If 1, 2, 3 and -1, 0, 1 are the direction rat | vedclass

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(C) The direction ratios of the normal to the plane $AOB$ are given by the cross product of the vectors representing the rays $OA$ and $OB$. Let $\vec

Vedclass · Accepted Answer

$\frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}$

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(C) The direction ratios of the normal to the plane $AOB$ are given by the cross product of the vectors representing the rays $OA$ and $OB$. Let $\vec{a} = \langle 1, 2, 3 angle$ and $\vec{b} = \langle -1, 0, 1 angle$. The normal vector $\vec{n}$ is given by $\vec{a} 	imes \vec{b}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ -1 & 0 & 1 \end{vmatrix} = \hat{i}(2-0) - \hat{j}(1+3) + \hat{k}(0+2) = 2\hat{i} - 4\hat{j} + 2\hat{k}$.Thus, the direction ratios of the normal are $\langle 2, -4, 2 angle$, which can be simplified to $\langle 1, -2, 1 angle$ or $\langle -1, 2, -1 angle$.The magnitude of the vector $\langle -1, 2, -1 angle$ is $\sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.The direction cosines are $\frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}$.

If $1, 2, 3$ and $-1, 0, 1$ are the direction ratios of the rays $OA$ and $OB$ respectively, then the direction cosines of a normal to the plane $AOB$ are

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