Let a⃗ = 2î - ĵ + 2k̂ and b⃗ = î + 2ĵ - k̂. Let a vector v⃗ be in the plane containing a⃗ and b⃗. If v⃗ is perpendicular to the vector 3î + 2ĵ - k̂ and its projection on a⃗ is 19 units, then | 2v⃗ |² is equal to _____________.

Solution

Maths sir · 4 steps

Write v⃗ = xa⃗ + yb⃗, then let the two conditions find x and y.

1. Write v. Any vector in the plane of a⃗ and b⃗ is v⃗ = xa⃗ + yb⃗ = (2x+y)î + (-x+2y)ĵ + (2x-y)k̂.
2. Perpendicular condition. Dot v⃗ with 3î + 2ĵ - k̂ and set it to zero: 3(2x+y) + 2(-x+2y) - (2x-y) = 2x + 8y = 0, so x = -4y.
3. Projection condition. Here |a⃗| = 3, so the projection condition is v⃗ · a⃗ = 19 × 3 = 57. With |a⃗|² = 9 and a⃗ · b⃗ = 2 - 2 - 2 = -2, this gives 9x - 2y = 57. Putting x = -4y gives -38y = 57, so y = -3/2 and x = 6.

   v⃗ = 6a⃗ - 3/2b⃗ = 21/2î - 9ĵ + 27/2k̂

4. So. |v⃗|² = (441 + 324 + 729)/4 = 1494/4, and |2v⃗|² = 4|v⃗|² = 1494. The answer is 1494.

The common slip: 373.5. Stopping at |v⃗|² and forgetting that doubling v⃗ multiplies the squared length by 4 gives 373.5, not 1494.

The idea to keep. Any vector in the plane of a⃗ and b⃗ is xa⃗ + yb⃗. Each condition gives one equation, and projection on a⃗ means v⃗ · a⃗ / |a⃗|.

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