Q78 / 150 · MHT CET 2026 (17 April Shift 1) · Mathematics
If $\theta$ is the angle between the lines $\dfrac{x-1}{2} = \dfrac{2y+3}{4};\ z = -2$ and $x = 1;\ \dfrac{y-1}{2} = \dfrac{z+1}{2}$, then
- A$\theta = \dfrac{\pi}{6}$
- Correct answerB$\theta = \dfrac{\pi}{3}$
- C$\theta = \dfrac{\pi}{4}$
- D$\theta = \dfrac{\pi}{2}$
Solution
The equation of the first line can be rewritten as $\dfrac{x-1}{2} = \dfrac{y+\dfrac{3}{2}}{2} = \dfrac{z+2}{0}$
The direction ratios of the first line are $2, 2, 0$.
The equation of the second line can be rewritten as $\dfrac{x-1}{0} = \dfrac{y-1}{2} = \dfrac{z+1}{2}$
The direction ratios of the second line are $0, 2, 2$.
Let $\theta$ be the angle between the two lines.
$\cos\theta = \dfrac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$\cos\theta = \dfrac{2(0) + 2(2) + 0(2)}{\sqrt{2^2+2^2+0^2}\sqrt{0^2+2^2+2^2}}$
$\cos\theta = \dfrac{4}{\sqrt{8}\sqrt{8}} = \dfrac{4}{8} = \dfrac{1}{2}$
$\theta = \dfrac{\pi}{3}$
Answer: $\theta = \dfrac{\pi}{3}$
The direction ratios of the first line are $2, 2, 0$.
The equation of the second line can be rewritten as $\dfrac{x-1}{0} = \dfrac{y-1}{2} = \dfrac{z+1}{2}$
The direction ratios of the second line are $0, 2, 2$.
Let $\theta$ be the angle between the two lines.
$\cos\theta = \dfrac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$\cos\theta = \dfrac{2(0) + 2(2) + 0(2)}{\sqrt{2^2+2^2+0^2}\sqrt{0^2+2^2+2^2}}$
$\cos\theta = \dfrac{4}{\sqrt{8}\sqrt{8}} = \dfrac{4}{8} = \dfrac{1}{2}$
$\theta = \dfrac{\pi}{3}$
Answer: $\theta = \dfrac{\pi}{3}$
Hence, the correct answer is option: 'B'.