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JEE Main 2023 · 25 Jan · Shift 1

JEE Main 2023 Mathematics question on Vector Algebra

Let a⃗,b⃗\vec{a}, \vec{b} and c⃗\vec{c} be three non zero vectors such that b⃗⋅c⃗=0\vec{b} \cdot \vec{c} = 0 and a⃗×(b⃗×c⃗)=b⃗−c⃗2\vec{a} \times (\vec{b} \times \vec{c}) = \dfrac{\vec{b} - \vec{c}}{2}. If d⃗\vec{d} be a vector such that b⃗⋅d⃗=a⃗⋅b⃗\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}, then (a⃗×b⃗)⋅(c⃗×d⃗)(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) is equal to
  1. A.
    34\dfrac{3}{4}
  2. B.
    14\dfrac{1}{4}
  3. C.
    −14-\dfrac{1}{4}
  4. D.
    12\dfrac{1}{2}
Show answer and solution

Correct option: B

Solution

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

This gives a·c = 1/2 and a·b = 1/2.

Now compute (a×b)·(c×d) using the identity (a×b)·(c×d) = (a·c)(b·d) - (a·d)(b·c).

Since b·c = 0, the second term vanishes. Also a·c = 1/2.

So (a×b)·(c×d) = (a·c)(b·d) = (1/2)(b·d).

Given b·d = a·b = 1/2, we get (1/2)(1/2) = 1/4.

[JEE Main 2023 · 25 Jan · Shift 1]

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