If a unit vector lies in the yz-plane and mak | vedclass

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(B) Let the unit vector be $\vec{r} = y\hat{j} + z\hat{k}$. Since it is a unit vector,$|\vec{r}| = \sqrt{y^2 + z^2} = 1$. The angle $\alpha$ between $

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$(0, \frac{\sqrt{3}}{2}, \frac{1}{2})$

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(B) Let the unit vector be $\vec{r} = y\hat{j} + z\hat{k}$. Since it is a unit vector,$|\vec{r}| = \sqrt{y^2 + z^2} = 1$. The angle $\alpha$ between $\vec{r}$ and the $y$-axis $(\hat{j})$ is given by $\cos(30^\circ) = \frac{\vec{r} \cdot \hat{j}}{|\vec{r}| |\hat{j}|} = \frac{y}{1 	imes 1} = y$. Thus,$y = \cos(30^\circ) = \frac{\sqrt{3}}{2}$. The angle $\beta$ between $\vec{r}$ and the $z$-axis $(\hat{k})$ is given by $\cos(60^\circ) = \frac{\vec{r} \cdot \hat{k}}{|\vec{r}| |\hat{k}|} = \frac{z}{1 	imes 1} = z$. Thus,$z = \cos(60^\circ) = \frac{1}{2}$. Since the vector lies in the $yz$-plane,the $x$-component is $0$. Therefore,the components are $(0, \frac{\sqrt{3}}{2}, \frac{1}{2})$.

If a unit vector lies in the $yz$-plane and makes angles of $30^\circ$ and $60^\circ$ with the positive $y$-axis and $z$-axis respectively,then its components along the coordinate axes are:

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