Explain the commutative law for vector addition.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Consider two vectors $\vec{A}$ and $\vec{B}$. According to the parallelogram law of vector addition,we can construct a parallelogram $OPRQ$ where $\vec{OP} = \vec{A}$ and $\vec{OR} = \vec{B}$.
From the triangle law of vector addition in $\Delta OPQ$:
$\vec{A} + \vec{B} = \vec{OP} + \vec{PQ} = \vec{OQ} \quad \dots (i)$
From the triangle law of vector addition in $\Delta ORQ$:
$\vec{B} + \vec{A} = \vec{OR} + \vec{RQ} = \vec{OQ} \quad \dots (ii)$
Since $\vec{PQ} = \vec{OR} = \vec{B}$ and $\vec{RQ} = \vec{OP} = \vec{A}$ in a parallelogram,comparing equations $(i)$ and $(ii)$,we get:
$\vec{A} + \vec{B} = \vec{B} + \vec{A}$
This proves that vector addition is commutative.

Explore More

Similar Questions

If the square of the resultant of two forces of equal magnitude is equal to three times their product,then the angle between them is ........ $^o$.

What should be the angle $\theta$ between two vectors $\vec{A}$ and $\vec{B}$ so that the magnitude of the resultant vector is minimum (in $^{\circ}$)?

Explain the analytical method for vector addition.

When $n$ vectors of different magnitudes are added,we get a null vector. Then the value of $n$ cannot be

Let $\overrightarrow C = \overrightarrow A + \overrightarrow B$. Which of the following statements are correct?
$(A)$ It is possible to have $|\overrightarrow C| < |\overrightarrow A|$ and $|\overrightarrow C| < |\overrightarrow B|$.
$(B)$ $|\overrightarrow C|$ is always greater than $|\overrightarrow A|$.
$(C)$ $|\overrightarrow C|$ may be equal to $|\overrightarrow A| + |\overrightarrow B|$.
$(D)$ $|\overrightarrow C|$ is never equal to $|\overrightarrow A| + |\overrightarrow B|$.

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