First slide

Planes in 3D

Question

If the equation of plane containing lines $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 1}{3}$ and $\frac{x - 3}{- 4} = \frac{z - 2}{2} = \frac{z - 1}{- 6}$ is $9 x + b y + c z = d$ then $(b + c + d)$ is

Moderate

A

$- 1$
B

$7$
C

$- 7$
D

$1$

Solution

The equation of the plane containing two parallel lines $\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}}$ and $\frac{x - x_{2}}{a_{1}} = \frac{y - y_{2}}{b_{1}} = \frac{z - z_{2}}{c_{1}}$ is $|\begin{matrix} x - x_{1} & y - y_{1} & z - z_{1} \\ x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \end{matrix}| = 0$
The given two lines are parallel, because their direction ratios are proportional with $〈2, - 1, 3〉$
Hence, the equation of the plane containing the both lines is $|\begin{matrix} x - 1 & y + 1 & z - 1 \\ 3 - 1 & 2 + 1 & 1 - 1 \\ 2 & - 1 & 3 \end{matrix}| = 0$
Simplifying
$\begin{matrix} (x - 1) (9) - (y + 1) (6) + (z - 1) (- 2 - 6) = 0 \\ 9 x - 9 - 6 y - 6 - 8 z + 8 = 0 \\ 9 x - 6 y - 8 z - 7 = 0 \\ 9 x - 6 y - 8 z = 7 \end{matrix}$
Compare this equation with $a x + b y + c z = d$
Hence, $b + c + d = - 6 - 8 + 7 = - 7$

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