First slide

Straight lines in 3D

Question

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

Moderate

A

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

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

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

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

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

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