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AIEEE 2003

JEE Main 2003 Mathematics question on Vector Algebra

If a⃗=(i^+j^+k^)\vec{a} = (\hat{i} + \hat{j} + \hat{k}), a⃗⋅b⃗=1\vec{a} \cdot \vec{b} = 1 and a⃗×b⃗=j^−k^\vec{a} \times \vec{b} = \hat{j} - \hat{k}, then b⃗\vec{b} is equal to
  1. A.
    i^−j^+k^\hat{i} - \hat{j} + \hat{k}
  2. B.
    2j^−k^2\hat{j} - \hat{k}
  3. C.
    i^\hat{i}
  4. D.
    2i^2\hat{i}
Show answer and solution

Correct option: C

Solution

Using the vector identity: b = (a·b)a/|a|² - a×(a×b)/|a|²

a×b = j - k, so a×(a×b) = (i+j+k)×(j-k) = i(1·(-1)-1·1) - j(1·(-1)-1·0) + k(1·1-1·0) = (-2i+j+k) ... let me compute directly:

(i+j+k)×(j-k) = |i j k; 1 1 1; 0 1 -1| = i(1·(-1)-1·1) - j(1·(-1)-1·0) + k(1·1-1·0) = i(-2) - j(-1) + k(1) = -2i+j+k

Now b = [(a·b)a - a×(a×b)] / |a|²

= [1·(i+j+k) - (-2i+j+k)] / 3

= [(i+j+k+2i-j-k)] / 3

= [3i] / 3 = i

So b = î, which is option 3.

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