The vector vec{r} whose magnitude is 3sqrt{2} | vedclass

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(D) Let the vector be $\vec{r} = xi + yj + zk$. The direction cosines are $l = \cos \alpha$, $m = \cos \beta$, $n = \cos \gamma$.Given $\beta = \frac{

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$\pm 3i + 3j$

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(D) Let the vector be $\vec{r} = xi + yj + zk$. The direction cosines are $l = \cos \alpha$, $m = \cos \beta$, $n = \cos \gamma$.Given $\beta = \frac{\pi}{4}$ and $\gamma = \frac{\pi}{2}$.Thus, $m = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $n = \cos(\frac{\pi}{2}) = 0$.Since $l^2 + m^2 + n^2 = 1$, we have $l^2 + (\frac{1}{\sqrt{2}})^2 + 0^2 = 1$, so $l^2 + \frac{1}{2} = 1$, which gives $l^2 = \frac{1}{2}$, so $l = \pm \frac{1}{\sqrt{2}}$.The unit vector $\hat{r} = li + mj + nk = \pm \frac{1}{\sqrt{2}}i + \frac{1}{\sqrt{2}}j + 0k$.The vector $\vec{r} = |\vec{r}| \hat{r} = 3\sqrt{2} (\pm \frac{1}{\sqrt{2}}i + \frac{1}{\sqrt{2}}j) = \pm 3i + 3j$.

The vector $\vec{r}$ whose magnitude is $3\sqrt{2}$ units and which makes angles of $\frac{\pi}{4}$ and $\frac{\pi}{2}$ with the positive $y$- and $z$-axes respectively is....

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