Let the foot of perpendicular of the point P(3,-2,-9) on the plane passing through the points (-1,-2,-3),(9,3,4),(9,-2,1) be Q(α, β, γ). Then the distance of Q from the origin is :

Solution

Maths sir · 4 steps

Get the plane's normal from a cross product, then slide from P along that normal until you land on the plane.

1. Find the normal. Take A(-1,-2,-3). Then AB⃗ = (10,5,7) and AC⃗ = (10,0,4). Their cross product is (20, 30, -50), so a normal is n⃗ = (2,3,-5).
2. Equation of plane. Through A: 2(x+1)+3(y+2)-5(z+3)=0, that is 2x+3y-5z-7=0. Check with (9,3,4): 18+9-20-7=0.

   (x-3)/2=(y+2)/3=(z+9)/-5 = -(2(3)+3(-2)-5(-9)-7)/(2²+3²+(-5)²) = -38/38 = -1

3. Locate Q. So Q = (3-2, -2-3, -9+5) = (1,-5,-4). It lies on the plane: 2-15+20-7=0.
4. So. OQ = √(1²+(-5)²+(-4)²) = √(1+25+16) = √42, which is option (c).

The idea to keep. Foot of perpendicular from (x₁,y₁,z₁) on ax+by+cz+d=0: each coordinate moves by a t, b t, c t with t = -(ax₁+by₁+cz₁+d)/(a²+b²+c²).

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