vec{r} is a vector perpendicular to the plane | vedclass

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(D) The vector $\vec{r}$ is perpendicular to the plane determined by $\vec{a} = 2 \hat{i} - \hat{j}$ and $\vec{b} = \hat{j} + 2 \hat{k}$. Thus, $\vec{

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$\frac{3 \sqrt{6}}{2}$

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(D) The vector $\vec{r}$ is perpendicular to the plane determined by $\vec{a} = 2 \hat{i} - \hat{j}$ and $\vec{b} = \hat{j} + 2 \hat{k}$. Thus, $\vec{r}$ is parallel to $\vec{a} 	imes \vec{b}$.$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 0 \ 0 & 1 & 2 \end{vmatrix} = \hat{i}(-2 - 0) - \hat{j}(4 - 0) + \hat{k}(2 - 0) = -2 \hat{i} - 4 \hat{j} + 2 \hat{k}$.Let $\vec{r} = \lambda(-2 \hat{i} - 4 \hat{j} + 2 \hat{k}) = 2\lambda(-\hat{i} - 2 \hat{j} + \hat{k})$.The projection of $\vec{r}$ on $\vec{v} = 2 \hat{i} + \hat{j} + 2 \hat{k}$ is given by $\frac{|\vec{r} \cdot \vec{v}|}{|\vec{v}|} = 1$.$|\vec{v}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = 3$.$\vec{r} \cdot \vec{v} = \lambda(-2 \hat{i} - 4 \hat{j} + 2 \hat{k}) \cdot (2 \hat{i} + \hat{j} + 2 \hat{k}) = \lambda(-4 - 4 + 4) = -4\lambda$.So, $\frac{|-4\lambda|}{3} = 1 \Rightarrow |\lambda| = \frac{3}{4}$.$|\vec{r}| = |\lambda| \sqrt{(-2)^2 + (-4)^2 + 2^2} = \frac{3}{4} \sqrt{4 + 16 + 4} = \frac{3}{4} \sqrt{24} = \frac{3}{4} 	imes 2 \sqrt{6} = \frac{3 \sqrt{6}}{2}$.

$\vec{r}$ is a vector perpendicular to the plane determined by the vectors $2 \hat{i}-\hat{j}$ and $\hat{j}+2 \hat{k}$. If the magnitude of the projection of $\vec{r}$ on the vector $2 \hat{i}+\hat{j}+2 \hat{k}$ is $1$, then $|\vec{r}|=$

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