If the lines (x-1)/2 = (y+1)/(-3) = (z-2)/2 and x = 0, (2y-4)/(-2) = (z+1)/(2-) are perpendicular, then value of is Option 1: 3/4 Option 2: 3/7 Option 3: 7/3 Option 4: 4/3

Answer: 7/3. Solution: The first line has direction ratios (2,-3,2). The second lies in x=0 and can be written (y-2)/(-1)=(z+1)/(2-), giving direction ratios (0,-1,2-). Perpendicularity requires the dot product to vanish: 0-(-3)+2(2-)=7-3=0, so =73.

Geometry

3D Geometry

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If the lines $\frac{x-1}{2} = \frac{y+1}{\lambda-3} = \frac{z-2}{2}$ and $x = 0, \frac{2y-4}{-2} = \frac{z+1}{2-\lambda}$ are perpendicular, then value of $\lambda$ is

$\frac{3}{4}$

$\frac{3}{7}$

$\frac{7}{3}$

$\frac{4}{3}$

✅ Correct Option: 3

The first line has direction ratios $(2,\lambda-3,2)$ . The second lies in $x=0$ and can be written $\frac{y-2}{-1}=\frac{z+1}{2-\lambda}$ , giving direction ratios $(0,-1,2-\lambda)$ . Perpendicularity requires the dot product to vanish: $0-(\lambda-3)+2(2-\lambda)=7-3\lambda=0$ , so $\lambda=\frac73$ .

2026: 18 May Shift 1

The shortest distance between the lines $\vec{r} = 2\hat{i}+3\hat{j}+4\hat{k}+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $\vec{r} = 5\hat{i}+5\hat{j}+5\hat{k}+\mu(-\hat{i}+\hat{j}-\hat{k})$ where $\lambda,\mu \in R$ (set of real numbers) is:

2025: 13 May Shift 2

The direction cosines of a line equally inclined with the co-ordinate axes are

2026: 21 May Shift 1

Match List-I with List-II

List-I (Straight Lines)	List-II (Direction Ratios)
(A) $\vec{r} = (\hat{i}+\hat{j}) + \lambda(2\hat{i}-\hat{j}+3\hat{k})$	(I) 2,1,-3
(B) $\frac{x}{2} = \frac{1-y}{-1} = \frac{z+1}{3}$	(II) -2,1,3
(C) $\frac{1-2x}{4} = \frac{y+1}{1} = \frac{2z+3}{6}$	(III) 2,-1,3
(D) $\vec{r} = (2\hat{i}-\hat{k}) + \lambda(2\hat{i}+\hat{j}-3\hat{k})$	(IV) 2,1,3

Choose the correct answer from the options given below:

2025: 29 May Shift 2

The direction cosines of a line which makes equal angles with co-ordinate axes are:

2022: 17 Aug Shift 2

The distance of plane $\vec{r}\cdot(6\hat{i} - 3\hat{j} - 2\hat{k}) + 1 = 0$ from origin is :

2022: 4 Aug Shift 1

The equation of line of intersection of the planes $x + y + 3z = 7$ and $x - y + 2z = 3$ is:

If the lines x−12=y+1λ−3=z−22\frac{x-1}{2} = \frac{y+1}{\lambda-3} = \frac{z-2}{2}2x−1​=λ−3y+1​=2z−2​ and x=0,2y−4−2=z+12−λx = 0, \frac{2y-4}{-2} = \frac{z+1}{2-\lambda}x=0,−22y−4​=2−λz+1​ are perpendicular, then value of λ\lambdaλ is

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CUET Mathematics Past Year Questions (Topic-Wise)

If the lines $\frac{x-1}{2} = \frac{y+1}{\lambda-3} = \frac{z-2}{2}$ and $x = 0, \frac{2y-4}{-2} = \frac{z+1}{2-\lambda}$ are perpendicular, then value of $\lambda$ is