The vector a⃗ = - î + 2ĵ + k̂ is rotated through a right angle, passing through the y-axis in its way and the resulting vector is b⃗. Then the projection of 3a⃗ + √2 b⃗ on c⃗ = 5î + 4ĵ + 3k̂ is :

Solution

Maths sir · 5 steps

Passing the y-axis puts b⃗ in the plane of a⃗ and ĵ, at right angles to a⃗.

1. Where b lives. The rotation sweeps through the y-axis, so b⃗ lies in the plane of a⃗ and ĵ. It is a right-angle rotation, so b⃗ ⊥ a⃗. A rotation does not change length, so |b⃗| = |a⃗| = √6.
2. Direction of b. Remove from ĵ its part along a⃗. Since a⃗ · ĵ = 2 and |a⃗|² = 6, this leaves ĵ - 2/6a⃗ = 1/3(î + ĵ - k̂). It is perpendicular to a⃗ and points towards the y-axis, which is the way a⃗ turns.
3. Fix the length. |î + ĵ - k̂| = √3, so b⃗ = √6/√3(î + ĵ - k̂) = √2(î + ĵ - k̂). Check: a⃗ · b⃗ = √2(-1 + 2 - 1) = 0.
4. Combine. 3a⃗ + √2 b⃗ = (-3î + 6ĵ + 3k̂) + (2î + 2ĵ - 2k̂) = -î + 8ĵ + k̂. Also |c⃗| = √(25 + 16 + 9) = 5√2.

   Projection = ((-î + 8ĵ + k̂)·(5î + 4ĵ + 3k̂))/5√2 = (-5 + 32 + 3)/5√2 = 30/5√2 = 3√2

5. So. The projection is 3√2, which is option (d).

The idea to keep. To rotate a⃗ by 90° towards a vector p⃗, take p⃗ - (a⃗·p⃗)/|a⃗|²a⃗ and scale it to length |a⃗|.

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