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Question: If lines \(\frac{x}{k}\) = 2 – y =\(\frac{z - 1}{2}\) and \(\frac{x + 1}{3} = \frac{y - 1}{k}\)= z +...

If lines xk\frac{x}{k} = 2 – y =z−12\frac{z - 1}{2} and x+13=y−1k\frac{x + 1}{3} = \frac{y - 1}{k}= z + 2 are perpendicular to each other, then k is equal to

A

1

B

2

C

–1

D

–2

Answer

–1

Explanation

Solution

x−0k=y−2−1=z−12\frac{x - 0}{k} = \frac{y - 2}{- 1} = \frac{z - 1}{2} and x+13=y−1k=z+21\frac{x + 1}{3} = \frac{y - 1}{k} = \frac{z + 2}{1}

\ k(3) + k(–1) + 2(1) = 0

2k = – 2 ̃ k = –1