Q68 / 155 · WBJEE 2020 · Mathematics

The unit vector in ZOX plane, making angles $45^{\circ}$ and $60^{\circ}$ respectively with $\vec{\alpha}=2 \hat{i}+2 \hat{j}-\hat{k}$ and $\vec{\beta}=j-\hat{k}$ is

  • A
    $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$
  • Correct answer
    B
    $\frac{1}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}}$
  • C
    $\frac{1}{\sqrt{2}} \hat{i}-\frac{1}{\sqrt{2}} \hat{j}$
  • D
    $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}$

Solution

Hint: Let the vector be $\vec{r}=x \hat{i}+z \hat{k} \Rightarrow|r|=1$ $\vec{r} \cdot \vec{a}=|r| |\vec{\alpha}| \cos 45^{\circ}$ $\therefore 2 x-z=\frac{3}{\sqrt{2}}$ $\vec{r} \cdot \vec{\beta}=|\vec{r}| |\vec{\beta}| \cos 60^{\circ}$ $z=-\frac{1}{\sqrt{2}}$ $\therefore \mathrm{x}=\frac{1}{\sqrt{2}}$ $\therefore \overrightarrow{\mathrm{r}}=\frac{1}{\sqrt{2}} \hat{i}-\frac{1}{\sqrt{2}} \hat{k}$

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

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