Let a⃗=î-ĵ+2 k̂ and let b⃗ be a vector such that a⃗ × b⃗=2 î-k̂ and a⃗ · b⃗=3. Then the projection of b⃗ on the vector a⃗-b⃗ is :

Solution

Maths sir · 4 steps

You never need b⃗ itself: dot and cross together give |b⃗|, and that is enough.

1. Length of b. Use the identity below with |a⃗|² = 1 + 1 + 4 = 6, a⃗ · b⃗ = 3 and |a⃗ × b⃗|² = 4 + 1 = 5.

   |a⃗|²|b⃗|² = (a⃗ · b⃗)² + |a⃗ × b⃗|² ⇒ 6|b⃗|² = 9 + 5 ⇒ |b⃗|² = 7/3

2. Numerator. The projection of b⃗ on a⃗ - b⃗ is (b⃗ · (a⃗ - b⃗))/(|a⃗ - b⃗|). The numerator is a⃗ · b⃗ - |b⃗|² = 3 - 7/3 = 2/3.
3. Denominator. |a⃗ - b⃗|² = |a⃗|² + |b⃗|² - 2 a⃗ · b⃗ = 6 + 7/3 - 6 = 7/3, so |a⃗ - b⃗| = √(7/3).
4. So. Projection = 2/3 × √(3/7) = 2/(√3 × √7) = 2/√21, which is option (a).

The idea to keep. When a question gives both a⃗ · b⃗ and a⃗ × b⃗, use |a⃗|²|b⃗|² = (a⃗ · b⃗)² + |a⃗ × b⃗|² to get |b⃗| without solving for b⃗.

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