Let a⃗ = î + 2ĵ + k̂ and b⃗ = 2î + ĵ - k̂ . Let c be a unit vector in the plane of the vectors a⃗ and b⃗ and be perpendicular to a⃗ . Then such a vector c is:

Solution

Maths sir · 3 steps

Being perpendicular to a is not enough; the vector must also lie in the plane, so check both conditions.

1. Two conditions. The vector c must satisfy c · a⃗ = 0. It must also lie in the plane of a⃗ and b⃗, which means it is perpendicular to the plane's normal a⃗ × b⃗.

   a⃗ × b⃗ = |î, ĵ, k̂; 1, 2, 1; 2, 1, -1| = -3î + 3ĵ - 3k̂

   Option (direction)	Dot with a⃗ = (1, 2, 1)	Dot with normal (1, -1, 1)	Verdict
   (a) (-1, 0, 1)	-1 + 0 + 1 = 0	-1 + 0 + 1 = 0	perpendicular to a⃗ and in the plane
   (b) (0, 1, -2)	0 + 2 - 2 = 0	0 - 1 - 2 = -3	not in the plane
   (c) (1, -1, 1)	1 - 2 + 1 = 0	1 + 1 + 1 = 3	this is the normal itself
   (d) (-1, 1, -1)	-1 + 2 - 1 = 0	-1 - 1 - 1 = -3	this is the normal itself

2. Direct check. Remove the part of b⃗ along a⃗. Here a⃗ · b⃗ = 3 and |a⃗|² = 6, so b⃗ - 1/2a⃗ = 3/2î - 3/2k̂. This points along ±(î - k̂), and its unit vector is ±1/√2(î - k̂).
3. So. The required unit vector is c = 1/√2(-î + k̂), which is option (a).

The idea to keep. In the plane of a⃗ and b⃗ means perpendicular to a⃗ × b⃗. To get the in-plane vector perpendicular to a⃗, use b⃗ - (a⃗ · b⃗)/|a⃗|²a⃗.

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