If A=(1,8,4) and B=(2,-3,1),then the directio | vedclass

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(B) Given,$A=(1,8,4)$ and $B=(2,-3,1)$.The vectors representing the points are $\vec{OA} = \hat{i} + 8\hat{j} + 4\hat{k}$ and $\vec{OB} = 2\hat{i} - 3

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$\frac{2 \sqrt{10}}{9}, \frac{7 \sqrt{10}}{90}, \frac{-19 \sqrt{10}}{90}$

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(B) Given,$A=(1,8,4)$ and $B=(2,-3,1)$.The vectors representing the points are $\vec{OA} = \hat{i} + 8\hat{j} + 4\hat{k}$ and $\vec{OB} = 2\hat{i} - 3\hat{j} + \hat{k}$.The normal vector $\vec{n}$ to the plane $AOB$ is given by the cross product $\vec{OA} 	imes \vec{OB}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 8 & 4 \ 2 & -3 & 1 \end{vmatrix} = \hat{i}(8 - (-12)) - \hat{j}(1 - 8) + \hat{k}(-3 - 16) = 20\hat{i} + 7\hat{j} - 19\hat{k}$.The magnitude of the normal vector is $|\vec{n}| = \sqrt{20^2 + 7^2 + (-19)^2} = \sqrt{400 + 49 + 361} = \sqrt{810} = 9\sqrt{10}$.The direction cosines are the components of the unit normal vector $\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{20}{9\sqrt{10}}\hat{i} + \frac{7}{9\sqrt{10}}\hat{j} - \frac{19}{9\sqrt{10}}\hat{k}$.Simplifying the components: $\frac{20}{9\sqrt{10}} = \frac{20\sqrt{10}}{90} = \frac{2\sqrt{10}}{9}$,$\frac{7}{9\sqrt{10}} = \frac{7\sqrt{10}}{90}$,and $-\frac{19}{9\sqrt{10}} = -\frac{19\sqrt{10}}{90}$.

If $A=(1,8,4)$ and $B=(2,-3,1)$,then the direction cosines of a normal to the plane $AOB$ are

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