Let a⃗=3 î+2 ĵ+k̂, b⃗=2 î-ĵ+3 k̂ and c⃗ be a vector such that (a⃗+b⃗) × c⃗=2(a⃗ × b⃗)+24 ĵ-6 k̂ and (a⃗-b⃗+î) · c⃗=-3. Then |c⃗|² is equal to ________.

Solution

Maths sir · 5 steps

Write c⃗ = (x, y, z) and turn both conditions into plain linear equations in x, y, z.

1. Known vectors. a⃗ + b⃗ = (5, 1, 4) and a⃗ - b⃗ + î = (2, 3, -2). Also a⃗ × b⃗ = (7, -7, -7), so the right side is (14, -14, -14) + (0, 24, -6) = (14, 10, -20).
2. Expand the cross product. With c⃗ = (x, y, z), (5, 1, 4) × (x, y, z) = (z - 4y, 4x - 5z, 5y - x). Match each component with (14, 10, -20).

   z - 4y = 14, 4x - 5z = 10, 5y - x = -20

3. Use the dot product. These three are not independent: a cross product with a⃗ + b⃗ only fixes c⃗ up to a multiple of a⃗ + b⃗. The dot condition gives the missing equation 2x + 3y - 2z = -3.
4. Solve. Put x = 5y + 20 and z = 4y + 14 into it: 2(5y + 20) + 3y - 2(4y + 14) = 5y + 12 = -3, so y = -3. Then x = 5 and z = 2, and 4x - 5z = 10 checks out.
5. So. c⃗ = 5î - 3ĵ + 2k̂, and |c⃗|² = 25 + 9 + 4 = 38.

The idea to keep. A cross product equation u⃗ × c⃗ = v⃗ gives only two independent equations, because adding any multiple of u⃗ to c⃗ changes nothing. The extra dot condition pins down c⃗.

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