Let the line L: (x-1)/2=(y+1)/-1=(z-3)/1 intersect the plane 2 x+y+3 z=16 at the point P. Let the point Q be the foot of perpendicular from the point R(1,-1,-3) on the line L. If α is the area of triangle P Q R, then α² is equal to __________.

Solution

Maths sir · 4 steps

Q is the foot of the perpendicular, so the triangle is right-angled at Q and the area is half of QP times QR.

1. Find P. A general point on L is (1+2t, -1-t, 3+t). Putting it in the plane gives 2(1+2t)+(-1-t)+3(3+t)=16, that is 10+6t=16, so t=1 and P=(3,-2,4).
2. Find Q. From R(1,-1,-3) to the general point, the vector is (2t, -t, 6+t). It must be perpendicular to the direction (2,-1,1): 4t+t+6+t=0, so t=-1 and Q=(-1,0,2).
3. Two legs. QR⃗=(2,-1,-5), so QR=√(4+1+25)=√30. QP⃗=(4,-2,2), so QP=√(16+4+4)=√24. The angle at Q is 90° because QP lies along L and QR is perpendicular to L.

   α=1/2 QP· QR=1/2√(24× 30)=1/2√720

4. So. α²=720/4= 180.

The common slip: 720. Forgetting the 1/2 in the area gives α² = 24 × 30 = 720, not 180.

The idea to keep. Whenever a triangle uses the foot of a perpendicular on a line and another point on that same line, it is right-angled at the foot. Half the product of the two legs gives the area.

Answer from the NTA final answer key. Solution by PrepWiser.