Find the equations of the straight line perpendicular to the lines $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}; \frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}$

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Solution

Condition for Perpendicularity $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$
$⟹ - 3 a + 2 b + c = 0 \dots (1)$
$⟹ a - 3 b + 2 c = 0 \dots (2)$
$⟹ (a, b, c) = \pm (1, 1, 1)$
let the line pass through some point $(p, q, r)$
$⟹ x - p = y - q = z - r$

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Similar questions

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

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

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

(1, 0, 2)

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

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

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

(a) 2 x - y + z = 3

(b) 4 x - y + 3 z = 1

(c) 4 x + y + 3 z = 1

(d) x + 7 y + z = 3

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