If the position vectors of the vertices $A, B$ and $C$ are $6i$,$6j$ and $k$ respectively with respect to the origin $O$,then the volume of the tetrahedron $OABC$ is

  • A
    $6$
  • B
    $3$
  • C
    $\frac{1}{6}$
  • D
    $\frac{1}{3}$

Explore More

Similar Questions

If $[\bar{a} \bar{b} \bar{c}] = 4$,then the volume (in cubic units) of the parallelepiped with $\bar{a} + 2 \bar{b}, \bar{b} + 2 \bar{c}$ and $\bar{c} + 2 \bar{a}$ as coterminal edges,is

$(\overrightarrow{a}+2 \overrightarrow{b}-\overrightarrow{c}) \cdot ((\overrightarrow{a}-\overrightarrow{b}) \times (\overrightarrow{a}-\overrightarrow{b}-\overrightarrow{c}))$ is equal to

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and $\lambda$ is a real number,then for what value of $\lambda$ does the equation $[\lambda(\vec{a} + \vec{b}), \lambda^2\vec{b}, \lambda\vec{c}] = [\vec{a}, \vec{b} + \vec{c}, \vec{b}]$ hold?

Difficult
View Solution

If the vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+2\hat{k}$ and $\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$ are coplanar, then $x=$

If $\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$ and $\vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$,then $[\vec{a} \vec{b} \vec{c}]$ depends on

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