The angle between the pair of lines given by r = i + 2j - 3k + (i - 2j + 2k) and r = 5i + j + k + (3i - 2j + 6k) is Option 1: ^(-1)(21/19) Option 2: ^(-1)(19/21) Option 3: ^(-1)(19/21) Option 4: ^(-1)(21/19)

Answer: ^(-1)(19/21). Solution: The angle between two lines is the angle between their respective direction vectors. From the given equations, we identify the direction vectors: * Vector 1 (b_1): i - 2j + 2k * Vector 2 (b_2): 3i - 2j + 6k The cosine of the angle is given by the formula: = (, b_1 * b_2, )/(, b_1, b_2, ) Calculate the Dot Product (b_1 * b_2): b_1 * b_2 = (1)(3) + (-2)(-2) + (2)(6) b_1 * b_2 = 3 + 4 + 12 = 19 Calculate the Magnitudes (, b_1, and, b_2, ): , b_1, = √(1^2 + (-2)^2 + 2^2) = √(1 + 4 + 4) = √9 = 3 , b_2, = √(3^2 + (-2)^2 + 6^2) = √(9 + 4 + 36) = √49 = 7 Substitute into the Cosine Formula: = 19/(3 x 7) = 19/21 = ^(-1)(19/21)

Geometry

3D Geometry

Medium

core

The angle between the pair of lines given by $\vec{r} = \hat{i} + 2\hat{j} - 3\hat{k} + \lambda (\hat{i} - 2\hat{j} + 2\hat{k})$ and $\vec{r} = 5\hat{i} + \hat{j} + \hat{k} + \mu (3\hat{i} - 2\hat{j} + 6\hat{k})$ is

$\cos^{-1}\left(\frac{21}{19}\right)$

$\sin^{-1}\left(\frac{19}{21}\right)$

$\cos^{-1}\left(\frac{19}{21}\right)$

$\sin^{-1}\left(\frac{21}{19}\right)$

✅ Correct Option: 3

The angle $\theta$ between two lines is the angle between their respective direction vectors. From the given equations, we identify the direction vectors:

Vector 1 ( $\vec{b}_1$ ): $\hat{i} - 2\hat{j} + 2\hat{k}$
Vector 2 ( $\vec{b}_2$ ): $3\hat{i} - 2\hat{j} + 6\hat{k}$

The cosine of the angle $\theta$ is given by the formula:

$\cos \theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1| |\vec{b}_2|}$

Calculate the Dot Product ( $\vec{b}_1 \cdot \vec{b}_2$ ):

$\vec{b}_1 \cdot \vec{b}_2 = (1)(3) + (-2)(-2) + (2)(6)$

$\vec{b}_1 \cdot \vec{b}_2 = 3 + 4 + 12 = 19$

Calculate the Magnitudes ( $|\vec{b}_1|$ and $|\vec{b}_2|$ ):

$|\vec{b}_1| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$
$|\vec{b}_2| = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$

Substitute into the Cosine Formula:

$\cos \theta = \frac{19}{3 \times 7}$

$\cos \theta = \frac{19}{21}$

$\theta = \cos^{-1}\left(\frac{19}{21}\right)$

2026: 29 May Shift 1

If 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 $\vec r=(a\hat i+4\hat j+5\hat k)+\mu(4\hat i+6\hat j+8\hat k)$ , $a \in N$ , where $\lambda$ and $\mu$ are parameters, is $\sqrt{\dfrac{5}{29}}$ , then $a$ is equal to

2026: 15 May Shift 1

If the lines $\dfrac{x+1}{3}=\dfrac{y+3}{5}=\dfrac{z+\lambda}{7}$ and $\dfrac{x-2}{1}=\dfrac{y-4}{3}=\dfrac{z-6}{5}$ are intersecting , then the value of ' $\lambda$ ' is

2023: 15 June Shift 2

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 :

2025: 3 June Shift 2

The angle at which the line, $\frac{x-1}{0} = \frac{2-y}{-1} = \frac{2z-3}{-2}$ is inclined with the positive direction of z-axis is

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:

2026: 18 May Shift 2

If the two lines given by

$x=py+q, z=ry+s$ and

$x=p'y+q', z=r'y+s'$

are perpendicular, then which of the following is equal to -1 ?

Solution

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