MCQ
$(\vec{a} \cdot \hat{i})^2+(\vec{a} \cdot \hat{j})^2+(\vec{a} \cdot \hat{k})^2$ is equal to
  • A
    1
  • B
    $|\vec{a}|$
  • C
    $-\vec{a}$
  • $|\vec{a}|^2$

Answer

Correct option: D.
$|\vec{a}|^2$
(d) : Let $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k} \Rightarrow(\vec{a} \cdot \hat{i})^2=x^2$
Similarly, $(\vec{a} \cdot \hat{j})^2=y^2$ and $(\vec{a} \cdot \hat{k})^2=z^2$
$
\therefore \quad(\vec{a} \cdot \hat{i})^2+(\vec{a} \cdot \hat{j})^2+(\vec{a} \cdot \hat{k})^2=x^2+y^2+z^2=|\vec{a}|^2
$

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