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The length of the perpendicular from P(1,0,2) on the line `(x+1)/(3)=(y-2)/(-2)=(z+1)/(-1)` isA. `(3sqrt6)/(2)`B. `(6sqrt3)/(5)`C. `3sqrt2`D. `2sqrt3` |
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Answer» Correct Answer - A `(x+1)/(3) = (y -2)/(-2) = (z+1)/(-1) = M` Now , any point on the given line is `M -= (3lambda -1,-2 lambda + 2, -lambda -1)` Direction ratio of `PM -=( 3 lambda -2, -2 ldmba +2, - lambda-3)` and ratio of the line (3,2,-1) . Since ,PM is perpendicular to the line, hence `3(3lambda -2)-2(2-2lambda) - 1(-lamda-3)=0" "[because a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2) =0]` `rArr " " 14lambda - 7 = 0 rArr=(1)/(2)` Hence , coordinate of `M = (3xx(1)/(2) -1,1-2xx (1)/(2) +2,-(1)/(2)-1) = ((1)/(2) , 1-(3)/(2))` Now , perpendicular length `Pm = sqrt((1-(1)/(2))^(2) +(0-1)^(2) +(2+(3)/(2))^(2))` ` = sqrt((1)/(4) +1+(49)/(4)) = sqrt((54)/(4)) =(3sqrt(6))/(2)` |
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