The magnitude of the projection of the vector | vedclass

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(D) Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.The vector perpendicular to the plane containing $\vec{a

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

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(D) Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.The vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ is given by $\vec{n} = \vec{a} 	imes \vec{b}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(3-1) + \hat{k}(2-1) = \hat{i} - 2\hat{j} + \hat{k}$.The magnitude of the projection of vector $\vec{v} = 2\hat{i} + 3\hat{j} + \hat{k}$ on vector $\vec{n}$ is given by $\frac{|\vec{v} \cdot \vec{n}|}{|\vec{n}|}$.$\vec{v} \cdot \vec{n} = (2)(1) + (3)(-2) + (1)(1) = 2 - 6 + 1 = -3$.$|\vec{n}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.Therefore,the magnitude of the projection is $\frac{|-3|}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \sqrt{\frac{9}{6}} = \sqrt{\frac{3}{2}}$.

The magnitude of the projection of the vector $2\hat{i} + 3\hat{j} + \hat{k}$ on the vector perpendicular to the plane containing the vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 3\hat{k}$ is

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