Let a⃗=4 î-ĵ+k̂, b⃗=11 î-ĵ+k̂ and c⃗ be a vector such that (a⃗+b⃗) × c⃗=c⃗ ×(-2 a⃗+3 b⃗). If (2 a⃗+3 b⃗) · c⃗=1670, then |c⃗|² is equal to:

Solution

Maths sir · 4 steps

Bring everything to one side of the cross product, and c turns out parallel to a known vector.

1. Flip the right side. Swapping the order of a cross product changes its sign, so c⃗ × (-2a⃗+3b⃗) = (2a⃗-3b⃗) × c⃗. The given condition becomes (a⃗+b⃗) × c⃗ - (2a⃗-3b⃗) × c⃗ = 0⃗.

   (4b⃗-a⃗) × c⃗ = 0⃗ ⇒ c⃗ = λ (4b⃗-a⃗)

2. Find the direction. 4b⃗-a⃗ = (44î-4ĵ+4k̂) - (4î-ĵ+k̂) = 40î-3ĵ+3k̂. So c⃗ = λ(40î-3ĵ+3k̂).
3. Use the dot product. 2a⃗+3b⃗ = 41î-5ĵ+5k̂. Then (2a⃗+3b⃗) · c⃗ = λ(1640+15+15) = 1670λ. This equals 1670, so λ = 1.
4. So. c⃗ = 40î-3ĵ+3k̂ and |c⃗|² = 1600+9+9 = 1618, option (b).

The idea to keep. When u⃗ × c⃗ = 0⃗ with u⃗ ≠ 0⃗, c⃗ must be λu⃗. Always flip c⃗ × v⃗ to -v⃗ × c⃗ before collecting terms.

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