A vector a⃗ = α î + 2ĵ + β k̂( α ,β ∈ R ) lies in the plane of the vectors, b⃗ = î + ĵ and c⃗ = î - ĵ + 4k̂. If a⃗ bisects the angle between b⃗ and c⃗, then:

Solution

Maths sir · 4 steps

Make b and c the same length first. Then their sum points straight along the angle bisector.

1. Equal lengths. |b⃗| = √2 and |c⃗| = √(1 + 1 + 16) = 3√2. So |c⃗| = 3|b⃗|, and 3b⃗ has the same length as c⃗.
2. Bisector direction. Two vectors of equal length form a rhombus, and the diagonal of a rhombus bisects the angle between its sides. So the internal bisector lies along 3b⃗ + c⃗, which also lies in the plane of b⃗ and c⃗.

   3b⃗ + c⃗ = (3î + 3ĵ) + (î - ĵ + 4k̂) = 4î + 2ĵ + 4k̂

3. Match the j-part. Take a⃗ = λ(4î + 2ĵ + 4k̂). The ĵ component must be 2, so λ = 1. That gives α = 4 and β = 4.
4. So. a⃗ · k̂ = β = 4, so a⃗ · k̂ - 4 = 0. The answer is (b).

The idea to keep. The bisector of the angle between two vectors is along the sum of their unit vectors. Scale them to equal length first, and you can skip the square roots.

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