What is the point of intersection of the line | 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 $\frac{x-3}{5-3} = \frac{y-4}

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 $\frac{x-3}{5-3} = \frac{y-4}{1-4} = \frac{z-1}{6-1} = \lambda$. This simplifies to $\frac{x-3}{2} = \frac{y-4}{-3} = \frac{z-1}{5} = \lambda$. Any point on this line is $(2\lambda + 3, -3\lambda + 4, 5\lambda + 1)$. Since the point lies on the $xy$-plane,its $z$-coordinate must be $0$. Setting $5\lambda + 1 = 0$,we get $\lambda = -\frac{1}{5}$. Substituting $\lambda = -\frac{1}{5}$ into the coordinates: $x = 2(-\frac{1}{5}) + 3 = -\frac{2}{5} + \frac{15}{5} = \frac{13}{5}$. $y = -3(-\frac{1}{5}) + 4 = \frac{3}{5} + \frac{20}{5} = \frac{23}{5}$. $z = 5(-\frac{1}{5}) + 1 = 0$. Thus,the point of intersection is $\left( \frac{13}{5}, \frac{23}{5}, 0 \right)$.

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

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