Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then the image of R in the plane P is :

Solution

Maths sir · 4 steps

Find the plane's normal with a cross product, then walk twice the perpendicular distance from R along it.

1. Normal to the plane. Take A(2,1,0). Then AB⃗ = (2,0,1) and AC⃗ = (3,-1,1). Their cross product is n⃗ = (0+1, 3-2, -2-0) = (1,1,-2).
2. Equation of P. Through A: (x-2)+(y-1)-2z=0, so x+y-2z=3. Check: (4,1,1) gives 4+1-2=3 and (5,0,1) gives 5+0-2=3.

   (x-2)/1=(y-1)/1=(z-6)/-2=(-2 (2+1-12-3))/(1²+1²+(-2)²)=-2(-12)/6=4

3. Read off the image. x = 2+4 = 6, y = 1+4 = 5, z = 6-8 = -2. Check: the midpoint of R and (6,5,-2) is (4,3,2), and 4+3-4=3, so it lies on P.
4. So. The image of R in the plane P is (6,5,-2), option (b).

The idea to keep. For the image of a point in ax+by+cz=d, move along the normal by -2(ax₁+by₁+cz₁-d)/(a²+b²+c²). The foot uses the same formula without the 2.

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