If a⃗ is a nonzero vector such that its projections on the vectors 2 î-ĵ+2 k̂, î+2 ĵ-2 k̂ and k̂ are equal, then a unit vector along a⃗ is :

Solution

Maths sir · 4 steps

Equal projections give two plane equations, and their cross product points straight along the vector.

1. Write the projections. Let a⃗ = xî + yĵ + zk̂. The projection of a⃗ on b⃗ is (a⃗·b⃗)/|b⃗|. The first two vectors both have length √(4+1+4) = 3, and k̂ has length 1.

   (2x - y + 2z)/3 = (x + 2y - 2z)/3 = z

2. Two conditions. Setting each fraction equal to z gives 2x - y - z = 0 and x + 2y - 5z = 0. So a⃗ is perpendicular to both (2, -1, -1) and (1, 2, -5).
3. Take the cross product. (2, -1, -1) × (1, 2, -5) = (5 + 2, -(-10 + 1), 4 + 1) = (7, 9, 5). Check: 14 - 9 - 5 = 0 and 7 + 18 - 25 = 0.
4. So. a⃗ is parallel to 7î + 9ĵ + 5k̂, whose length is √(49 + 81 + 25) = √155. The direction -(7î + 9ĵ + 5k̂) also works, but it is not among the options, so the answer is (d).

The idea to keep. Divide each dot product by the length of its vector before equating. Two linear conditions on a⃗ make it perpendicular to two vectors, so their cross product gives its direction.

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