The equation of the line passing through (1, | vedclass

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Question

(A) Let the direction ratios of the required line be $(a, b, c)$.The direction ratios of the first line are $(1, 2, 4)$ and the second line are $(2, 2

Vedclass · Accepted Answer

$\frac{x-1}{6} = \frac{2-y}{7} = \frac{z-3}{2}$

Vedclass · Answer

(A) Let the direction ratios of the required line be $(a, b, c)$.The direction ratios of the first line are $(1, 2, 4)$ and the second line are $(2, 2, 1)$.Since the required line is perpendicular to both,its direction vector is the cross product of the direction vectors of the given lines:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 4 \ 2 & 2 & 1 \end{vmatrix} = \hat{i}(2-8) - \hat{j}(1-8) + \hat{k}(2-4) = -6\hat{i} + 7\hat{j} - 2\hat{k}$.Thus,the direction ratios are proportional to $(-6, 7, -2)$ or $(6, -7, 2)$.The line passes through $(1, 2, 3)$,so its equation is $\frac{x-1}{6} = \frac{y-2}{-7} = \frac{z-3}{2}$.This can be rewritten as $\frac{x-1}{6} = \frac{2-y}{7} = \frac{z-3}{2}$.

The equation of the line passing through $(1, 2, 3)$ and perpendicular to the lines $x-1 = \frac{y+2}{2} = \frac{z+4}{4}$ and $\frac{x-1}{2} = \frac{y-2}{2} = z+3$ is

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