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

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Solution

$\begin{matrix} \frac{x - 2}{a} = \frac{y + 1}{b} = \frac{z + 1}{c} 4 a + b + c = 0 a - 2 b + c = 0 \frac{a}{3} = \frac{- b}{3} = \frac{c}{- 9} \frac{a}{1} = \frac{b}{- 1} = \frac{c}{- 3} E q u a t i o n o f l i n e : \frac{x - 2}{1} = \frac{y + 2}{- 1} = \frac{z + 1}{- 3} \end{matrix}$

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Similar questions

(2, - 1, - 1)

4 x + y + z + 2 = 0

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

(1, 1, 1)

x - 2 y + z = 2, 4 x + 3 y - z + 1 = 0

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

x - y + z + 2 = 0

(- 1, 3, - 2)

\frac{x}{1} = \frac{y}{2} = \frac{z}{3}

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

2 x - y + z = 3

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

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

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