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

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Solution

Let the equation of plane passing through ( $3, 2, 0$ ) is $a (x - 3) + b (y - 2) + c (z - 0) = 0$
$\begin{matrix} \frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4} p a s s t h r o u g h (3, 2, 0) a n d h a s d i r e c t i o n r a t i o s a s 1, 5, 4 i f t h e e q u a t i o n o f p l a n e c o n t a i n s t h i s, i t s h o u l d p a s s f r o m (3, 6, 4) a n d p a r a l l e l t o (1, 5, 4) H e n c e a (3 - 3) + b (6 - 2) + c (4) = 0 4 b + 4 c = 0 b + c = 0 a n d a + 5 b + 4 c = 0 o n s o l v i n g t h e s e t w o e q u a t i o n s a = c & b = - c H e n c e, e q u a t i o n o f p l a n e w i l l b e c (x - 3) - c (y - 2) + c z = 0 x - 3 - y + 2 + z = 0 x - y + z - 1 = 0 x - y + z = 1 \end{matrix}$

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