Geometry

3D Geometry

Hard

core

If the straight lines $x = 1 + s$ , $y = -3 - \lambda s$ , $z = 1 + \lambda s$ and $x = \frac{t}{2}$ , $y = 1 + t$ , $z = 2 - t$ with parameters $s$ and $t$ respectively, are coplanar, then $\lambda$ is equal to :

$-2$

$-1$

✅ Correct Option: 2

Direction vectors: $\vec{d_1}=(1,-\lambda,\lambda)$ , $\vec{d_2}=(1,2,-2)$ .

Vector between points: $(-1,4,1)$ .

Coplanarity: $\begin{vmatrix} -1 & 4 & 1 \\ 1 & -\lambda & \lambda \\ 1 & 2 & -2 \end{vmatrix} = 0$ .

This gives $5\lambda + 10 = 0 \Rightarrow \lambda = -2$ .

2026: 21 May Shift 2

Match the LIST-I with LIST-II

LIST-I (Line)		LIST-II (Direction cosiness)
A. $\vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(3\hat{i} + 2\hat{k})$		I. $\left\langle \dfrac{2}{\sqrt{5}}, \dfrac{1}{\sqrt{5}}, 0 \right\rangle$
B. $\dfrac{x-1}{2} = \dfrac{3-y}{1} = \dfrac{2z-1}{4}$		II. $\left\langle -\dfrac{1}{3}, \dfrac{2}{3}, -\dfrac{2}{3} \right\rangle$
C. $\dfrac{x+1}{1} = \dfrac{2y+1}{1}, z = -1$		III. $\left\langle \dfrac{3}{\sqrt{13}}, 0, \dfrac{2}{\sqrt{13}} \right\rangle$
D. $\dfrac{1-x}{1} = \dfrac{y}{2} = \dfrac{1-z}{2}$		IV. $\left\langle \dfrac{2}{3}, -\dfrac{1}{3}, \dfrac{2}{3} \right\rangle$

Choose the correct answer from the options given below:

2022: 7 Aug Shift 2

Equation of y-axis in space, in vector form is :

2022: 4 Aug Shift 1

The points of trisection of the segment joining the points (1, 0, 2) and (1, 3, 2) are:

A. $(1, \frac{3}{2}, \frac{4}{3})$

B. (1, 1, 2)

C. (1, 2, 2)

D. $(1, \frac{3}{2}, 2)$

Choose the correct answer from the options given below:

2025: 14 May Shift 1

The shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{4} = \frac{y-4}{6} = \frac{z-5}{8}$ is equal to

2026: 19 May Shift 2

If the lines $\frac{x-1}{2} = \frac{y-m}{3} = \frac{z-3}{4}$ and $\frac{x-4}{5} = \frac{y-1}{2} = z$ are interesecting, then the value of $m$ , is

2022: 17 Aug Shift 2

The equation of the plane, parallel to the plane $3x + 4y - 12z = 3$ and passes through $(1, 1, -1)$ , is :

If the straight lines x=1+sx = 1 + sx=1+s, y=−3−λsy = -3 - \lambda sy=−3−λs, z=1+λsz = 1 + \lambda sz=1+λs and x=t2x = \frac{t}{2}x=2t​, y=1+ty = 1 + ty=1+t, z=2−tz = 2 - tz=2−t with parameters sss and ttt respectively, are coplanar, then λ\lambdaλ is equal to :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If the straight lines $x = 1 + s$ , $y = -3 - \lambda s$ , $z = 1 + \lambda s$ and $x = \frac{t}{2}$ , $y = 1 + t$ , $z = 2 - t$ with parameters $s$ and $t$ respectively, are coplanar, then $\lambda$ is equal to :