The plane, passing through the points (0,-1,2) and (-1,2,1) and parallel to the line passing through (5,1,-7) and (1,-1,-1), also passes through the point :

Solution

Maths sir · 3 steps

The normal is perpendicular to both directions in the plane, so get it with one cross product.

1. Directions in the plane. With A(0,-1,2) and B(-1,2,1), the plane contains AB⃗ = (-1,3,-1). It is parallel to the line through (5,1,-7) and (1,-1,-1), whose direction (-4,-2,6) simplifies to (2,1,-3).

   n⃗ = (-1,3,-1)×(2,1,-3) = (-8,-5,-7)

2. Equation of the plane. Take the normal as (8,5,7). Putting in A gives 0 - 5 + 14 = 9, so the plane is 8x + 5y + 7z = 9. As a check, B gives -8 + 10 + 7 = 9.

   Option	8x + 5y + 7z	Equals 9?
   (a) (0,5,-2)	0 + 25 - 14 = 11	No
   (b) (2,0,1)	16 + 0 + 7 = 23	No
   (c) (1,-2,1)	8 - 10 + 7 = 5	No
   (d) (-2,5,0)	-16 + 25 + 0 = 9	Yes

3. So. Only (-2,5,0) satisfies 8x + 5y + 7z = 9, so the answer is (d).

The idea to keep. A plane containing one direction and parallel to another has its normal along their cross product. Check the normal by dotting it with both directions: each result must be 0.

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