Ou0sxBmmwtNWG6HOVU7k9k2k5e374ev | maths | 3d-geometry | plane-in-space | Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point
(2, 1, 6). Then the image of R in the plane P is :
| [{"identifier": "A", "content": "(4, 3, 2) "}, {"identifier": "B", "content": "(6, 5, - 2) "}, {"identifier": "C", "content": "(3, 4, -2)"}, {"identifier": "D", "content": "(6, 5, 2)"}] | ["B"] | null | Plane passing through (2, 1, 0), (4, 1, 1) and
(5, 0, 1) is
<br><br>$$\left| {\matrix{
{x - 2} & {y - 1} & {z - 0} \cr
{4 - 2} & {1 - 1} & {1 - 0} \cr
{5 - 2} & {0 - 1} & {1 - 0} \cr
} } \right|$$ = 0
<br><br>$$ \Rightarrow $$ x + y – 2z = 3
<br><br>$$ \therefore $$ Image of ... | mcq | jee-main-2020-online-7th-january-morning-slot |
T0VNJtB7BfAmHEDeS51klregikd | maths | 3d-geometry | plane-in-space | The equation of the plane passing through the point (1, 2, -3) and perpendicular to the
planes <br/><br>3x + y - 2z = 5 and 2x - 5y - z = 7, is :</br> | [{"identifier": "A", "content": "6x - 5y + 2z + 10 =0"}, {"identifier": "B", "content": "3x - 10y - 2z + 11 = 0"}, {"identifier": "C", "content": "6x - 5y - 2z - 2 = 0"}, {"identifier": "D", "content": "11x + y + 17z + 38 = 0"}] | ["D"] | null | Given, equation of planes are
<br/><br/>3x + y - 2z = 5
<br/><br/>2x - 5y - z = 7
<br/><br/>and point ( 1, 2, 3).
<br/><br/>Normal vector of required plane = n = $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & 1 & { - 2} \cr
2 & { - 5} & { - 1} \cr
} } \right|$$
<br/><br/>= $${\w... | mcq | jee-main-2021-online-24th-february-morning-slot |
t1IiOet4hCn5Cjas1c1klt7cbt1 | maths | 3d-geometry | plane-in-space | A plane passes through the points A(1, 2, 3), B(2, 3, 1) and C(2, 4, 2). If O is the origin and P is (2, $$-$$1, 1), then the projection of $$\overrightarrow {OP} $$ on this plane is of length : | [{"identifier": "A", "content": "$$\\sqrt {{2 \\over 7}} $$"}, {"identifier": "B", "content": "$$\\sqrt {{2 \\over 5}} $$"}, {"identifier": "C", "content": "$$\\sqrt {{2 \\over 3}} $$"}, {"identifier": "D", "content": "$$\\sqrt {{2 \\over 11}} $$"}] | ["D"] | null | A(1, 2, 3), B(2, 3, 1), C(2, 4, 2), O(0, 0, 0)<br><br>Equation of plane passing through A, B, C will be<br><br>$$\left| {\matrix{
{x - 1} & {y - 2} & {z - 3} \cr
{2 - 1} & {3 - 2} & {1 - 3} \cr
{2 - 1} & {4 - 2} & {2 - 3} \cr
} } \right| = 0$$<br><br>$$ \Rightarrow \left| {\matr... | mcq | jee-main-2021-online-25th-february-evening-slot |
MgtSscZbWDByJ9eKdx1kluh6hjj | maths | 3d-geometry | plane-in-space | Consider the three planes<br/><br/>P<sub>1</sub> : 3x + 15y + 21z = 9,<br/><br/>P<sub>2</sub> : x $$-$$ 3y $$-$$ z = 5, and <br/><br/>P<sub>3</sub> : 2x + 10y + 14z = 5<br/><br/>Then, which one of the following is true? | [{"identifier": "A", "content": "P<sub>1</sub> and P<sub>2</sub> are parallel."}, {"identifier": "B", "content": "P<sub>1</sub>, P<sub>2</sub> and P<sub>3</sub> all are parallel."}, {"identifier": "C", "content": "P<sub>1</sub> and P<sub>3</sub> are parallel."}, {"identifier": "D", "content": "P<sub>2</sub> and P<sub>3... | ["C"] | null | P<sub>1</sub> : 3x + 15y + 21z = 9,<br><br>P<sub>2</sub> : x $$-$$ 3y $$-$$ z = 5<br><br>P<sub>3</sub> : x + 5y + 7z = 5/2
<br><br>$$ \therefore $$ P<sub>1</sub> and P<sub>3</sub> are parallel. | mcq | jee-main-2021-online-26th-february-morning-slot |
XBPH37cUetzOYSGHad1kluh8adb | maths | 3d-geometry | plane-in-space | If (1, 5, 35), (7, 5, 5), (1, $$\lambda$$, 7) and (2$$\lambda$$, 1, 2) are coplanar, then the sum of all possible values of $$\lambda$$ is : | [{"identifier": "A", "content": "$$ - {{44} \\over 5}$$"}, {"identifier": "B", "content": "$$ - {{39} \\over 5}$$"}, {"identifier": "C", "content": "$${{44} \\over 5}$$"}, {"identifier": "D", "content": "$${{39} \\over 5}$$"}] | ["C"] | null | A(1, 5, 35), B(7, 5, 5), C(1, $$\lambda$$, 7), D(2$$\lambda$$, 1, 2)<br><br>$$\overrightarrow {AB} $$ = 6$$\widehat i$$ $$-$$ 30$$\widehat k$$,
<br><br>$$\overrightarrow {BC} $$ = $$-$$6$$\widehat i$$ ($$\lambda$$ $$-$$ 5)$$\widehat j$$ + 2$$\widehat k$$,
<br><br>$$\overrightarrow {CD} $$ = (2$$\lambda$$ $$-$$ 1)$$\wi... | mcq | jee-main-2021-online-26th-february-morning-slot |
c11i6Yyeu5j7HAV9um1kluwweez | maths | 3d-geometry | plane-in-space | If the mirror image of the point (1, 3, 5) with respect to the plane <br/><br/>4x $$-$$ 5y + 2z = 8 is ($$\alpha$$, $$\beta$$, $$\gamma$$), then 5($$\alpha$$ + $$\beta$$ + $$\gamma$$) equals : | [{"identifier": "A", "content": "39"}, {"identifier": "B", "content": "41"}, {"identifier": "C", "content": "47"}, {"identifier": "D", "content": "43"}] | ["C"] | null | Image of (1, 3, 5) in the plane 4x $$-$$ 5y + 2z = 8 is ($$\alpha$$, $$\beta$$, $$\gamma$$)<br><br>$$ \Rightarrow {{\alpha - 1} \over 4} = {{\beta - 3} \over { - 5}} = {{\gamma - 5} \over 2} = - {{(4(1) - 5(3) + 2(5) - 8)} \over {{4^2} + {5^2} + {2^2}}} = {2 \over 5}$$<br><br>$$ \therefore $$ $$\alpha = 1 + 4\left... | mcq | jee-main-2021-online-26th-february-evening-slot |
KTmD8ld7JILBD51VmY1kmhxegkb | maths | 3d-geometry | plane-in-space | If for a > 0, the feet of perpendiculars from the points A(a, $$-$$2a, 3) and B(0, 4, 5) on the plane lx + my + nz = 0 are points C(0, $$-$$a, $$-$$1) and D respectively, then the length of line segment CD is equal to : | [{"identifier": "A", "content": "$$\\sqrt {41} $$"}, {"identifier": "B", "content": "$$\\sqrt {55} $$"}, {"identifier": "C", "content": "$$\\sqrt {31} $$"}, {"identifier": "D", "content": "$$\\sqrt {66} $$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265417/exam_images/r4nxivfvhrrbvtvvir7b.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Morning Shift Mathematics - 3D Geometry Question 204 English Explanation">
<br><br>Let ... | mcq | jee-main-2021-online-16th-march-morning-shift |
LOKOg3XFtIRmJyfvtU1kmiwwdp4 | maths | 3d-geometry | plane-in-space | If (x, y, z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0), (0, 42, 0) and (0, 0, 42), then the value of the expression <br/>$$3 + {{x - 11} \over {{{(y - 19)}^2}{{(z - 12)}^2}}} + {{y - 19} \over {{{(x - 11)}^2}{{(z - 12)}^2}}} + {{z - 12} \over {{{(x - 11)}^2}{{(y - 19)}^2}}} - {... | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "39"}, {"identifier": "C", "content": "$$-$$45"}, {"identifier": "D", "content": "0"}] | ["A"] | null | From intercept from, equation of plane is x + y + z = 42<br><br>$$ \Rightarrow $$ (x $$-$$ 11) + (y $$-$$ 19) + (z $$-$$ 12) = 0<br><br>let a = x $$-$$ 11, b = y $$-$$ 19, c = z $$-$$ 12<br><br>a + b + c = 0<br><br>Now, given expression is<br><br>$$3 + {a \over {{b^2}{c^2}}} + {b \over {{a^2}{c^2}}} + {c \over {{a^2}{b... | mcq | jee-main-2021-online-16th-march-evening-shift |
eUwQOrxgcwYlcgZH1a1kmjaqpo0 | maths | 3d-geometry | plane-in-space | The equation of the plane which contains the y-axis and passes through the point (1, 2, 3) is : | [{"identifier": "A", "content": "x + 3z = 0"}, {"identifier": "B", "content": "3x $$-$$ z = 0"}, {"identifier": "C", "content": "x + 3z = 10"}, {"identifier": "D", "content": "3x + z = 6"}] | ["B"] | null | Let the equation of the plane is a (x $$-$$ 1) + b(y $$-$$ 2) + c(z $$-$$ 3) = 0<br><br>Y-axis lies on it.<br><br>D.R.'s of y-axis are 0, 1, 0<br><br>$$ \therefore $$ 0.a + 1.b + 0.c = 0 $$ \Rightarrow $$ b = 0<br><br>$$ \therefore $$ Equation of plane is a(x $$-$$ 1) + c(z $$-$$ 3) = 0<br><br>x = 0, z = 0 also satisfy... | mcq | jee-main-2021-online-17th-march-morning-shift |
lRRAykFzsnTCagmm7r1kmko3iaz | maths | 3d-geometry | plane-in-space | Let P be an arbitrary point having sum of the squares of the distances from the planes x + y + z = 0, lx $$-$$ nz = 0 and x $$-$$ 2y + z = 0, equal to 9. If the locus of the point P is x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> = 9, then the value of l $$-$$ n is equal to _________. | [] | null | 0 | Let point P is ($$\alpha$$, $$\beta$$, $$\gamma$$)<br><br>$${\left( {{{\alpha + \beta + \gamma } \over {\sqrt 3 }}} \right)^2} + {\left( {{{l\alpha - n\gamma } \over {\sqrt {{l^2} + {n^2}} }}} \right)^2} + {\left( {{{\alpha - 2\beta + \gamma } \over {\sqrt 6 }}} \right)^2} = 9$$<br><br>Locus is $${{{{(x + y + z)}^... | integer | jee-main-2021-online-17th-march-evening-shift |
yDMFPVyWe7YnyrRy3A1kmllwl3e | maths | 3d-geometry | plane-in-space | Let the plane ax + by + cz + d = 0 bisect the line joining the points (4, $$-$$3, 1) and (2, 3, $$-$$5) at the right angles. If a, b, c, d are integers, then the <br/>minimum value of (a<sup>2</sup> + b<sup>2</sup> + c<sup>2</sup> + d<sup>2</sup>) is _________. | [] | null | 28 | Normal of plane = $$\overrightarrow {PQ} = - 2\widehat i + 6\widehat j - 6\widehat k$$<br><br>a = $$-$$2, b = 6, c = $$-$$6<br><br>& equation of plane is <br><br>$$-$$2x + 6y $$-$$ 6z + d = 0<br><br>$$ M(3,0, - 2)$$ is the midpoint of the line which present on the plane
<br>which satisfy the plane<br><br>$$ \the... | integer | jee-main-2021-online-18th-march-morning-shift |
rnZCDGXIib2fJxI3oU1kmlm0roj | maths | 3d-geometry | plane-in-space | The equation of the planes parallel to the plane x $$-$$ 2y + 2z $$-$$ 3 = 0 which are at unit distance from the point (1, 2, 3) is ax + by + cz + d = 0. If (b $$-$$ d) = k(c $$-$$ a), then the positive value of k is : | [] | null | 4 | The equation of the planes parallel to the plane x $$-$$ 2y + 2z $$-$$ 3 = 0
<br><br>$$x - 2y + 2z + \lambda = 0$$<br><br>Now given<br><br>$$d = {{\left| {1 - 4 + 6 + \lambda } \right|} \over {\sqrt 9 }} = 1$$<br><br>$$\left| {\lambda + 3} \right| = 3$$<br><br>$$\lambda + 3 = \pm 3 \Rightarrow \lambda = 0, - 6$$<b... | integer | jee-main-2021-online-18th-march-morning-shift |
wimTMaIOTLKy7lhiU31kmm3y91x | maths | 3d-geometry | plane-in-space | Let the mirror image of the point (1, 3, a) with respect to the plane $$\overrightarrow r .\left( {2\widehat i - \widehat j + \widehat k} \right) - b = 0$$ be ($$-$$3, 5, 2). Then, the value of | a + b | is equal to ____________. | [] | null | 1 | <p>Given equation of plane in vector form is $$\overrightarrow r \,.\,(2\widehat i - \widehat j + \widehat k) - b = 0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3b0s96m/cc9e3c18-ff43-4aab-b277-327a534afd5b/f4fd09e0-d659-11ec-9a06-bd4ec5b93eb4/file-1l3b0s96n.png?format=png" data-orsrc="ht... | integer | jee-main-2021-online-18th-march-evening-shift |
1ks090yaw | maths | 3d-geometry | plane-in-space | Let the plane passing through the point ($$-$$1, 0, $$-$$2) and perpendicular to each of the planes 2x + y $$-$$ z = 2 and x $$-$$ y $$-$$ z = 3 be ax + by + cz + 8 = 0. Then the value of a + b + c is equal to : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "4"}] | ["D"] | null | Normal of required plane $$\left( {2\widehat i + \widehat j - \widehat k} \right) \times \left( {\widehat i - \widehat j - \widehat k} \right)$$<br><br>$$ = - 2\widehat i + \widehat j - 3\widehat k$$<br><br>Equation of plane <br><br>$$ - 2(x + 1) + 1(y - 0) - 3(z + 2) = 0$$<br><br>$$ - 2x + y - 3z - 8 = 0$$<br><br>$$2... | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktd1xchr | maths | 3d-geometry | plane-in-space | Let P be the plane passing through the point (1, 2, 3) and the line of intersection of the planes $$\overrightarrow r \,.\,\left( {\widehat i + \widehat j + 4\widehat k} \right) = 16$$ and $$\overrightarrow r \,.\,\left( { - \widehat i + \widehat j + \widehat k} \right) = 6$$. Then which of the following points does NO... | [{"identifier": "A", "content": "(3, 3, 2)"}, {"identifier": "B", "content": "(6, $$-$$6, 2)"}, {"identifier": "C", "content": "(4, 2, 2)"}, {"identifier": "D", "content": "($$-$$8, 8, 6)"}] | ["C"] | null | $$(x + y + 4z - 16) + \lambda ( - x + y + z - 6) = 0$$<br><br>Passes through (1, 2, 3)<br><br>$$ - 1 + \lambda ( - 2) \Rightarrow \lambda = - {1 \over 2}$$<br><br>$$2(x + y + 4z - 16) - ( - x + y + z - 6) = 0$$<br><br>$$3x + y + 7z - 26 = 0$$ | mcq | jee-main-2021-online-26th-august-evening-shift |
1kto1zhhj | maths | 3d-geometry | plane-in-space | Let the acute angle bisector of the two planes x $$-$$ 2y $$-$$ 2z + 1 = 0 and 2x $$-$$ 3y $$-$$ 6z + 1 = 0 be the plane P. Then which of the following points lies on P? | [{"identifier": "A", "content": "$$\\left( {3,1, - {1 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - 2,0, - {1 \\over 2}} \\right)$$"}, {"identifier": "C", "content": "(0, 2, $$-$$4)"}, {"identifier": "D", "content": "(4, 0, $$-$$2)"}] | ["B"] | null | $${P_1}:x - 2y - 2z + 1 = 0$$<br><br>$${P_2}:2x - 3y - 6z + 1 = 0$$<br><br>$$\left| {{{x - 2y - 2z + 1} \over {\sqrt {1 + 4 + 4} }}} \right| = \left| {{{2x - 3y - 6z + 1} \over {\sqrt {{2^2} + {3^2} + {6^2}} }}} \right|$$<br><br>$${{x - 2y - 2z + 1} \over 3} = \pm {{2x - 3y - 6z + 1} \over 7}$$<br><br>Since $${a_1}{a_... | mcq | jee-main-2021-online-1st-september-evening-shift |
1l544kl7n | maths | 3d-geometry | plane-in-space | <p>If the mirror image of the point (2, 4, 7) in the plane 3x $$-$$ y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal to :</p> | [{"identifier": "A", "content": "54"}, {"identifier": "B", "content": "50"}, {"identifier": "C", "content": "$$-$$6"}, {"identifier": "D", "content": "$$-$$42"}] | ["C"] | null | <p>We know mirror image of point (x<sub>1</sub>, y<sub>1</sub>, z<sub>1</sub>) in the plane ax + by + cz = d</p>
<p>$${{x - {x_1}} \over a} = {{y - {y_1}} \over b} = {{z - {z_1}} \over c} = {{ - 2(a{x_1} + b{y_1} + c{z_1} - d)} \over {{a^2} + {b^2} + {c^2}}}$$</p>
<p>Here given point (2, 4, 7) and plane $$3x - y + 4z =... | mcq | jee-main-2022-online-29th-june-morning-shift |
1l5461cy3 | maths | 3d-geometry | plane-in-space | <p>Let d be the distance between the foot of perpendiculars of the points P(1, 2, $$-$$1) and Q(2, $$-$$1, 3) on the plane $$-$$x + y + z = 1. Then d<sup>2</sup> is equal to ___________.</p> | [] | null | 26 | <p>Foot of perpendicular from P</p>
<p>$${{x - 1} \over { - 1}} = {{y - 2} \over 1} = {{z + 1} \over 1} = {{ - ( - 1 + 2 - 1 - 1)} \over 3}$$</p>
<p>$$ \Rightarrow p' \equiv \left( {{2 \over 3},{7 \over 3},{{ - 2} \over 3}} \right)$$</p>
<p>and foot of perpendicular from Q</p>
<p>$${{x - 2} \over { - 1}} = {{y + 1} \ov... | integer | jee-main-2022-online-29th-june-morning-shift |
1l55ini45 | maths | 3d-geometry | plane-in-space | <p>Let the plane ax + by + cz = d pass through (2, 3, $$-$$5) and is perpendicular to the planes <br/>2x + y $$-$$ 5z = 10 and 3x + 5y $$-$$ 7z = 12. If a, b, c, d are integers d > 0 and gcd (|a|, |b|, |c|, d) = 1, then the value of a + 7b + c + 20d is equal to :</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "20"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "22"}] | ["D"] | null | <p>Equation of pane through point (2, 3, $$-$$5) and perpendicular to planes 2x + y $$-$$ 5z = 10 and 3x + 5y $$-$$ 7z = 12 is</p>
<p>$$\left| {\matrix{
{x - 2} & {y - 3} & {z + 5} \cr
2 & 1 & { - 5} \cr
3 & 5 & { - 7} \cr
} } \right| = 0$$</p>
<p>$$\therefore$$ Equation of plane is $$(x - 2)( - 7 + 25... | mcq | jee-main-2022-online-28th-june-evening-shift |
1l566zqfs | maths | 3d-geometry | plane-in-space | <p>The acute angle between the planes P<sub>1</sub> and P<sub>2</sub>, when P<sub>1</sub> and P<sub>2</sub> are the planes passing through the intersection of the planes $$5x + 8y + 13z - 29 = 0$$ and $$8x - 7y + z - 20 = 0$$ and the points (2, 1, 3) and (0, 1, 2), respectively, is :</p> | [{"identifier": "A", "content": "$${\\pi \\over 3}$$"}, {"identifier": "B", "content": "$${\\pi \\over 4}$$"}, {"identifier": "C", "content": "$${\\pi \\over 6}$$"}, {"identifier": "D", "content": "$${\\pi \\over 12}$$"}] | ["A"] | null | <p>Family of Plane's equation can be given by</p>
<p>$$(5 + 8\lambda )x + (8 - 7\lambda )y + (13 + \lambda )z - (29 + 20\lambda ) = 0$$</p>
<p>P<sub>1</sub> passes through (2, 1, 3)</p>
<p>$$ \Rightarrow (10 + 16\lambda ) + (8 - 7\lambda ) + (39 + 3\lambda ) - (29 + 20\lambda ) = 0$$</p>
<p>$$ \Rightarrow - 8\lambda ... | mcq | jee-main-2022-online-28th-june-morning-shift |
1l57p9hh3 | maths | 3d-geometry | plane-in-space | <p>Let the mirror image of the point (a, b, c) with respect to the plane 3x $$-$$ 4y + 12z + 19 = 0 be (a $$-$$ 6, $$\beta$$, $$\gamma$$). If a + b + c = 5, then 7$$\beta$$ $$-$$ 9$$\gamma$$ is equal to ______________.</p> | [] | null | 137 | <p>$${{x - a} \over 3} = {{y - b} \over { - 4}} = {{z - c} \over {12}} = {{ - 2(3a - 4b + 12c + 19)} \over {{3^2} + {{( - 4)}^2} + {{12}^2}}}$$</p>
<p>$${{x - a} \over 3} = {{y - b} \over { - 4}} = {{z - c} \over {12}} = {{ - 6a + 8b - 24c - 38} \over {169}}$$</p>
<p>$$(x,y,z) \equiv (a - 6,\,\beta ,\gamma )$$</p>
<p>$... | integer | jee-main-2022-online-27th-june-morning-shift |
1l58g6a8s | maths | 3d-geometry | plane-in-space | <p>If the plane $$2x + y - 5z = 0$$ is rotated about its line of intersection with the plane $$3x - y + 4z - 7 = 0$$ by an angle of $${\pi \over 2}$$, then the plane after the rotation passes through the point :</p> | [{"identifier": "A", "content": "(2, $$-$$2, 0)"}, {"identifier": "B", "content": "($$-$$2, 2, 0)"}, {"identifier": "C", "content": "(1, 0, 2)"}, {"identifier": "D", "content": "($$-$$1, 0, $$-$$2)"}] | ["C"] | null | <p>$${P_1}:2x + y - 52 = 0$$, $${P_2}:3x - y + 4z - 7 = 0$$</p>
<p>Family of planes P<sub>1</sub> and P<sub>2</sub></p>
<p>$$P:{P_1} + \lambda {P_2}$$</p>
<p>$$\therefore$$ $$P:(2 + 3\lambda )x + (1 - \lambda )y + ( - 5 + 4\lambda )z - 7\lambda = 0$$</p>
<p>$$\because$$ $$P \bot {P_1}$$</p>
<p>$$\therefore$$ $$4 + 6\l... | mcq | jee-main-2022-online-26th-june-evening-shift |
1l59k5k64 | maths | 3d-geometry | plane-in-space | <p>Let p be the plane passing through the intersection of the planes $$\overrightarrow r \,.\,\left( {\widehat i + 3\widehat j - \widehat k} \right) = 5$$ and $$\overrightarrow r \,.\,\left( {2\widehat i - \widehat j + \widehat k} \right) = 3$$, and the point (2, 1, $$-$$2). Let the position vectors of the points X and... | [{"identifier": "A", "content": "X and X + Y are on the same side of P"}, {"identifier": "B", "content": "Y and Y $$-$$ X are on the opposite sides of P"}, {"identifier": "C", "content": "X and Y are on the opposite sides of P"}, {"identifier": "D", "content": "X + Y and X $$-$$ Y are on the same side of P"}] | ["C"] | null | <p>Let the equation of required plane</p>
<p>$$\pi :(x + 3y - z - 5) + \lambda (2x - y + z - 3) = 0$$</p>
<p>$$\because$$ (2, 1, $$-$$2) lies on it so, $$2 + \lambda ( - 2) = 0$$</p>
<p>$$ \Rightarrow \lambda = 1$$</p>
<p>Hence, $$\pi :3x + 2y - 8 = 0$$</p>
<p>$$\because$$ $${\pi _x} = - 9$$, $${\pi _y} = 5$$, $${\pi... | mcq | jee-main-2022-online-25th-june-evening-shift |
1l5bagla7 | maths | 3d-geometry | plane-in-space | <p>Let the points on the plane P be equidistant from the points ($$-$$4, 2, 1) and (2, $$-$$2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is :</p> | [{"identifier": "A", "content": "$${\\pi \\over 6}$$"}, {"identifier": "B", "content": "$${\\pi \\over 4}$$"}, {"identifier": "C", "content": "$${\\pi \\over 3}$$"}, {"identifier": "D", "content": "$${5\\pi \\over 12}$$"}] | ["C"] | null | <p>Let P(x, y, z) be any point on plane P<sub>1</sub></p>
<p>Then $${(x + 4)^2} + {(y - 2)^2} + {(z - 1)^2} = {(x - 2)^2} + {(y + 2)^2} + {(z - 3)^2}$$</p>
<p>$$ \Rightarrow 12x - 8y + 4z + 4 = 0$$</p>
<p>$$ \Rightarrow 3x - 2y + z + 1 = 0$$</p>
<p>And $${P_2}:2x + y + 3z = 0$$</p>
<p>$$\therefore$$ angle between P<sub... | mcq | jee-main-2022-online-24th-june-evening-shift |
1l6f2mwg1 | maths | 3d-geometry | plane-in-space | <p>A plane $$E$$ is perpendicular to the two planes $$2 x-2 y+z=0$$ and $$x-y+2 z=4$$, and passes through the point $$P(1,-1,1)$$. If the distance of the plane $$E$$ from the point $$Q(a, a, 2)$$ is $$3 \sqrt{2}$$, then $$(P Q)^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "21"}, {"identifier": "D", "content": "33"}] | ["C"] | null | <p>First plane, $${P_1} = 2x - 2y + z = 0$$, normal vector $$ \equiv {\overline n _1} = (2, - 2,1)$$</p>
<p>Second plane, $${P_2} \equiv x - y + 2z = 4$$, normal vector $$ \equiv {\overline n _2} = (1, - 1,2)$$</p>
<p>Plane perpendicular to P<sub>1</sub> and P<sub>2</sub> will have normal vector $${\overline n _3}$$</p... | mcq | jee-main-2022-online-25th-july-evening-shift |
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