The equation of the line passing through the points $(3, 2, 4)$ and $(4, 5, 2)$ is

  • A
    $\frac{x + 3}{1} = \frac{y + 2}{3} = \frac{z + 4}{-2}$
  • B
    $\frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z - 4}{-2}$
  • C
    $\frac{x + 3}{7} = \frac{y + 2}{7} = \frac{z + 4}{6}$
  • D
    $\frac{x - 3}{7} = \frac{y - 2}{7} = \frac{z - 4}{6}$

Explore More

Similar Questions

The point $P$ lies on the line $AB$,where $A \equiv (2, 4, 5)$ and $B \equiv (1, 2, 3)$. If the $z$-coordinate of point $P$ is $3$,then its $y$-coordinate is:

If the image of the point $P(1, -2, 1)$ with respect to the line passing through the points $B(1, 1, 2)$ and $C(2, 2, 1)$ is $R(l, m, n)$, then $l^2 + m^2 + n^2 =$

The Cartesian equation of the line passing through the point $\bar{i}-2 \bar{j}+\bar{k}$ and parallel to the vector $\bar{i}+\bar{j}+3 \bar{k}$ is

The lines $\frac{x - 1}{2} = \frac{y - 1}{2} = \frac{z - 3}{0}$ and $\frac{x - 2}{0} = \frac{y - 3}{0} = \frac{z - 4}{1}$ are:

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

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo