Q124 / 225 · JEE Main 2005 · Mathematics
For any vectora a the value of $(\vec{a} \times \hat{i})^2+(\vec{a} \times \hat{j})^2+(\vec{a} \times \hat{k})^2$ is equal to
- A$3 \vec{a}^2$
- B$\vec{a}^2$
- Correct answerC$2 \vec{a}^2$
- D$4 \vec{a}^2$
Solution
Let $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}$
$$
\begin{aligned}
& \vec{a} \times \hat{\mathbf{i}}=z \hat{\mathbf{j}}-y \hat{\mathbf{k}} \\
& \Rightarrow(\vec{a} \times \hat{\mathbf{i}})^2=y^2+z^2
\end{aligned}
$$
similarly $(\vec{a} \times \hat{j})^2=x^2+z^2$
and $(\vec{a} \times \hat{k})^2=x^2+y^2 \Rightarrow(\vec{a} \times \hat{i})^2=y^2+z^2$
similarly $(\vec{a} \times \hat{j})^2=x^2+z^2$
and $(\vec{a} \times \hat{k})^2=x^2+y^2$
$$
\Rightarrow(\vec{a} \times \hat{i})^2+(\vec{a} \times \hat{j})^2+(\vec{a} \times \hat{k})^2=2\left(x^2+y^2+z^2\right)=2 \vec{a}^2 .
$$
Hence, the correct answer is option: 'C'.