The direction ratios of the lines OA and OB a | vedclass

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Question

(A) The normal vector $\vec{n}$ to the plane $AOB$ is given by the cross product of the vectors $\vec{OA}$ and $\vec{OB}$.Given $\vec{OA} = (1, -2, -1

Vedclass · Accepted Answer

$\left( \frac{4}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{-2}{\sqrt{29}} \right)$

Vedclass · Answer

(A) The normal vector $\vec{n}$ to the plane $AOB$ is given by the cross product of the vectors $\vec{OA}$ and $\vec{OB}$.Given $\vec{OA} = (1, -2, -1)$ and $\vec{OB} = (3, -2, 3)$.$\vec{n} = \vec{OA} 	imes \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & -1 \ 3 & -2 & 3 \end{vmatrix} = \hat{i}(-6 - 2) - \hat{j}(3 + 3) + \hat{k}(-2 + 6) = -8\hat{i} - 6\hat{j} + 4\hat{k}$.Dividing by $-2$ to simplify,the normal vector is proportional to $(4, 3, -2)$.The magnitude of the vector $(4, 3, -2)$ is $\sqrt{4^2 + 3^2 + (-2)^2} = \sqrt{16 + 9 + 4} = \sqrt{29}$.The direction cosines are $\left( \frac{4}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{-2}{\sqrt{29}} ight)$.

The direction ratios of the lines $OA$ and $OB$ are $1, -2, -1$ and $3, -2, 3$. Then the direction cosines of the normal of plane $AOB$,where $O$ is the origin,are

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