Find the vector equation of the through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to plane $x - y + z = 0$ . Hence find whether the plane thus contains the line $\frac{x - 2}{5} = \frac{y - 3}{4} = \frac{z}{5}$ or not.

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Solution

The equation of through line of intersection of planes
$x + y + z = 0, 2 x + 3 y + 4 z - 5 = 0$
$\begin{matrix} i s x + y + z + λ (2 x + 3 y + 4 z - 5) = 0 - - - - (1) (1 + 2 λ) x + (1 + 3 λ) y + (1 + 4 λ) z - 5 λ = 0 A b o v e p l a n e i s^{'} ⊥^{'} t o x - y + z = 0 (1 + 2 λ) (1) + (1 + 3 λ) (- 1) + (1 + 4 λ) (1) = 0 1 + 2 λ - 1 - 3 λ + 1 + 4 λ = 0 3 λ + 1 = 0 λ = - \frac{1}{3} i n - - - - (1) x + y + z - \frac{1}{3} (2 x + 3 y + 4 z - 5) = 0 3 x + 3 y + 3 z - 2 x - 3 y - 4 z + 5 = 0 x - 2 + 5 = 0 i s r e g e q n o f p l a n e G i v e n l i n e \frac{x - 5}{5} = \frac{y - 3}{4} = \frac{z}{5} p a s s e s t h r o u g h (2, 3, 0) (2, 3, 0) d o e s n^{'} t s a t i s f l y x - z + 5 = 0 l i n e d o e s n^{'} t l i n e o n t h e p l a n e, b u t p a r a l l e l t o t h e p l a n e (1) (5) + 0 (4) + (- 1) 5 = 0 D i s n o r m a l t o p l a n e i n ⊥ n o l i n e \end{matrix}$

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Similar questions

Find the equation of the plane through the line of intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x-y+z=0.

x + y + z = 1

2 x + 3 y + 4 z = 5

x - y + z = 0

x + y + z = 1

2 x + 3 y + 4 z = 5

x - y + z = 0

2 x + y - z = 3, 5 x - 3 y + 4 z + 9 = 0

\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z - 5}{5} .

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