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1999

JEE Main 1999 Mathematics question on Vector Algebra

Let a⃗=2i^+j^+k^\vec{a} = 2\hat{i} + \hat{j} + \hat{k}, b⃗=i^+2j^−k^\vec{b} = \hat{i} + 2\hat{j} - \hat{k} and a unit vector c⃗\vec{c} be coplanar. If c⃗\vec{c} is perpendicular to a⃗\vec{a}, then c⃗\vec{c} is equal to
  1. A.
    12(−j^+k^)\frac{1}{\sqrt{2}}(-\hat{j} + \hat{k})
  2. B.
    13(−i^−j^−k^)\frac{1}{\sqrt{3}}(-\hat{i} - \hat{j} - \hat{k})
  3. C.
    15(i^−2j^)\frac{1}{\sqrt{5}}(\hat{i} - 2\hat{j})
  4. D.
    15(i^−j^−k^)\frac{1}{\sqrt{5}}(\hat{i} - \hat{j} - \hat{k})
Show answer and solution

Correct option: A

Solution

Since c is coplanar with a and b, write c = αa + βb = (2α+β)i + (α+2β)j + (α-β)k.

Apply c⊥a: c·a = 0 → 2(2α+β) + (α+2β) + (α-β) = 0 → 4α+2β+α+2β+α-β = 0 → 6α+3β = 0 → β = -2α.

So c = α[(2-2)i + (1-4)j + (1+2)k] = α[0i - 3j + 3k].

Normalizing: |c| = |α|·3√2 = 1 → |α| = 1/(3√2), giving c = ±(1/√2)(-j+k).

This matches option 1: (1/√2)(-j+k).

[1999]

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