If the vertices of a triangle are $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$,then the area of the triangle is:

  • A
    $\sqrt{6}$
  • B
    $2\sqrt{6}$
  • C
    $3\sqrt{6}$
  • D
    $4\sqrt{6}$

Explore More

Similar Questions

Let $\bar{a}$ and $\bar{b}$ be two vectors such that $|\bar{a}|=1$,$|\bar{b}|=4$,and $\bar{a} \cdot \bar{b}=2$. If $\bar{c}=(2 \bar{a} \times \bar{b})-3 \bar{b}$,then the angle between $\bar{b}$ and $\bar{c}$ is

If $a = i + j + k$,$b = i + 3j + 5k$,and $c = 7i + 9j + 11k$,then the area of the parallelogram having diagonals $a + b$ and $b + c$ is

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{c}=\hat{j}-\hat{k}$,$\vec{a} \times \vec{b}=\vec{c}$ and $\vec{a} \cdot \vec{b}=1$,then $\vec{b}$ is equal to:

Find a unit vector perpendicular to each of the vectors $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b}),$ where $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}+3\hat{k}.$

If $(\bar{i}+\bar{j}+\bar{k})$,$(\bar{i}+2\bar{j}+3\bar{k})$ and $(2\bar{i}-\bar{j}+\bar{k})$ are the position vectors of the vertices $A$,$B$ and $C$ of $\triangle ABC$ respectively,then the vector equation of the altitude through $A$ 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