3D Geometry- JEE Advanced Previous Year Questions with Solutions

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Q. (A) Let $\mathrm{P}(3,2,6)$ be a point in space and $\mathrm{Q}$ be a point on the line $\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\mu(-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+5 \hat{\mathrm{k}}) .$ Then the value of $\mu$ for which the vector $\overline{\mathrm{PQ}}$ is parallel to the plane $x-4 y+3 z=1$ is –

(A) $\frac{1}{4}$

(B) $-\frac{1}{4}$

(C) $\frac{1}{8}$

(D) $-\frac{1}{8}$

(B) A line with positive direction cosines passes through the point P (2, –1, 2) and makes equal angles with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals –

(A) 1

(B) $\sqrt{2}$

(C) $\sqrt{3}$

(D) 2

(C) Let $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ be points with integer coordinates satisfying the system of homogeneous equations $: 3 x-y-z=0 ;-3 x+z=0 ;-3 x+2 y+z=0 .$ Then the number of such points for which $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2} \leq 100$ is

[JEE 2009, 3+3+4]

Sol. ( (a) $A ;(b) C ;(c) 7$ )


Q. (A) Equation of the plane containing the straight line $\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$

(A) x + 2y – 2z = 0

(B) 3x + 2y – 2z = 0

(C) x – 2y + z = 0

(D) 5x + 2y – 4z = 0

(B) If the distance of the point $\mathrm{P}(1,-2,1)$ from the plane $\mathrm{x}+2 \mathrm{y}-2 \mathrm{z}=\alpha$ where $\alpha>0,$ is $5,$ then the foot of the perpendicular from $P$ to the plane is-

(A) $\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)$

(B) $\left(\frac{4}{3},-\frac{4}{3}, \frac{1}{3}\right)$

(C) $\left(\frac{1}{3}, \frac{2}{3}, \frac{10}{3}\right)$

(D) $\left(\frac{2}{3},-\frac{1}{3}, \frac{5}{2}\right)$

(C) If the distance between the plane Ax – 2y + z = d and the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt{6},$ then $|d|$ is

(D) Match the statements in Column-I with the values in Column-II.

[JEE 2010, 3+5+3+(2+2+2+2)]

Sol. ((A) $\mathrm{C} ;(\mathrm{B}) \mathrm{A} ;(\mathrm{C}) 6 ;(\mathrm{D})(\mathrm{A}) \mathrm{t}(\mathrm{B}) \mathrm{p}, \mathrm{r}(\mathrm{C}) \mathrm{q}(\mathrm{D}) \mathrm{r}$ )

(a) Normal vector to the plane containing the

lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is

$\hat{n}=\left|\begin{array}{lll}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {3} & {4} & {2} \\ {4} & {2} & {3}\end{array}\right|=8 \hat{i}-\hat{j}-10 \hat{k}$

Let direction ratios of required plane be a, b, c.

Now 8a – b – 10c = 0

and $2 \mathrm{a}+3 \mathrm{b}+4 \mathrm{c}=0\left(\because \text { plane contains the line } \frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{4}\right)$

$\Rightarrow \frac{a}{1}=\frac{b}{-2}=\frac{c}{1}$

$\cdot$ plane contains the line, which passes through origin, hence origin lies on a plane.

$\Rightarrow$ equation of required plane is $x-2 y+z=0$

(b) $\quad \because \quad\left|\frac{1-4-2-\alpha}{3}\right|=5$

$\Rightarrow \alpha=10,-20$

$\Rightarrow \alpha=10 \because \alpha>0$


Q. (A) The point P is the intersection of the straight line joining the points Q(2,3,5) and R(1,–1,4) with the plane 5x – 4y – z = 1. If S is the foot of the perpendicular drawn from the point T(2,1,4) to QR, then the length of the line segment PS is –

(A) $\frac{1}{\sqrt{2}}$

(B) $\sqrt{2}$

(C) 2

(D) $2 \sqrt{2}$

(B) The equation of a plane passing through the line of intersection of the planes x + 2y $+3 \mathrm{z}=2$ and $\mathrm{x}-\mathrm{y}+\mathrm{z}=3$ and at a distance $\frac{2}{\sqrt{3}}$ from the point $(3,1,-1)$ is

(A) $5 x-11 y+z=17$

(B) $\sqrt{2} x+y=3 \sqrt{2}-1$

(C) $x+y+z=\sqrt{3}$

(D) $x-\sqrt{2} y=1-\sqrt{2}$

(C) If the straight lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar, then the plane(s) containing these two lines is(are)

(A) y + 2z = –1

(B) y + z = –1

(C) y – z = –1

(D) y – 2z = –1

[JEE 2012, 3+3+4]

Sol. ((a) $A ;(b) A ;(c) B, C$)

(a) Line QR :

$\frac{x-2}{1}=\frac{y-3}{4}=\frac{z-5}{1}=\lambda$

Any point on line QR :

$(\lambda+2,4 \lambda+3, \lambda+5)$

$\therefore$ Point of intersection with plane :

$5 \lambda+10-16 \lambda-12-\lambda-5=1$

$\Rightarrow \lambda=-\frac{2}{3}$

$\therefore \mathrm{P}\left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right)$


Q. Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x$ $+\mathrm{y}+\mathrm{z}=3 .$ The feet of perpendiculars lie on the line

(A) $\frac{x}{5}=\frac{y-1}{8}=\frac{z-2}{-13}$

(B) $\frac{x}{2}=\frac{y-1}{3}=\frac{z-2}{-5}$

(C) $\frac{x}{4}=\frac{y-1}{3}=\frac{z-2}{-7}$

(D) $\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$

[JEE-Advanced 2013, 2]

Sol. (D)


Q. A line $\ell$ passing through the origin is perpendicular to the lines

$\ell_{1}:(3+\mathrm{t}) \hat{\mathrm{i}}+(-1+2 \mathrm{t}) \hat{\mathrm{j}}+(4+2 \mathrm{t}) \hat{\mathrm{k}},-\infty<\mathrm{t}<\infty$

$\ell_{2}:(3+2 s) \hat{\mathrm{i}}+(3+2 \mathrm{s}) \hat{\mathrm{j}}+(2+\mathrm{s}) \hat{\mathrm{k}},-\infty<\mathrm{s}<\infty$ Then, the coordinate(s) of the point(s) on $\ell_{2}$

at a distance of $\sqrt{17}$ from the point of intersection of $\ell$ and $\ell_{1}$ is (are) –

(A) $\left(\frac{7}{3}, \frac{7}{3}, \frac{5}{3}\right)$

(B) (–1,–1,0)

(C) (1,1,1)

(D) $\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$

[JEE-Advanced 2013, 4, (–1)]

Sol. (B,D)


Q. Two lines $L_{1}: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$ and $L_{2}: x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha}$ are coplanar. Then $\alpha$ can take value(s)

(A) 1              (B) 2             (C) 3                  (D) 4

[JEE-Advanced 2013, 3, (–1)]

Sol. (A,D)

$\mathrm{L}_{1}: \frac{\mathrm{x}-5}{0}=\frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2}$

$\mathrm{L}_{2}: \frac{\mathrm{x}-\alpha}{0}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}$

for lines to be coplanar

$\left|\begin{array}{ccc}{5-\alpha} & {0} & {0} \\ {0} & {3-\alpha} & {-2} \\ {0} & {-1} & {2-\alpha}\end{array}\right|=0$

$\Rightarrow \quad(5-\alpha)((3-\alpha)(2-\alpha)-2)=0$

$\Rightarrow \quad(5-\alpha)\left(\alpha^{2}-5 \alpha+4\right)=0$

$\Rightarrow \quad \alpha=1,4,5$


Q. Consider the lines $\mathrm{L}_{1}: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}+3}{1}, \mathrm{L}_{2}: \frac{\mathrm{x}-4}{1}=\frac{\mathrm{y}+3}{1}=\frac{\mathrm{z}+3}{2}$ and the planes $\mathrm{P}_{1}: 7 \mathrm{x}+\mathrm{y}+2 \mathrm{z}=3, \mathrm{P}_{2}: 3 \mathrm{x}+5 \mathrm{y}-6 \mathrm{z}=4 .$ Let $\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{d}$ be the

equation of the plane passing through the point of intersection of lines $L_{1}$ and $\mathrm{L}_{2}$ and perpendicular to planes $\mathrm{P}_{1}$ and $\mathrm{P}_{2} .$ Match List-I with List-II and select the correct answer using the code given below the lists.

[JEE-Advanced 2013, 3, (–1)]

Sol. (A)

For point of intersection of $L_{1}$ and $L_{2}$

$\left\{\begin{array}{l}{2 \lambda+1=\mu+4} \\ {-\lambda=\mu-3} \\ {\lambda-3=2 \mu-3}\end{array}\right.$

$\Rightarrow \mu=1$

$\Rightarrow$ point of intersction is $(5,-2,-1)$

Now, vector normal to the plane is $\overrightarrow{\mathrm{n}}_{1} \times \overrightarrow{\mathrm{n}}_{2}=\left|\begin{array}{ccc}{\hat{\mathrm{i}}} & {\hat{\mathrm{j}}} & {\hat{\mathrm{k}}} \\ {7} & {1} & {2} \\ {3} & {5} & {-6}\end{array}\right|$

$=-16(\hat{i}-3 \hat{j}-2 \hat{k})$

Let equation of required plane be $x-3 y-2 z=\alpha$

$\because$ it passes through $(5,-2,-1)$

$\therefore \alpha=13$

$\Rightarrow$ equation of plane is $x-3 y-2 z=13$


Q. From a point $\mathrm{P}(\lambda, \lambda, \lambda),$ perpendiculars $\mathrm{PQ}$ and $\mathrm{PR}$ are drawn respectively on the lines $\mathrm{y}=$ $\mathrm{x}, \mathrm{z}=1$ and $\mathrm{y}=-\mathrm{x}, \mathrm{z}=-1 .$ If $\mathrm{P}$ is such that $\angle \mathrm{QPR}$ is a right angle, then the possible value(s) of $\lambda$ is (are)

(A) $\sqrt{2}$              (B) 1            (C) –1              (D) $-\sqrt{2}$

[JEE(Advanced)-2014, 3]

Sol. (C)

Line $\mathrm{L}_{1}$ given by $\mathrm{y}=\mathrm{x} ; \mathrm{z}=1$ can be expressed as

$\mathrm{L}_{1}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}-1}{0}$

Similarly $\mathrm{L}_{2}(\mathrm{y}=-\mathrm{x} ; \mathrm{z}=-1)$ can be expressed as

$\mathrm{L}_{2}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}+1}{0}$

Let any point $\mathrm{Q}(\alpha, \alpha, 1)$ on $\mathrm{L}_{1}$ and $\mathrm{R}(\beta,-\beta,-1)$ on $\mathrm{L}_{2}$

Given that $\mathrm{PQ}$ is perpendicular to $\mathrm{L}_{1}$

$\Rightarrow(\lambda-\alpha) .1+(\lambda-\alpha) \cdot 1+(\lambda-1) \cdot 0=0 \Rightarrow \lambda=\alpha$

$\therefore \mathrm{Q}(\lambda, \lambda, 1)$

Similarly PR is perpendicular to L $_{2}$ $(\lambda-\beta) \cdot 1+(\lambda+\beta)(-1)+(\lambda+1) \cdot 0=0 \Rightarrow \beta=0$

$\therefore \mathrm{R}(0,0,-1)$

Now as given

$\Rightarrow \overrightarrow{\mathrm{PR}} \cdot \overrightarrow{\mathrm{PQ}}=0$

$0 . \lambda+0 . \lambda+(\lambda-1)(\lambda+1)=0$

$\lambda \neq 1$ as $\mathrm{P} \& \mathrm{Q}$ are different points

$\Rightarrow \lambda=-1$


Q. In $\mathbb{D}^{3},$ consider the planes $P_{1}: y=0$ and $P_{2}: x+z=1 .$ Let $P_{3}$ be a plane, different from $\mathrm{P}_{1}$ and $\mathrm{P}_{2},$ which passes through the intersection of $\mathrm{P}_{1}$ and $\mathrm{P}_{2} .$ If the distance of the point $(0,1,0)$ from $P_{3}$ is 1 and the distance of a point $(\alpha, \beta, \gamma)$ from $P_{3}$ is $2,$ then which of the following relations is (are) true?

(A) $2 \alpha+\beta+2 \gamma+2=0$

(B) $2 \alpha-\beta+2 \gamma+4=0$

(C) $2 \alpha+\beta-2 \gamma-10=0$

(D) $2 \alpha-\beta+2 \gamma-8=0$

[JEE 2015, 4M, –2M]

Sol. (B,D)


Q. In $\square^{3},$ let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P_{1}: x+2 y-z+1=0$ and $P_{2}: 2 x-y+$ $\mathrm{z}-1=0 .$ Let $\mathrm{M}$ be the locus of the feet of the perpendiculars drawn from the points on L to the plane $P_{1} .$ Which of the following points lie(s) on M?

[JEE 2015, 4M, –2M]

Sol. (A,B)

Straight line ‘L’ is parallel to line of intersection of plane $\mathrm{P}_{1} \&$ plane $\mathrm{P}_{2}$

$\therefore$ Equation of line $^{\prime} \mathbf{L}^{\prime}$ is

$\frac{x}{1}=\frac{y}{-3}=\frac{z}{-5}=\lambda$

$\frac{\alpha-\lambda}{1}=\frac{\beta+3 \lambda}{2}=\frac{\gamma+5 \lambda}{-1}=\mathrm{k}$

$\left.\begin{array}{l}{\alpha=\mathrm{k}+\lambda} \\ {\beta=2 \mathrm{k}-3 \lambda} \\ {\mathrm{y}=-\mathrm{k}-5 \lambda}\end{array}\right\}$ …(1)

satisfying in plane $\mathrm{P}_{1}$

$\mathrm{k}+\lambda+4 \mathrm{k}-6 \lambda+\mathrm{k}+5 \lambda+1=0$

$6 k=-1$

putting in ( 1) required locus is

$\mathrm{x}=-\frac{1}{6}+\lambda$

$y=-\frac{1}{3}-3 \lambda$

$z=\frac{1}{6}-5 \lambda$

Now check the options.


Q. Consider a pyramid OPQRS located in the first octant $(\mathrm{x} \geq 0, \mathrm{y} \geq 0, \mathrm{z} \geq 0)$ with $\mathrm{O}$ as origin, and OP and OR along the x-axis and the y-axis, respectively. The base OPQR of the pyramid is a square with $\mathrm{OP}=3 .$ The point $\mathrm{S}$ is directly above the mid-point $\mathrm{T}$ of diagonal OQ such that TS $=3 .$ Then-

(A) the acute angle between $\mathrm{OQ}$ and $\mathrm{OS}$ is $\frac{\mathrm{K}}{3}$

(B) the equaiton of the plane containing the triangle $\mathrm{OQS}$ is $\mathrm{x}-\mathrm{y}=0$

(C) the length of the perpendicular from $P$ to the plane containing the triangle OQS is $\frac{3}{\sqrt{2}}$

(D) the perpendicular distance from $\mathrm{O}$ to the straight line containing RS is $\sqrt{\frac{15}{2}}$

[JEE(Advanced) 2016]

Sol. (B,C,D)


Q. Let $P$ be the image of the point $(3,1,7)$ with respect to the plane $x-y+z=3 .$ Then the equation of the plane passing through $\mathrm{P}$ and containing the straight line $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{1}$ is

(A) x + y – 3z = 0

(B) 3x + z = 0

(C) x – 4y + 7z = 0

(D) 2x – y = 0

[JEE(Advanced) 2016]

Sol. (C)

$\therefore x-4 y+7 z=0$


Q. The equation of the plane passing through the point (1,1,1) and perpendicular to the planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7, is-

(A) 14x + 2y + 15z = 31

(B) 14x + 2y – 15z = 1

(C) –14x + 2y + 15z = 3

(D) 14x – 2y + 15z = 27

[JEE(Advanced) 2017]

Sol. (A)


Q. Let $P_{1}: 2 x+y-z=3$ and $P_{2}: x+2 y+z=2$ be two planes. Then, which of the following statement(s) is (are) TRUE ?

(A) The line of intersection of $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ has direction ratios $1,2,-1$

(B) The line $\frac{3 x-4}{9}=\frac{1-3 y}{9}=\frac{z}{3}$ is perpendicular to the line of intersection of $P_{1}$ and $P_{2}$

(C) The acute angle between $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ is $60^{\circ}$

(D) If $P_{3}$ is the plane passing through the point $(4,2,-2)$ and perpendicular to the line of intersection of $\mathrm{P}_{1}$ and $\mathrm{P}_{2},$ then the distance of the point $(2,1,1)$ from the plane $\mathrm{P}_{2}$ is $\frac{2}{\sqrt{3}}$

[JEE(Advanced) 2018]

Sol. (C,D)

D.C. of line of intersection $(a, b, c)$

$\begin{aligned} \Rightarrow \quad & 2 \mathrm{a}+\mathrm{b}-\mathrm{c}=0 \\ & \mathrm{a}+2 \mathrm{b}+\mathrm{c}=0 \end{aligned}$

$\frac{a}{1+2}=\frac{b}{-1-2}=\frac{c}{4-1}$

$\therefore \quad \mathrm{D} . \mathrm{C} .$ is $(1,-1,1)$

B) $\frac{3 x-4}{9}=\frac{1-3 y}{9}=\frac{z}{3}$

$\Rightarrow \quad \frac{x-4 / 3}{3}=\frac{y-1 / 3}{-3}=\frac{z}{3}$

$\Rightarrow \quad$ lines are parallel.

(C) Acute angle between $\mathrm{P}_{1}$ and $\mathrm{P}_{2}=\cos ^{-1}\left(\frac{2 \times 1+1 \times 2-1 \times 1}{\sqrt{6} \sqrt{6}}\right)$ \[ =\cos ^{-1}\left(\frac{3}{6}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ} \]

(D) Plane is given by $(x-4)-(y-2)+(z+2)=0$ \[ \Rightarrow \quad x-y+z=0 \]

Distance of $(2,1,1)$ from plane $=\frac{2-1+1}{\sqrt{3}}=\frac{2}{\sqrt{3}}$


Q. Consider the cube in the first octant with sides $\mathrm{OP}, \mathrm{OQ}$ and $\mathrm{OR}$ of length $1,$ along the x-axis,

y-axis and z-axis, respectively, where $\mathrm{O}(0,0,0)$ is the origin. Let $\mathrm{S}\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$ be the centre

of the cube and $T$ be the vertex of the cube opposite to the origin $O$ such that $S$ lies on the diagonal OT.If $\overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{SP}}, \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{SQ}}, \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{SR}}$ and $\overrightarrow{\mathrm{t}}=\overrightarrow{\mathrm{ST}},$ then the value of $|(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}) \times(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{t}})|$ is

[JEE(Advanced) 2018]

Sol. 8


Q. Let P be a point in the first octant, whose image Q in the plane x + y = 3 (that is, the line segment PQ is perpendicular to the plane x + y = 3 and the mid-point of PQ lies in the plane x + y = 3) lies on the z-axis. Let the distance of P from the x-axis be 5. If R is the image of P in the xy-plane, then the length of PR is

Sol. 8


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