First slide

Straight lines in 3D

Question

The line of intersection of the planes $x - y + 6 z = 0$ and $2 x + 3 y - z = 5$ is

Very Easy

A

$\frac{x}{2} = \frac{y - 1}{3} = \frac{z + 2}{6}$
B

$\frac{x - 1}{17} = \frac{y + 1}{13} = \frac{z}{3}$
C

$\frac{x - 1}{17} = \frac{y - 1}{12} = \frac{z}{4}$
D

$\frac{x - 1}{- 17} = \frac{y - 1}{13} = \frac{z}{5}$

Solution

To get a point on the line of intersection of given two planes, substitute $z = 0$ in the equations
Hence, $x - y = 0 and 2 x + 3 y = 5$
it gives, a point on the line of intersection of two planes as $P (1, 1, 0)$
The direction ratios o the line is proportional to the vector $|\begin{matrix} i & j & k \\ 1 & - 1 & 6 \\ 2 & 3 & - 1 \end{matrix}| = i (- 17) - j (- 13) + k (5)$
Therefore, the equation of the line of intersection of two planes is $\frac{x - 1}{- 17} = \frac{y - 1}{13} = \frac{z}{5}$

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