Give the equations to find the magnitude and direction of the resultant vector of two vectors $\vec{A}$ and $\vec{B}$ that are inclined at an angle $\theta$ to each other.

  • A
    Magnitude: $R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$,Direction: $\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta}$
  • B
    Magnitude: $R = \sqrt{A^2 + B^2 - 2AB \cos \theta}$,Direction: $\tan \alpha = \frac{B \sin \theta}{A - B \cos \theta}$
  • C
    Magnitude: $R = A + B$,Direction: $\alpha = 0$
  • D
    Magnitude: $R = \sqrt{A^2 + B^2}$,Direction: $\tan \alpha = \frac{B}{A}$

Explore More

Similar Questions

What vector must be added to the two vectors $\hat{i} - 2\hat{j} + 2\hat{k}$ and $2\hat{i} + \hat{j} - \hat{k}$ so that the resultant may be a unit vector along the $X$-axis?

The sum of the magnitudes of two forces acting at a point is $16 \,N$. If their resultant is normal to the smaller force and has a magnitude of $8 \,N$, then the forces are:

Explain the subtraction of vectors.

Vectors $\vec{A}$ and $\vec{B}$ make angles of $20^\circ$ and $110^\circ$ with the $x$-axis,respectively. The magnitudes of these vectors are $5 \ m$ and $12 \ m$,respectively. The angle that their resultant vector makes with the $x$-axis is:

Difficult
View Solution

Is it possible that $\vec{A} + \vec{B} = \vec{A} - \vec{B}$?

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