The value of k for which the lines (2x - 3)/4 = (3 - y)/k = (z - 2)/(-2) and (x - 2)/1 = y/4 = (5 - z)/3 are perpendicular to each other is: Option 1: 5/2 Option 2: -5/2 Option 3: 2 Option 4: -2

Answer: 2. Solution: The lines need to be rewritten in standard form (x - x_0)/a = (y - y_0)/b = (z - z_0)/c, where (a, b, c) are the direction ratios. For Line 1: (2x - 3)/4 = (3 - y)/k = (z - 2)/(-2) (2(x - 3/2))/4 = (-(y - 3))/k = (z - 2)/(-2) (x - 3/2)/2 = (y - 3)/(-k) = (z - 2)/(-2) Direction ratios of Line 1: (a_1, b_1, c_1) = (2,\ -k,\ -2) For Line 2: (x - 2)/1 = y/4 = (5 - z)/3 (x - 2)/1 = (y - 0)/4 = (-(z - 5))/3 = (z - 5)/(-3) Direction ratios of Line 2: (a_2, b_2, c_2) = (1,\ 4,\ -3) For two lines to be perpendicular, the dot product of their direction ratios must equal zero: a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 (2)(1) + (-k)(4) + (-2)(-3) = 0 2 - 4k + 6 = 0 8 - 4k = 0 4k = 8 k = 2

Geometry

3D Geometry

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The value of k for which the lines $\frac{2x - 3}{4} = \frac{3 - y}{k} = \frac{z - 2}{-2}$ and $\frac{x - 2}{1} = \frac{y}{4} = \frac{5 - z}{3}$ are perpendicular to each other is:

$\frac{5}{2}$

$-\frac{5}{2}$

-2

✅ Correct Option: 3

The lines need to be rewritten in standard form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$ , where $(a, b, c)$ are the direction ratios.

For Line 1: $\frac{2x - 3}{4} = \frac{3 - y}{k} = \frac{z - 2}{-2}$

$\frac{2(x - \frac{3}{2})}{4} = \frac{-(y - 3)}{k} = \frac{z - 2}{-2}$

$\frac{x - \frac{3}{2}}{2} = \frac{y - 3}{-k} = \frac{z - 2}{-2}$

Direction ratios of Line 1: $(a_1, b_1, c_1) = (2,\ -k,\ -2)$

For Line 2: $\frac{x - 2}{1} = \frac{y}{4} = \frac{5 - z}{3}$

$\frac{x - 2}{1} = \frac{y - 0}{4} = \frac{-(z - 5)}{3} = \frac{z - 5}{-3}$

Direction ratios of Line 2: $(a_2, b_2, c_2) = (1,\ 4,\ -3)$

For two lines to be perpendicular, the dot product of their direction ratios must equal zero:

$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$

$(2)(1) + (-k)(4) + (-2)(-3) = 0$

$2 - 4k + 6 = 0$

$8 - 4k = 0$

$4k = 8$

$k = 2$

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: 30 May Shift 2

The coordinates of the image of the point P (5, 4, 2) in the line $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \mu(2\hat{i} + 3\hat{j} - \hat{k})$ , where $\lambda$ is a parameter, is

2022: 6 Aug Shift 2

The equation of plane passing through the point of (3, 2, 0) and containing the line $\frac{x-2}{2} = \frac{y+3}{4} = \frac{z-1}{1}$ is

2025: 27 May Shift 1

Cosine of the acute angle between the lines $\frac{x-3}{2} = \frac{y-2}{1} = \frac{z-5}{2}$ and $\frac{x-1}{6} = \frac{y-3}{-3} = \frac{z+6}{2}$ is

2025: 3 June Shift 1

The shortest distance between the lines $\vec{r} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = 2\hat{i} + \hat{j} - \hat{k} + \mu(4\hat{i} - 2\hat{j} + 2\hat{k})$ is

2026: 15 May Shift 1

Match the LIST-I with LIST-II

	LIST-I (Equation of Line)		LIST-II (Direction Ratio of line)
A.	$\dfrac{2-x}{1} = \dfrac{2y+1}{1} = \dfrac{z}{1}$	I.	$2, 1, 1$
B.	$x = 2y+3,\ z = y+1$	II.	$1, -2, 2$
C.	$\dfrac{x-1}{1} = \dfrac{y+1}{2},\ z = 1$	III.	$-2, 1, 2$
D.	$\vec{r} = \hat{i} + \hat{j} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})$	IV.	$1, 2, 0$

Choose the correct answer from the options given below:

The value of k for which the lines 2x−34=3−yk=z−2−2\frac{2x - 3}{4} = \frac{3 - y}{k} = \frac{z - 2}{-2}42x−3​=k3−y​=−2z−2​ and x−21=y4=5−z3\frac{x - 2}{1} = \frac{y}{4} = \frac{5 - z}{3}1x−2​=4y​=35−z​ are perpendicular to each other is:

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CUET Mathematics Past Year Questions (Topic-Wise)

The value of k for which the lines $\frac{2x - 3}{4} = \frac{3 - y}{k} = \frac{z - 2}{-2}$ and $\frac{x - 2}{1} = \frac{y}{4} = \frac{5 - z}{3}$ are perpendicular to each other is: