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Find the value of k so that the lines `(1-x)/(3)=(7y-14)/(2k) =(z-3)/(2) " and " (7-7x)/(3k) =(5-y)/(1)=(6-z)/(5)` are at right angles. |
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Answer» The given equation in standard form are `(x-1)/(-3) =(y-2)/((2k)/(7)) =(z-3)/(2) rArr (x-1)/(-21) =(y-2)/(2k) =(z-3)/(14)` `(x-1)/((-3k)/(7)) =(y-5)/(-1) =(z-6)/(-5) rArr (x-1)/(-3k) =(y-5)/(-7) =(z-6)/(-35)` Here `a_(1)=-21 , b_(1) =2k ,c_(1) =14 " and " a_(2) =-3k , b_(2) =-7 ,c_(2) =-35` Since the given lines are at right angles we have , `a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0` `rArr (-21) xx (-3k) +(2k) xx(-7)+14 xx(-35) =0` `rArr 63k -14k-490 =0 rArr 49k =490 rArr k=10` Hence k=10 |
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