A plane passes through the points A (1, 2, 3) | vedclass

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(C) The vectors lying on the plane are $\vec{AB} = (2-1)\hat{i} + (3-2)\hat{j} + (1-3)\hat{k} = \hat{i} + \hat{j} - 2\hat{k}$ and $\vec{AC} = (2-1)\ha

Vedclass · Accepted Answer

$\sqrt{\frac{2}{11}}$

Vedclass · Answer

(C) The vectors lying on the plane are $\vec{AB} = (2-1)\hat{i} + (3-2)\hat{j} + (1-3)\hat{k} = \hat{i} + \hat{j} - 2\hat{k}$ and $\vec{AC} = (2-1)\hat{i} + (4-2)\hat{j} + (2-3)\hat{k} = \hat{i} + 2\hat{j} - \hat{k}$.The normal vector $\vec{n}$ to the plane is given by $\vec{AB} 	imes \vec{AC} = \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -2 \ 1 & 2 & -1\end{array}ight| = \hat{i}(-1+4) - \hat{j}(-1+2) + \hat{k}(2-1) = 3\hat{i} - \hat{j} + \hat{k}$.The vector $\vec{OP} = 2\hat{i} - \hat{j} + \hat{k}$.Let $	heta$ be the angle between $\vec{OP}$ and the normal $\vec{n}$. Then $\cos 	heta = \frac{|\vec{OP} \cdot \vec{n}|}{|\vec{OP}| |\vec{n}|} = \frac{|(2)(3) + (-1)(-1) + (1)(1)|}{\sqrt{2^2 + (-1)^2 + 1^2} \sqrt{3^2 + (-1)^2 + 1^2}} = \frac{|6 + 1 + 1|}{\sqrt{6} \sqrt{11}} = \frac{8}{\sqrt{66}}$.Since $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{64}{66} = \frac{2}{66} = \frac{1}{33}$,we have $\sin 	heta = \sqrt{\frac{1}{33}}$.The length of the projection of $\vec{OP}$ on the plane is $|\vec{OP}| \sin 	heta = \sqrt{6} 	imes \sqrt{\frac{1}{33}} = \sqrt{\frac{6}{33}} = \sqrt{\frac{2}{11}}$.

$A$ plane passes through the points $A (1, 2, 3)$,$B (2, 3, 1)$,and $C (2, 4, 2)$. If $O$ is the origin and $P$ is $(2, -1, 1)$,then the projection of $\overline{OP}$ on this plane is of length .... .

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