The equation of the line passing through the | vedclass

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(B) Let the direction ratios of the required line be $\langle a, b, c \rangle$. Since the line is perpendicular to the lines with direction ratios $\l

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

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(B) Let the direction ratios of the required line be $\langle a, b, c \rangle$. Since the line is perpendicular to the lines with direction ratios $\langle 1, 2, 3 \rangle$ and $\langle -3, 2, 5 \rangle$,we have: $a + 2b + 3c = 0$ $-3a + 2b + 5c = 0$ Using the cross product method to find the direction ratios: $\frac{a}{(2)(5) - (3)(2)} = \frac{b}{(3)(-3) - (1)(5)} = \frac{c}{(1)(2) - (2)(-3)}$ $\frac{a}{10 - 6} = \frac{b}{-9 - 5} = \frac{c}{2 + 6}$ $\frac{a}{4} = \frac{b}{-14} = \frac{c}{8}$ Simplifying,we get $\langle a, b, c \rangle = \langle 2, -7, 4 \rangle$. The line passes through $(-1, 3, -2)$,so the equation is: $\frac{x - (-1)}{2} = \frac{y - 3}{-7} = \frac{z - (-2)}{4}$ $\frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4}$

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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