The magnitude of the projection of the vector | vedclass

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

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$\frac{1}{\sqrt{6}}$ units

Vedclass · Answer

(B) Let $\vec{a} = 2\hat{i}+\hat{j}+\hat{k}$,$\vec{b} = \hat{i}+\hat{j}+\hat{k}$,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 magnitude of the projection of $\vec{a}$ on $\vec{n}$ is given by $\left| \frac{\vec{a} \cdot \vec{n}}{|\vec{n}|} ight|$.$\vec{a} \cdot \vec{n} = (2)(1) + (1)(-2) + (1)(1) = 2 - 2 + 1 = 1$.$|\vec{n}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.Therefore,the required projection is $\frac{1}{\sqrt{6}}$ units.

The magnitude of the projection of the vector $2\hat{i}+\hat{j}+\hat{k}$ 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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