If $a, b, c, d$ are the position vectors of the points $A, B, C$ and $D$ respectively,referred to the same origin $O$,such that no three of these points are collinear and $a + c = b + d$,then the quadrilateral $ABCD$ is a

  • A
    Square
  • B
    Rhombus
  • C
    Rectangle
  • D
    Parallelogram

Explore More

Similar Questions

Let $P, Q, R$ and $S$ be the points on the plane with position vectors $-2 \hat{i}-\hat{j}, 4 \hat{i}, 3 \hat{i}+3 \hat{j}$ and $-3 \hat{i}+2 \hat{j}$ respectively. The quadrilateral $PQRS$ must be a

In a right-angled triangle $ABC$,if the hypotenuse $AB = p$,then $\overline{AB} \cdot \overline{AC} + \overline{BC} \cdot \overline{BA} + \overline{CA} \cdot \overline{CB} = ......$

If $\overline{a}, \overline{b}, \overline{c}$ are three vectors such that $|\overline{a}+\overline{b}+\overline{c}|=1$,$\overline{c}=\lambda(\overline{a} \times \overline{b})$ and $|\overline{a}|=\frac{1}{\sqrt{3}}, |\overline{b}|=\frac{1}{\sqrt{2}}, |\overline{c}|=\frac{1}{\sqrt{6}}$,then the angle between $\bar{a}$ and $\bar{b}$ is

Show that $|\vec{a}| \vec{b}+|\vec{b}| \vec{a}$ is perpendicular to $|\vec{a}| \vec{b}-|\vec{b}| \vec{a},$ for any two nonzero vectors $\vec{a}$ and $\vec{b}.$

If $\overrightarrow{x}$ and $\overrightarrow{y}$ are two non-zero vectors such that $|\overrightarrow{x}+\overrightarrow{y}|=|\overrightarrow{x}|$ and $2\overrightarrow{x}+\lambda\overrightarrow{y}$ is perpendicular to $\overrightarrow{y}$,then the value of $\lambda$ 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