Q49 / 160 · TS EAMCET 2017 · Mathematics

If a is a unit vector, then $|\mathbf{a} \times \hat{\mathbf{i}}|^2+|\mathbf{a} \times \hat{\mathbf{j}}|^2+|\mathbf{a} \times \hat{\mathbf{k}}|^2=$

  • Correct answer
    A
    2
  • B
    4
  • C
    1
  • D
    0

Solution

We have, $$ \begin{aligned} & |\mathbf{a} \times \hat{\mathbf{i}}|^2+|\mathbf{a} \times \hat{\mathbf{j}}|^2+|\mathbf{a} \times \hat{\mathbf{k}}|^2 \\ & (|\mathbf{a}||\hat{\mathbf{i}}| \sin \alpha)^2+(|\mathbf{a}||\hat{\mathbf{j}}| \sin \beta)^2+(|\mathbf{a}||\hat{\mathbf{k}}| \sin \gamma)^2 \\ & =\sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma \quad[\because|\mathbf{a}|=|\hat{\mathbf{i}}|=|\hat{\mathbf{j}}|=|\hat{\mathbf{k}}|=1] \\ & =1-\cos ^2 \alpha+1-\cos ^2 \beta+1-\cos ^2 \gamma \\ & =3-\left(\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma\right) \\ & =3-1=2 \quad\left[\because \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1\right] \end{aligned} $$

Hence, the correct answer is option: 'A'.

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