If foot of the perpendicular of P(2,-3,1) on the line x+12=y−33=z+2−1 is Q(a,b,c) then find the value of 14(a+b+c)
Given line is x+12=y−33=z+2−1=r....(1)PointP≡(2,−3,1)Co−ordinates of foot of the perpendicular on line (1) may be taken as Q≡(2r−1,3r+3,−r−2)We get direction ratios of PQ=2r−3,3r+6,−r−3Direction ratios of line segment are 2,3,−1(from(1)Since PQperpendiculartoAB∴2(2r−3)+3(3r+6)−1(−r−3)=0or,14r+15=0∴r=−1514∴Q≡(2r−1,3r+3,−r−2)=(−227,−314,−1314)=(a,b,c)(given)⇒−14(a+b+c)=−14(−227−1314,−1314)=44+3+13=60