The equation of the line passing through $(1, - 1, 2)$ and parallel to the line of intersection of the planes $2 x - y + z + 1 = 0 and x - y + z + 2 = 0$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{x - 1}{0} = \frac{y + 1}{- 1} = \frac{z - 2}{3}$

$\frac{x}{1} = \frac{y}{- 1} = \frac{z}{2}$

$\frac{x - 1}{0} = \frac{y - 1}{- 1} = \frac{z + 2}{3}$

$\frac{x - 1}{0} = \frac{y + 1}{1} = \frac{z - 2}{1}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The line of intersection of two planes is parallel to the vector which is cross product of two normal vectors to the plane and it is equal to parallel to the line of intersection of the planes $|\begin{matrix} i & j & k \\ 2 & - 1 & 1 \\ 1 & - 1 & 1 \end{matrix}| = i (0) - j (1) + k (- 1)$

The direction ratio are proportional to $⟨ 0, 1, 1 ⟩$

Therefore, the equation of the line passing through the point $(1, - 1, 2)$ and having direction ratios $⟨ 0, 1, 1 ⟩$ is $\frac{x - 1}{0} = \frac{y + 1}{1} = \frac{z - 2}{1}$