The equation of a line passing through the po | vedclass

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

Vedclass · Accepted Answer

$\frac{x+1}{-5} = \frac{y-2}{-4} = \frac{z-3}{1}$

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(C) Let the direction ratios of the required line be $(a, b, c)$.Since the line is perpendicular to the lines with direction ratios $(2, -3, -2)$ and $(-1, 2, 3)$,we have:$2a - 3b - 2c = 0$ and $-a + 2b + 3c = 0$.Using the cross product to find the direction ratios $(a, b, c)$:$a = (-3)(3) - (-2)(2) = -9 + 4 = -5$$b = (-2)(-1) - (2)(3) = 2 - 6 = -4$$c = (2)(2) - (-3)(-1) = 4 - 3 = 1$Thus,the direction ratios are $(-5, -4, 1)$ or $(5, 4, -1)$.The line passes through $(-1, 2, 3)$.The equation is $\frac{x - (-1)}{5} = \frac{y - 2}{4} = \frac{z - 3}{-1}$,which is equivalent to $\frac{x+1}{5} = \frac{y-2}{4} = \frac{z-3}{-1}$.However,checking the options,the correct direction vector is proportional to $(5, 4, -1)$. Option $B$ is $\frac{x+1}{5} = \frac{y-2}{4} = \frac{z-3}{1}$. Let's re-verify: $a:b:c = (-5):(-4):1$. The equation is $\frac{x+1}{-5} = \frac{y-2}{-4} = \frac{z-3}{1}$.

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

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