The magnitude of the projection of the vector | vedclass

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(A) Let $\vec{a} = 2\hat{i}+3\hat{j}+\hat{k}$,$\vec{b} = \hat{i}+\hat{j}+\hat{k}$,and $\vec{c} = \hat{i}+2\hat{j}+3\hat{k}$.The vector perpendicular t

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

Vedclass · Answer

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

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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