Let a unit vector u=x î+y ĵ+z k̂ make angles π/2, π/3 and (2 π)/3 with the vectors 1/√2 î+1/√2 k̂, 1/√2 ĵ+1/√2 k̂ and 1/√2 î+1/√2 ĵ respectively. If v⃗=1/√2 î+1/√2 ĵ+1/√2 k̂ then |u-v⃗|² is equal to

Solution

Maths sir · 4 steps

Turn each angle into a dot product; the three equations add up to x+y+z=0, and that settles it.

1. Angles to equations. All four vectors here are unit vectors, so u·a = cos θ. The angles π/2, π/3, 2π/3 give (x+z)/√2 = 0, (y+z)/√2 = 1/2 and (x+y)/√2 = -1/2.
2. Add them up. So x+z = 0, y+z = 1/√2 and x+y = -1/√2. Adding all three gives 2(x+y+z) = 0, so x+y+z = 0. This gives u = -1/√2î + 1/√2k̂, and |u| = 1 as required.

   |u-v⃗|² = |u|² + |v⃗|² - 2 u·v⃗

3. Put in numbers. |u|² = 1 and |v⃗|² = 1/2+1/2+1/2 = 3/2. Also u·v⃗ = (x+y+z)/√2 = 0.
4. So. |u-v⃗|² = 1 + 3/2 - 0 = 5/2, which is option (b).

The idea to keep. For |p⃗-q⃗|², expand it as |p⃗|²+|q⃗|²-2p⃗·q⃗. Often you only need the dot product, not every component.

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