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If the lines `(x-1)/(-3)=(y-2)/(2k)=(z-3)/2`and `(x-1)/(3k)=(y-1)/1=(z-6)/(-5)`are perpendicular, find the value of k. |
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Answer» We have, `(x-1)/(-3)=(y-2)/(2k)=(z-3)/(2)" "...(1)` and ` (x-1)/(3k)=(y-5)/(1)=(x-6)/(-5)" "...(2)` Now, directin ratios of line (1) are -3, 2k, and line (2) are k,1, -5. If lines are perpendicular, than `(-3)(3k)+(2k)(1)+(2)(-5)=0` `-9k+2k-10=0` `-7k=10` `k=-10/7` |
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