The point of intersection of the line joining | vedclass

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(B) The equation of the line passing through $(x_1, y_1, z_1) = (3, 4, 1)$ and $(x_2, y_2, z_2) = (5, 1, 6)$ is given by the section formula. Any poin

Vedclass · Accepted Answer

$\left( \frac{13}{5}, \frac{23}{5}, 0 \right)$

Vedclass · Answer

(B) The equation of the line passing through $(x_1, y_1, z_1) = (3, 4, 1)$ and $(x_2, y_2, z_2) = (5, 1, 6)$ is given by the section formula. Any point on this line can be represented as $\left( \frac{5\lambda + 3}{\lambda + 1}, \frac{1\lambda + 4}{\lambda + 1}, \frac{6\lambda + 1}{\lambda + 1} \right)$.Since the point lies on the $xy$-plane,its $z$-coordinate must be $0$.Setting the $z$-coordinate to $0$: $\frac{6\lambda + 1}{\lambda + 1} = 0 \Rightarrow 6\lambda + 1 = 0 \Rightarrow \lambda = -\frac{1}{6}$.Substituting $\lambda = -\frac{1}{6}$ into the coordinates:$x = \frac{5(-\frac{1}{6}) + 3}{-\frac{1}{6} + 1} = \frac{-\frac{5}{6} + 3}{\frac{5}{6}} = \frac{13/6}{5/6} = \frac{13}{5}$.$y = \frac{1(-\frac{1}{6}) + 4}{-\frac{1}{6} + 1} = \frac{-\frac{1}{6} + 4}{\frac{5}{6}} = \frac{23/6}{5/6} = \frac{23}{5}$.Thus,the point of intersection is $\left( \frac{13}{5}, \frac{23}{5}, 0 \right)$.

The point of intersection of the line joining the points $(3, 4, 1)$ and $(5, 1, 6)$ and the $xy$-plane is

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