1995

JEE Main 1995 Mathematics question on Vector Algebra

If a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are non-coplanar unit vectors such that a⃗×(b⃗×c⃗)=(b⃗+c⃗)2\vec{a} \times (\vec{b} \times \vec{c}) = \dfrac{(\vec{b} + \vec{c})}{\sqrt{2}}, then the angle between a⃗\vec{a} and b⃗\vec{b} is
  1. A.
    3π4\dfrac{3\pi}{4}
  2. B.
    π4\dfrac{\pi}{4}
  3. C.
    π2\dfrac{\pi}{2}
  4. D.
    π\pi
Show answer and solution

Correct option: A

Solution

Using the BAC-CAB rule: a×(b×c) = (a·c)b - (a·b)c = (b+c)/√2.

So (a·c) = 1/√2 and -(a·b) = 1/√2, giving a·b = -1/√2.

Since all are unit vectors, a·b = |a||b|cosθ = cosθ = -1/√2, so θ = 3π/4.

[1995]

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