Three particles P, Q and R are moving along the vectors A⃗=î+ĵ, B⃗=ĵ+k̂ and C⃗=-î+ĵ respectively. They strike on a point and start to move in different directions. Now particle P is moving normal to the plane which contains vector A⃗ and B⃗. Similarly particle Q is moving normal to the plane which contains vector A⃗ and C⃗. The angle between the direction of motion of P and Q is cos⁻¹(1/√x). Then the value of x is _______________.

Solution

Physics sir · 3 steps

Normal to a plane means cross product: find both normals, then take the angle between them.

1. Direction of P. A vector normal to the plane of A⃗ and B⃗ is A⃗ × B⃗. With A⃗ = (1, 1, 0) and B⃗ = (0, 1, 1), this gives n⃗₁ = î - ĵ + k̂, so |n⃗₁| = √3.
2. Direction of Q. Similarly, Q moves along A⃗ × C⃗. With C⃗ = (-1, 1, 0), this gives n⃗₂ = (1 × 1 - 1 × (-1))k̂ = 2k̂, so |n⃗₂| = 2.

   cos θ = (n⃗₁ · n⃗₂)/(|n⃗₁| |n⃗₂|) = ((1)(0) + (-1)(0) + (1)(2))/(√3 × 2) = 1/√3

3. So. The angle is cos⁻¹(1/√3). Matching this with cos⁻¹(1/√x) gives x = 3.

The idea to keep. Whenever a direction is normal to the plane of two vectors, take their cross product. To find the angle between two such normals, use the dot product formula.

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