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Find the value of \( k \) such that the line \( \frac{x-2}{3}=\frac{y+1}{2}=\frac{z-k}{1} \) - lies in the plane \( x - y - z +8=0 \) |
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Answer» Given line is \(\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-k}{1}\) = λ (Let) ----- (1) ∴ coefficient of arbitrary point on line (1) are (3λ + 2, 2λ - 1, λ + k). Given that line (1) lies on the plane x - y - z + 8 = 0. ∴ (3λ + 2) - (2λ - 1) - (λ + k) + 8 = 0 \(\Rightarrow\) 3λ - 2λ - λ + 2 + 1 - k + 8 = 0 \(\Rightarrow\) k = 2 + 1 + 8 = 11. Hence, for k = 1 given that line lies on plane x - y - z + 8 = 0. |
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