Let P1 be the plane 3 xy7 z11 and P2 be the plane passing th

Question

Let P_1 be the plane 3 x-y-7 z=11 and P_2 be the plane passing through the points (2,-1,0),(2,0,-1), and (5,1,1). If the foot of the perpendicular drawn from the point (7,4,-1) on the line of intersection of the planes P_1 and P_2 is (alpha, beta, gamma), then alpha+beta+gamma is equal to ___________.

DivineJEE · Accepted Answer

Equation of plane P_2 passing through (2,-1,0),(2,0, -1) and (5,1,1) is |\ccc x-5 y-1 z-1 \ 3 2 1 \ 3 1 2 \|=0 \ => (x-5)(4-1)-(y-1)(6-3) +(z-1)(3-6)=0 \ => 3 x-15-3 y+3-3 z+3=0 \ => 3 x-3 y-3 z-9=0 \ => x-y-z=3 .......(i) Now, direction ratios of line of intersection of P_1 and P_2 is |\ccc hat i hat j hat k \ 1 -1 -1 \ 3 -1 -7 \| \ =hat i(7-1)-hat j(-7+3)+hat k(-1+3) \ =6 hat i+4 hat j+2 hat k At z=0, x-y=3 [from (i)] 3 x-y=11 On solving, we get x=4 and y=1 So, equation of line is (x-4 / 6)=(y-1 / 4)=(z-2 / 6)=k Therefore (alpha, beta, gamma)=(6 k+4,4 k+1,2 k) \ => (6)(alpha-7)+4(beta-4)+2(gamma+1)=0 \ => 6(6 k+4-7)+4(4 k+1-4)+2(2 k+1)=0 \ => 36 k-18+16 k-12+4 k+4=0 \ => 56 k=26 => k=(1 / 2) \ So, alpha=7, beta=3 and gamma=1 \ Therefore alpha+beta+gamma=7+3+1=11

Explanation

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