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(A) The direction vectors of the two given lines are $\vec{b_1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $\vec{b_2} = -3 \hat{i} + 2 \hat{j} + 5 \hat{k}

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$\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + \lambda(4 \hat{i} - 14 \hat{j} + 8 \hat{k})$

Vedclass · Answer

(A) The direction vectors of the two given lines are $\vec{b_1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $\vec{b_2} = -3 \hat{i} + 2 \hat{j} + 5 \hat{k}$.Since the required line is perpendicular to both,its direction vector $\vec{v}$ is given by the cross product $\vec{b_1} 	imes \vec{b_2}$.$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ -3 & 2 & 5 \end{vmatrix} = \hat{i}(10 - 6) - \hat{j}(5 - (-9)) + \hat{k}(2 - (-6)) = 4 \hat{i} - 14 \hat{j} + 8 \hat{k}$.The line passes through $(1, 2, 3)$,so its vector equation is $\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + \lambda(4 \hat{i} - 14 \hat{j} + 8 \hat{k})$.

The equation of the line passing through the point $(1, 2, 3)$ and perpendicular to the lines $\frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3}$ and $\bar{r} = \lambda(-3 \hat{i} + 2 \hat{j} + 5 \hat{k})$ is

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