The vector equation of the line passing through $P(1, 2, 3)$ and $Q(2, 3, 4)$ is

  • A
    $(\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})$
  • B
    $(\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} - \hat{k})$
  • C
    $(\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$
  • D
    $(\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 6\hat{j} + 12\hat{k})$

Explore More

Similar Questions

Show that the points $A(2, 3, -4)$,$B(1, -2, 3)$,and $C(3, 8, -11)$ are collinear.

The distance of the point $A(7, -2, 11)$ from the line $\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$ measured along the line $\frac{x-7}{2} = \frac{y+2}{-3} = \frac{z-11}{6}$ is:

The perpendicular distance from the point $P(3, 5, 2)$ to the line $L$ passing through the point $2\hat{i} + \hat{j}$ and parallel to the vector $\hat{i} + 5\hat{j} + 2\hat{k}$ is

The direction cosines of the line which is perpendicular to the lines $\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}$ and $\frac{x+5}{1}=\frac{y+3}{2}=\frac{z-6}{-2}$ are

The length of the perpendicular from the point $P(2, -1, 4)$ to the straight line $\frac{x + 3}{10} = \frac{y - 2}{-7} = \frac{z}{1}$ 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