Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. Which one of the following is correct?

  • A
    $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}=\vec{0}$
  • B
    $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a} \neq \vec{0}$
  • C
    $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{a} \times \vec{c}=\vec{0}$
  • D
    $\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}$ are mutually perpendicular

Explore More

Similar Questions

$\vec{u}, \vec{v}, \vec{w}$ are three unit vectors. Let $\vec{p}=\vec{u}+\vec{v}+\vec{w}$ and $\vec{q}=\vec{u} \times(\vec{v} \times \vec{w})$. If $\vec{p} \cdot \vec{u}=\frac{3}{2}, \vec{p} \cdot \vec{v}=\frac{7}{4}, |\vec{p}|=2$ and $\vec{v}=K \vec{q}$,then $K=$

Let $ABC$ be a triangle of area $15 \sqrt{2}$ and the vectors $\overrightarrow{AB}=\hat{i}+2 \hat{j}-7 \hat{k}$,$\overrightarrow{BC}=a \hat{i}+b \hat{j}+c \hat{k}$ and $\overrightarrow{AC}=6 \hat{i}+d \hat{j}-2 \hat{k}$,where $d>0$. Then the square of the length of the largest side of the triangle $ABC$ is:

Let $\vec{a} = \alpha \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - \alpha \hat{k}$,where $\alpha > 0$. If the projection of $\vec{a} \times \vec{b}$ on the vector $\vec{c} = -\hat{i} + 2 \hat{j} - 2 \hat{k}$ is $30$,then $\alpha$ is equal to:

Find the vector perpendicular to the vectors $4 i-j+3 k$ and $-2 i+j-2 k$ whose magnitude is $9$.

If the vertices of $\Delta ABC$ are $A=(2,3,5), B=(-1,3,2), C=(3,5,-2)$,then the area of the $\Delta ABC$ (in sq. units) 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