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(A) Let $\vec{a} = 2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b} = 4 \hat{i}-\hat{j}+3 \hat{k}$.$A$ vector perpendicular to both $\vec{a}$ and $\vec{b}$ i

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$-2 \hat{i}+4 \hat{j}+4 \hat{k}$

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(A) Let $\vec{a} = 2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b} = 4 \hat{i}-\hat{j}+3 \hat{k}$.$A$ vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by the cross product $\vec{r} = \vec{a} 	imes \vec{b}$.$\vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 2 \ 4 & -1 & 3 \end{vmatrix}$$= \hat{i}(-3 - (-2)) - \hat{j}(6 - 8) + \hat{k}(-2 - (-4))$$= \hat{i}(-1) - \hat{j}(-2) + \hat{k}(2) = -\hat{i} + 2 \hat{j} + 2 \hat{k}$.The magnitude of $\vec{r}$ is $|\vec{r}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.$A$ unit vector in the direction of $\vec{r}$ is $\hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{-\hat{i} + 2 \hat{j} + 2 \hat{k}}{3}$.$A$ vector of magnitude $6$ in the direction of $\vec{r}$ is $6 \hat{r} = 6 \left( \frac{-\hat{i} + 2 \hat{j} + 2 \hat{k}}{3} ight) = 2(-\hat{i} + 2 \hat{j} + 2 \hat{k}) = -2 \hat{i} + 4 \hat{j} + 4 \hat{k}$.Note: The vector $-6 \hat{r} = 2 \hat{i} - 4 \hat{j} - 4 \hat{k}$ is also a valid answer.

Find a vector of magnitude $6,$ which is perpendicular to both the vectors $\vec{a} = 2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b} = 4 \hat{i}-\hat{j}+3 \hat{k}$.

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