If $\vec{a}$ is any vector in space,then

  • A
    $\vec{a} = (\vec{a} \cdot \hat{i})\hat{i} + (\vec{a} \cdot \hat{j})\hat{j} + (\vec{a} \cdot \hat{k})\hat{k}$
  • B
    $\vec{a} = (\vec{a} \times \hat{i}) + (\vec{a} \times \hat{j}) + (\vec{a} \times \hat{k})$
  • C
    $\vec{a} = \hat{j}(\vec{a} \cdot \hat{i}) + \hat{k}(\vec{a} \cdot \hat{j}) + \hat{i}(\vec{a} \cdot \hat{k})$
  • D
    $\vec{a} = (\vec{a} \times \hat{i}) \times \hat{i} + (\vec{a} \times \hat{j}) \times \hat{j} + (\vec{a} \times \hat{k}) \times \hat{k}$

Explore More

Similar Questions

The position vectors of the points $A, B, C$ are $(2\hat{i}+\hat{j}-\hat{k}), (3\hat{i}-2\hat{j}+\hat{k})$ and $(\hat{i}+4\hat{j}-3\hat{k})$ respectively. These points

If the points $P, Q$ and $R$ have the position vectors $\hat{i}-2 \hat{j}+3 \hat{k}$,$-2 \hat{i}+3 \hat{j}+2 \hat{k}$ and $-8 \hat{i}+13 \hat{j}$ respectively,then these points are

$M$ and $N$ are the midpoints of the diagonals $AC$ and $BD$ respectively of quadrilateral $ABCD$,then $\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}=$

The magnitude of the projection of the vector $\vec{a} = -\hat{i} + 2\hat{j} - \hat{k}$ on the unit vector $\hat{i}$ is . . . . . . .

$a, b, c$ are non-coplanar vectors. If $a+3 b+4 c=x(a-2 b+3 c)+y(a+5 b-2 c)+z(6 a+14 b+4 c)$,then $x+y+z=$

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