Q122 / 160 · TS EAMCET 2008 · Physics

The component of vector $\overrightarrow{\mathbf{A}}=a_x \hat{\mathbf{i}}+a_y \hat{\mathbf{j}}+a_z \hat{\mathbf{k}}$ along the direction of $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ is

  • A
    $a_x-a_y+a_z$
  • B
    $a_x-a_y$
  • Correct answer
    C
    $\left(a_x-a_y\right) / \sqrt{2}$
  • D
    $\left(a_x+a_y+a_z\right)$

Solution

Let $\overrightarrow{\mathbf{B}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$ Then component of vector $\overrightarrow{\mathbf{A}}$ along $\overrightarrow{\mathbf{B}}$ $\begin{aligned} & =\frac{\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}}}{|\overrightarrow{\mathbf{B}}|} \\ & =\frac{\left(a_x \hat{\mathbf{i}}+a_y \hat{\mathbf{j}}+a_z \hat{\mathbf{k}}\right) \cdot(\hat{\mathbf{i}}-\hat{\mathbf{j}})}{|\hat{\mathbf{i}}-\hat{\mathbf{j}}|} \\ & =\frac{a_x-a_y}{\sqrt{2}} \end{aligned}$

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

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