Let a⃗=3 î+ĵ-k̂ and c⃗=2 î-3 ĵ+3 k̂. If b⃗ is a vector such that a⃗=b⃗ × c⃗ and |b⃗|²=50, then |72-| b⃗+c⃗|² | is equal to __________.

Solution

Maths sir · 5 steps

You cannot find b⃗ itself, and you do not need to: Lagrange's identity hands you b⃗·c⃗.

1. Magnitudes first. |a⃗|² = 9 + 1 + 1 = 11 and |c⃗|² = 4 + 9 + 9 = 22. Since a⃗ = b⃗ × c⃗, we have |b⃗ × c⃗|² = 11.
2. Lagrange's identity. For any two vectors, the squared cross product and the squared dot product add up to the product of the squared lengths.

   |b⃗ × c⃗|² + (b⃗·c⃗)² = |b⃗|² |c⃗|²

3. Find the dot product. 11 + (b⃗·c⃗)² = 50 × 22 = 1100, so (b⃗·c⃗)² = 1089 and b⃗·c⃗ = ± 33.
4. Expand the square. |b⃗+c⃗|² = |b⃗|² + |c⃗|² + 2 b⃗·c⃗ = 50 + 22 ± 66 = 72 ± 66.
5. So. 72 - |b⃗+c⃗|² = ∓ 66, and the modulus removes the sign ambiguity. The answer is 66.

The common slip: 33. Dropping the factor 2 in |b⃗+c⃗|² = |b⃗|² + |c⃗|² + 2 b⃗·c⃗ gives 33, not 66.

The idea to keep. Whenever a cross product and both lengths are known, use |u⃗×v⃗|² + (u⃗·v⃗)² = |u⃗|²|v⃗|² to get the dot product without finding the vectors.

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