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

The shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is

The image of the point $ (1,6,3) $ in the line $ \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} $ is

Let in a $\triangle ABC$,the length of the side $AC$ be $6$,the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle ABC$ is

Let $L$ be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let $S$ be the set of all points $(a, b, c)$ on $L$,whose distance from the point $P(-1, -1, -9)$ is $7$. Then $\sum_{(a,b,c)\in S} (a+b+c)$ is equal to:

If the lines $\frac{x-1}{5}=\frac{y+1}{3}=\frac{3-z}{\lambda}$ and $\frac{x+1}{4}=\frac{1-3y}{15}=z+1$ are perpendicular to each other,then $\lambda=$

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