First slide

Planes in 3D

Question

$The plane which passes through the point (3, 2, 0) and the line \frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4} is$

Moderate

A

$x - y + z = 1$
B

$x + y + z = 5$
C

$x + 2 y - z = 1$
D

$2 x - y + z = 5$

Solution

$The required plane is |\begin{array}{ccc} x - 3 & y - 6 & z - 4 \\ 3 - 3 & 2 - 6 & 0 - 4 \\ 1 & 5 & 4 \end{array}| = 0$
$or |\begin{array}{ccc} x - 3 & y - z - 2 & z - 4 \\ 0 & 0 & - 4 \\ 1 & 1 & 4 \end{array}| = 0$
$(Operating C_{2} \to C_{2} - C_{3})$
$\begin{array}{l} or 4 (x - 3 - y + z + 2) = 0 \\ or x - y + z = 1 \end{array}$

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