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