Q89 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
The measure of the angle between the lines $x=k+1, \quad y=2 k-1, \quad z=2 k+3, \quad k \in R \quad$ and $\quad \dfrac{x-1}{2}=\dfrac{y+1}{1}=\dfrac{z-1}{-2}$ is
- A$\sin ^{-1}\left(\dfrac{4}{3}\right)$
- B$\cos ^{-1}\left(\dfrac{4}{9}\right)$
- C$\sin ^{-1}\left(\dfrac{\sqrt{5}}{3}\right)$
- Correct answerD$\dfrac{\pi}{2}$
Solution
The first line is given by $x = k+1$, $y = 2k-1$, $z = 2k+3$. This can be written in symmetric form as $\dfrac{x-1}{1} = \dfrac{y+1}{2} = \dfrac{z-3}{2} = k$. The direction vector of this line is $\vec{v_1} = \hat{i} + 2\hat{j} + 2\hat{k}$.
The second line is given by $\dfrac{x-1}{2} = \dfrac{y+1}{1} = \dfrac{z-1}{-2}$. The direction vector of this line is $\vec{v_2} = 2\hat{i} + \hat{j} - 2\hat{k}$.
The angle $\theta$ between the two lines is given by $\cos \theta = \dfrac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.
Calculating the dot product: $\vec{v_1} \cdot \vec{v_2} = (1)(2) + (2)(1) + (2)(-2) = 2 + 2 - 4 = 0$.
Since the dot product is $0$, the lines are perpendicular to each other. Therefore, the angle between the lines is $\dfrac{\pi}{2}$.
Answer: $\dfrac{\pi}{2}$
The second line is given by $\dfrac{x-1}{2} = \dfrac{y+1}{1} = \dfrac{z-1}{-2}$. The direction vector of this line is $\vec{v_2} = 2\hat{i} + \hat{j} - 2\hat{k}$.
The angle $\theta$ between the two lines is given by $\cos \theta = \dfrac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.
Calculating the dot product: $\vec{v_1} \cdot \vec{v_2} = (1)(2) + (2)(1) + (2)(-2) = 2 + 2 - 4 = 0$.
Since the dot product is $0$, the lines are perpendicular to each other. Therefore, the angle between the lines is $\dfrac{\pi}{2}$.
Answer: $\dfrac{\pi}{2}$
Hence, the correct answer is option: 'D'.