The direction cosines of the normal to the pl | vedclass

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(A) Let the two vectors be $\vec{b_1} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b_2} = 4\hat{i} + 5\hat{j} - 3\hat{k}$.The normal vector $\vec{n}$ to

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$\frac{-11}{\sqrt{179}}, \frac{7}{\sqrt{179}}, \frac{-3}{\sqrt{179}}$

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(A) Let the two vectors be $\vec{b_1} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b_2} = 4\hat{i} + 5\hat{j} - 3\hat{k}$.The normal vector $\vec{n}$ to the plane containing these lines is given by the cross product $\vec{b_1} 	imes \vec{b_2}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ 4 & 5 & -3 \end{vmatrix} = \hat{i}(-6 - 5) - \hat{j}(-3 - 4) + \hat{k}(5 - 8) = -11\hat{i} + 7\hat{j} - 3\hat{k}$.The magnitude of the normal vector is $|\vec{n}| = \sqrt{(-11)^2 + 7^2 + (-3)^2} = \sqrt{121 + 49 + 9} = \sqrt{179}$.The direction cosines are given by $\frac{a}{|\vec{n}|}, \frac{b}{|\vec{n}|}, \frac{c}{|\vec{n}|}$, where $a, b, c$ are the components of the normal vector.Thus, the direction cosines are $\frac{-11}{\sqrt{179}}, \frac{7}{\sqrt{179}}, \frac{-3}{\sqrt{179}}$.

The direction cosines of the normal to the plane containing the lines having direction ratios $1, 2, 1$ and $4, 5, -3$ are

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