Acute angle between the lines x/3 = y/4 = z/5 and (x-1)/4 = (y+1)/(-3) = (z+10)/5 is: Option 1: ()/2 Option 2: ()/3 Option 3: ()/6 Option 4: ()/4

Answer: ()/3. Solution: For any line written as (x-x_0)/a = (y-y_0)/b = (z-z_0)/c, the direction vector is (a, b, c). From Line 1: x/3 = y/4 = z/5 Direction vector: b_1 = (3, 4, 5) From Line 2: (x-1)/4 = (y+1)/(-3) = (z+10)/5 Direction vector: b_2 = (4, -3, 5) The dot product of the two direction vectors: b_1 * b_2 = (3)(4) + (4)(-3) + (5)(5) = 12 - 12 + 25 = 25 The magnitude of b_1:, b_1, = √(3^2 + 4^2 + 5^2) = √(9 + 16 + 25) = √50 = 5√2 The magnitude of b_2:, b_2, = √(4^2 + (-3)^2 + 5^2) = √(16 + 9 + 25) = √50 = 5√2 The acute angle between the lines is given by: = (, b_1 * b_2, )/(, b_1, x, b_2, ) = (, 25, )/((5√2)(5√2)) = 25/(25 x 2) = 25/50 = 1/2 = 1/2 = ()/3 Therefore, the acute angle between the lines is ()/3.

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Acute angle between the lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{5}$ and $\frac{x-1}{4} = \frac{y+1}{-3} = \frac{z+10}{5}$ is:

$\frac{\pi}{2}$

$\frac{\pi}{3}$

$\frac{\pi}{6}$

$\frac{\pi}{4}$

✅ Correct Option: 2

For any line written as $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$ , the direction vector is $(a, b, c)$ .

From Line 1: $\frac{x}{3} = \frac{y}{4} = \frac{z}{5}$

Direction vector: $\vec{b_1} = (3, 4, 5)$

From Line 2: $\frac{x-1}{4} = \frac{y+1}{-3} = \frac{z+10}{5}$

Direction vector: $\vec{b_2} = (4, -3, 5)$

The dot product of the two direction vectors:

$\vec{b_1} \cdot \vec{b_2} = (3)(4) + (4)(-3) + (5)(5)$

$= 12 - 12 + 25$

$= 25$

The magnitude of $\vec{b_1}$ :

$|\vec{b_1}| = \sqrt{3^2 + 4^2 + 5^2}$

$= \sqrt{9 + 16 + 25}$

$= \sqrt{50}$

$= 5\sqrt{2}$

The magnitude of $\vec{b_2}$ :

$|\vec{b_2}| = \sqrt{4^2 + (-3)^2 + 5^2}$

$= \sqrt{16 + 9 + 25}$

$= \sqrt{50}$

$= 5\sqrt{2}$

The acute angle between the lines is given by:

$\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| \times |\vec{b_2}|}$

$\cos \theta = \frac{|25|}{(5\sqrt{2})(5\sqrt{2})}$

$= \frac{25}{25 \times 2}$

$= \frac{25}{50}$

$= \frac{1}{2}$

$\cos \theta = \frac{1}{2}$

$\theta = \frac{\pi}{3}$

Therefore, the acute angle between the lines is $\frac{\pi}{3}$ .

2026: 30 May Shift 1

Let $L_1$ and $L_2$ lines such that $L_1:\vec{r}=\hat{i}+\hat{j}+\lambda(2\hat{i}-\hat{j}+\hat{k})$ , $L_2:\vec{r}=2\hat{i}+\hat{j}-\hat{k}+\mu(4\hat{i}-2\hat{j}+2\hat{k})$ and $\theta$ be the acute angle between $L_1$ and $L_2$ , then

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	$\cos\theta$	I.	$\dfrac{\sqrt{66}}{6}$
B.	$\sin\theta$	II.	$3$
C.	the shortest distance (in units) between $L_1$ and $L_2$	III.	$1$
D.	(sum of squares of direction cosines of $L_1$ ) $+2$	IV.	$0$

Choose the correct answer from the options given below:

2025: 30 May Shift 1

The shortest distance between the lines $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$ and $\vec{r} = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \mu(4\hat{i} + 6\hat{j} + 8\hat{k})$ is equal to

2026: 12 May Shift 2

For a given line L: $\dfrac{x}{2}=\dfrac{y+1}{3}=\dfrac{-z-3}{-5}$ , which of the following is/are True ?

(A) The direction ratio of line L is (2, 3, -5)

(B) The perpendicular distance of the point (1, 1, 3) on the line L is $\sqrt{3}$

(D) The image of point (1, 1, 3) with respect to line L is (3, 3, 1)

Choose the correct answer from the options given below:

2026: 30 May Shift 2

The equation of line passing through the points (2, -1, 3) and (4, 2, 1) is

(A) $\frac{x-2}{2} = \frac{y+1}{3} = \frac{z-3}{-2}$

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

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

Choose the correct answer from the options given below:

2026: 18 May Shift 1

Equation of a line passing through (1, 2, -3) which is parallel to the line $\frac{x+2}{3} = \frac{y-3}{-4} = \frac{z+5}{5}$ is

2025: 15 May Shift 2

If the lines $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{\lambda}$ and $\frac{x}{1} = \frac{y}{2\lambda} = \frac{z}{3}$ are perpendicular to each other, then $\lambda$ is equal to

Acute angle between the lines x3=y4=z5\frac{x}{3} = \frac{y}{4} = \frac{z}{5}3x​=4y​=5z​ and x−14=y+1−3=z+105\frac{x-1}{4} = \frac{y+1}{-3} = \frac{z+10}{5}4x−1​=−3y+1​=5z+10​ is:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Acute angle between the lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{5}$ and $\frac{x-1}{4} = \frac{y+1}{-3} = \frac{z+10}{5}$ is: