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(A) The equation of the line passing through points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{

Vedclass · Accepted Answer

$\left(\frac{17}{3}, 0, \frac{23}{3}\right)$

Vedclass · Answer

(A) The equation of the line passing through points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.The line passing through $(5, 1, 6)$ and $(3, 4, 1)$ is given by $\frac{x-5}{3-5} = \frac{y-1}{4-1} = \frac{z-6}{1-6}$.This simplifies to $\frac{x-5}{-2} = \frac{y-1}{3} = \frac{z-6}{-5} = k$.Any point on this line is of the form $(5-2k, 3k+1, 6-5k)$.Since the line crosses the $ZX$-plane,the $y$-coordinate must be $0$.Setting $3k+1 = 0$,we get $k = -\frac{1}{3}$.Substituting $k = -\frac{1}{3}$ into the coordinates:$x = 5 - 2(-\frac{1}{3}) = 5 + \frac{2}{3} = \frac{17}{3}$.$y = 3(-\frac{1}{3}) + 1 = 0$.$z = 6 - 5(-\frac{1}{3}) = 6 + \frac{5}{3} = \frac{23}{3}$.Thus,the required point is $\left(\frac{17}{3}, 0, \frac{23}{3}\right)$.

Find the coordinates of the point where the line passing through $(5, 1, 6)$ and $(3, 4, 1)$ crosses the $ZX$-plane.

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