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If the lines (x-1)/(-3)=(y-2)/(2k)=(z-3...

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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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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GURUKUL PUBLICATION - MAHARASHTRA PREVIOUS YEAR PAPERS- MARCH 2016-SECTION II
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