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1995

JEE Main 1995 Mathematics question on Vector Algebra

Let a⃗=i^−j^\vec{a} = \hat{i} - \hat{j}, b⃗=j^−k^\vec{b} = \hat{j} - \hat{k}, c⃗=k^−i^\vec{c} = \hat{k} - \hat{i}. If d⃗\vec{d} is a unit vector such that a⃗⋅d⃗=0=[b⃗ c⃗ d⃗]\vec{a} \cdot \vec{d} = 0 = [\vec{b}\ \vec{c}\ \vec{d}], then d⃗\vec{d} equals
  1. A.
    ±i^+j^−2k^6\pm \dfrac{\hat{i} + \hat{j} - 2\hat{k}}{\sqrt{6}}
  2. B.
    ±i^+j^−k^3\pm \dfrac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}}
  3. C.
    ±i^+j^+k^3\pm \dfrac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}
  4. D.
    ±k^\pm \hat{k}
Show answer and solution

Correct option: A

Solution

Since [b c d] = 0, vector d is coplanar with b and c, so d = λb + μc = λ(j−k) + μ(k−i) = −μi + λj + (μ−λ)k.

Applying a·d = 0: (i−j)·(−μi + λj + (μ−λ)k) = −μ − λ = 0, giving μ = −λ.

So d = λ(i + j − 2k). Normalizing: |i + j − 2k| = √6, thus d = ±(i + j − 2k)/√6.

[1995]

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