Find the coordinates of the point where the l | vedclass

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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 given by $\frac{x-x_{1}}{x_{2}-x_{1}} = \fr

Vedclass · Accepted Answer

$\left(0, \frac{17}{2}, -\frac{13}{2}\right)$

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 given by $\frac{x-x_{1}}{x_{2}-x_{1}} = \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{z-z_{1}}{z_{2}-z_{1}}$.Substituting the given points $(5, 1, 6)$ and $(3, 4, 1)$ into the formula:$\frac{x-5}{3-5} = \frac{y-1}{4-1} = \frac{z-6}{1-6}$$\Rightarrow \frac{x-5}{-2} = \frac{y-1}{3} = \frac{z-6}{-5} = k$ (where $k$ is a constant).Expressing $x, y, z$ in terms of $k$:$x = 5 - 2k$$y = 3k + 1$$z = 6 - 5k$Any point on this line is of the form $(5 - 2k, 3k + 1, 6 - 5k)$.The equation of the $YZ$-plane is $x = 0$.Since the line crosses the $YZ$-plane,we set the $x$-coordinate to $0$:$5 - 2k = 0 \Rightarrow k = \frac{5}{2}$.Now,substitute $k = \frac{5}{2}$ into the expressions for $y$ and $z$:$y = 3\left(\frac{5}{2}\right) + 1 = \frac{15}{2} + 1 = \frac{17}{2}$.$z = 6 - 5\left(\frac{5}{2}\right) = 6 - \frac{25}{2} = \frac{12 - 25}{2} = -\frac{13}{2}$.Thus,the coordinates of the required point are $\left(0, \frac{17}{2}, -\frac{13}{2}\right)$.

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

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