Let a line pass through two distinct points P(-2,-1,3) and Q, and be parallel to the vector 3 î+2 ĵ+2 k̂. If the distance of the point Q from the point R(1,3,3) is 5 , then the square of the area of △ P Q R is equal to :

Solution

Maths sir · 4 steps

Write Q with one parameter along the line, let QR = 5 fix it, then a cross product gives the area.

1. Write Q generally. Q lies on the line through P along (3, 2, 2), so Q = (-2+3λ, -1+2λ, 3+2λ) for some λ ≠ 0.
2. Use QR = 5. QR⃗ = (3-3λ, 4-2λ, -2λ), so QR² = 17λ² - 34λ + 25 = 25. This gives 17λ(λ - 2) = 0. Since Q is different from P, λ = 2 and Q = (4, 3, 7).
3. Cross product. PQ⃗ = (6, 4, 4) and PR⃗ = (3, 4, 0). Their cross product is (-16, 12, 12), whose squared length is 256 + 144 + 144 = 544.

   (Area)² = 1/4|PQ⃗ × PR⃗|² = 544/4 = 136

4. So. The square of the area of △ PQR is 136, option (c). As a check, PR = QR = 5 and PQ = √68, so the height to PQ is √(25 - 17) = √8 and the area is 1/2√68√8 = √136.

The idea to keep. A point on a line in 3D is just one parameter. Write it in terms of λ, use the given condition to find λ, and the area of a triangle is half the length of the cross product.

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