JEE Main 2023 · 30 January · morning · Single correct · +4 / −1
Let a unit vector O P make angles α, β, γ with the positive directions of the co-ordinate axes OX, OY, OZ respectively, where β ∈(0, π/2). If OP is perpendicular to the plane through points (1,2,3),(2,3,4) and (1,5,7), then which one of the following is true?
- (a) α ∈(π/2, π) and γ ∈(π/2, π)
- (b) α ∈(0, π/2) and γ ∈(π/2, π)
- (c) α ∈(π/2, π) and γ ∈(0, π/2)
- (d) α ∈(0, π/2) and γ ∈(0, π/2)
Answer and solution
Answer (a) α ∈(π/2, π) and γ ∈(π/2, π)
Solution
Maths sir · 4 steps
Find the plane's normal with a cross product, then flip its sign so that cos β comes out positive.
- Two vectors in plane. Take A(1,2,3), B(2,3,4), C(1,5,7). Then AB⃗ = (1, 1, 1) and AC⃗ = (0, 3, 4) both lie in the plane.
AB⃗ × AC⃗ = |î, ĵ, k̂; 1, 1, 1; 0, 3, 4| = î - 4ĵ + 3k̂
- Two possible directions. The unit normal is ±1/√26(1, -4, 3). Its components are exactly cos α, cos β, cos γ.
- Use the condition. Since β ∈ (0, π/2), cos β > 0. That forces the minus sign: OP = 1/√26(-1, 4, -3).
- So. cos α = -1/√26 < 0 and cos γ = -3/√26 < 0, so both α and γ lie in (π/2, π). The answer is (a).
The idea to keep. A normal to a plane comes with two signs. The sign of each direction cosine decides whether its angle is acute or obtuse, so use the given condition to fix the sign first.
Answer from the NTA final answer key. Solution by PrepWiser.