The length of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$ is (in $\text{ units}$)

  • A
    $4$
  • B
    $5$
  • C
    $6$
  • D
    $7$

Explore More

Similar Questions

Find the position vector of the image of the point with position vector $\vec{P} = 2\hat{i} + \hat{j} + 3\hat{k}$ in the line whose vector equation is $\vec{r} = \hat{j} - 2\hat{k} + \lambda(\hat{i} + \hat{j} - \hat{k})$.

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

The distance from the point $-i + 2j + 6k$ to the straight line passing through the point $(2, 3, -4)$ and parallel to the vector $6i + 3j - 4k$ is

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is

The shortest distance between the skew lines $\vec{r}=(-\hat{i}-2 \hat{j}-3 \hat{k})+t(3 \hat{i}-2 \hat{j}-2 \hat{k})$ and $\vec{r}=(7 \hat{i}+4 \hat{k})+s(\hat{i}-2 \hat{j}+2 \hat{k})$ 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