The equation of the line passing through the | vedclass

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(B) Let the direction ratios of the required line be $a, b, c$. Since the line is perpendicular to the lines with direction ratios $(1, 2, 3)$ and $(-

Vedclass · Accepted Answer

$\frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4}$

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(B) Let the direction ratios of the required line be $a, b, c$. Since the line is perpendicular to the lines with direction ratios $(1, 2, 3)$ and $(-3, 2, 5)$,we have: $a + 2b + 3c = 0$  --- $(i)$ $-3a + 2b + 5c = 0$ --- $(ii)$ Using the cross-multiplication method to solve for $a, b, c$: $\frac{a}{(2)(5) - (3)(2)} = \frac{b}{(3)(-3) - (1)(5)} = \frac{c}{(1)(2) - (2)(-3)}$ $\frac{a}{10 - 6} = \frac{b}{-9 - 5} = \frac{c}{2 + 6}$ $\frac{a}{4} = \frac{b}{-14} = \frac{c}{8}$ Dividing by $2$,we get the direction ratios as $(2, -7, 4)$. The line passes through the point $(-1, 3, -2)$. Therefore,the equation of the line is $\frac{x - (-1)}{2} = \frac{y - 3}{-7} = \frac{z - (-2)}{4}$,which simplifies to $\frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4}$.

The equation of the line passing through the point $(-1, 3, -2)$ and perpendicular to each of the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5}$ is

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