The point of intersection of the line joining | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

If the two lines $l_{1}: \frac{x-2}{3}=\frac{y+1}{-2}, z=2$ and $l_{2}: \frac{x-1}{1}=\frac{2y+3}{\alpha}=\frac{z+5}{2}$ are perpendicular,then the angle between the lines $l_{2}$ and $l_{3}: \frac{1-x}{3}=\frac{2y-1}{-4}=\frac{z}{4}$ is

The equation of the line that passes through the origin and is parallel to the $X$-axis is . . . . . . .

The shortest distance between the lines $r = (3t - 4)\hat{i} - 2\hat{j} - (1 + 2t)\hat{k}$ and $r = (6 + s)\hat{i} + (2 - 2s)\hat{j} + 2(1 + s)\hat{k}$ is

The Cartesian equation of the line passing through the point $(-3,0,1)$ and perpendicular to vectors $\hat{i}-2\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ is

Let $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module