Find the equation of the plane through the point $(1, 2, 4)$ and parallel to the line $3 x + 2 y - z - 4 = 0, x - 2 y - 2 z - 5 = 0$

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Solution

The required plane is of the form
$\begin{matrix} 3 x + 2 y - z - 4 + λ (x - 2 y - 2 z - 5) = 0 - - - - (1) T h e a b o v e p l a n e p a s s e s t h r o u g h (1, 2, 4) . 3 + 4 - 4 - 4 + λ (1 - 4 - 8 - 5) = 0 λ = \frac{- 1}{16} i n - - - - (1) 3 x + 2 y - z - 4 \frac{- 1}{16} (x - 2 y - 2 z - 5) = 0 47 x + 34 y - 14 z - 59 = 0 \end{matrix}$

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Find the equation of the plane through the point (1,2,4) and parallel to the line 3x+2y−z−4=0,x−2y−2z−5=0

The required plane is of the form3x+2y−z−4+λ(x−2y−2z−5)=0−−−−(1)Theaboveplanepassesthrough(1,2,4).3+4−4−4+λ(1−4−8−5)=0λ=−116in−−−−(1)3x+2y−z−4−116(x−2y−2z−5)=047x+34y−14z−59=0

Find the equation of the plane through the point $(1, 2, 4)$ and parallel to the line $3 x + 2 y - z - 4 = 0, x - 2 y - 2 z - 5 = 0$