Find the coordinates of the point where the l | vedclass

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

Question

(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.

Similar Questions

Let the area of the triangle formed by the lines $x+2=y-1=z$,$\frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}$ and $\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}$ be $A$. Then $A^2$ is equal to . . . . . . .

The angle between the lines $\frac{x-2}{2} = \frac{y-3}{-2} = \frac{z-5}{1}$ and $\frac{x-2}{1} = \frac{y-3}{2} = \frac{z-5}{2}$ is $ . . . . . . $. (in $^{\circ}$)

Find the shortest distance between the lines whose vector equations are $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})$ and $\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$.

Let the point,on the line passing through the points $P(1, -2, 3)$ and $Q(5, -4, 7)$,farther from the origin and at a distance of $9$ units from the point $P$,be $(\alpha, \beta, \gamma)$. Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:

On a line with direction cosines $l, m, n$,$A(x_1, y_1, z_1)$ is a fixed point. If $B = (x_1 + 4kl, y_1 + 4km, z_1 + 4kn)$ and $C = (x_1 + kl, y_1 + km, z_1 + kn)$ where $k > 0$,then the ratio in which the point $B$ divides the line segment joining $A$ and $C$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module