In a regular hexagon $ABCDEF$,$\overrightarrow{AE} = $

  • A
    $\overrightarrow{AC} + \overrightarrow{AF} + \overrightarrow{AB}$
  • B
    $\overrightarrow{AC} + \overrightarrow{AF} - \overrightarrow{AB}$
  • C
    $\overrightarrow{AC} + \overrightarrow{AB} - \overrightarrow{AF}$
  • D
    None of these

Explore More

Similar Questions

Find a vector of magnitude $5$ units,and parallel to the resultant of the vectors $\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}.$

If the points with position vectors $10\hat{i} + 3\hat{j}$,$12\hat{i} - 5\hat{j}$,and $a\hat{i} + 11\hat{j}$ are collinear,then the value of $a$ is:

If $\overline{AB} \parallel \overline{CD}$ and the position vectors of the four points $A, B, C,$ and $D$ are $2i + j, i - 3j, 3i + 2j,$ and $i + \lambda j$ respectively,then the value of $\lambda$ is .....

Let $\vec{u}, \vec{v}, \vec{w}$ be vectors such that $\vec{u} + \vec{v} + \vec{w} = \vec{0}$. If $|\vec{u}| = 3$,$|\vec{v}| = 4$,and $|\vec{w}| = 5$,then $\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}$ is:

Difficult
View Solution

If the position vector of one end of the line segment $AB$ is $2\hat{i} + 3\hat{j} - \hat{k}$ and the position vector of its midpoint is $3\,(\hat{i} + \hat{j} + \hat{k}),$ then the position vector of the other end 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