If the lines $ \frac{x-1}{-3}=\frac{y-2}{-2 k}=\frac{z-3}{2} $ and $ \frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5} $ are perpendicular, what would be the value of $ k $ and the equation of plane containing these lines?

Question

If the lines $ \frac{x-1}{-3}=\frac{y-2}{-2 k}=\frac{z-3}{2} $ and $ \frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5} $ are perpendicular, what would be the value of $ k $ and the equation of plane containing these lines?, $ k=4 $ and $ -22 x+19 y=-96 $, $ k=2 $ and $ -56 y+5 z=-23 $, $ k=2 $ and $ -22 x+19 y+5 z=31 $, None of these

MARKS App · Accepted Answer

The given lines are x − 1 − 3 = y − 2 − 2 k = z − 3 2 . . . . ( 1 ) x − 1 k = y − 2 1 = z − 3 5 . . . . . . . ( 2 ) For ( 1 ) and ( 2 ) to be ⊥ , we must have − 3 k − 2 k + 10 = 0 ⇒ − 5 k + 10 ⇒ k = 2. The equation of lines becomes x − 1 − 3 = y − 2 − 4 = z − 3 2 , x − 1 2 = y − 2 1 = z − 3 5 Now, the equation of plane containing these two lines is x - x 1 y - y 1 z - z 1 l 1 m 1 n 1 l 2 m 2 n 2 = 0 where x 1 = 1 , y 1 = 2 , z 1 = 3 , l 1 = − 3 , m 1 = − 4 , n 1 = 2 l 2 = 2 , m 2 = 1 , n 2 = 5 ∵ x - 1 y - 2 z - 3 - 3 - 4 2 2 1 5 = 0 ⇒ x − 1 − 20 − 2 − y − 2 − 15 − 4 + z − 3 − 3 + 8 = 0 ⇒ − 22 x + 22 + 19 y − 38 + 5 z − 15 = 0 ⇒ 22 x − 19 y − 5 z + 31 = 0 .

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