The unit vector in ZOX plane, making angles 4 | vedclass

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(B) Let the unit vector in the $ZOX$ plane be $\vec{r} = x \hat{i} + z \hat{k}$.Since it is a unit vector, $|\vec{r}|^2 = x^2 + z^2 = 1$.Given $\vec{\

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}} \hat{i}-\frac{1}{\sqrt{2}} \hat{k}$

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(B) Let the unit vector in the $ZOX$ plane be $\vec{r} = x \hat{i} + z \hat{k}$.Since it is a unit vector, $|\vec{r}|^2 = x^2 + z^2 = 1$.Given $\vec{\alpha} = 2 \hat{i} + 2 \hat{j} - \hat{k}$, we have $|\vec{\alpha}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4+4+1} = 3$.The angle between $\vec{r}$ and $\vec{\alpha}$ is $45^{\circ}$, so $\vec{r} \cdot \vec{\alpha} = |\vec{r}| |\vec{\alpha}| \cos 45^{\circ}$.$(x \hat{i} + z \hat{k}) \cdot (2 \hat{i} + 2 \hat{j} - \hat{k}) = 1 \cdot 3 \cdot \frac{1}{\sqrt{2}} \Rightarrow 2x - z = \frac{3}{\sqrt{2}}$.Given $\vec{\beta} = \hat{j} - \hat{k}$, we have $|\vec{\beta}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$.The angle between $\vec{r}$ and $\vec{\beta}$ is $60^{\circ}$, so $\vec{r} \cdot \vec{\beta} = |\vec{r}| |\vec{\beta}| \cos 60^{\circ}$.$(x \hat{i} + z \hat{k}) \cdot (0 \hat{i} + 1 \hat{j} - 1 \hat{k}) = 1 \cdot \sqrt{2} \cdot \frac{1}{2} \Rightarrow -z = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \Rightarrow z = -\frac{1}{\sqrt{2}}$.Substituting $z$ into $2x - z = \frac{3}{\sqrt{2}}$, we get $2x - (-\frac{1}{\sqrt{2}}) = \frac{3}{\sqrt{2}} \Rightarrow 2x = \frac{2}{\sqrt{2}} = \sqrt{2} \Rightarrow x = \frac{1}{\sqrt{2}}$.Thus, $\vec{r} = \frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{k}$.

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 $\vec{\beta}=\hat{j}-\hat{k}$ is

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