# Is It Possible to Add Two Vectors of Unequal Magnitudes and Get Zero? is It Possible to Add Three Vectors of Equal Magnitudes and Get Zero? - Physics

Is it possible to add two vectors of unequal magnitudes and get zero? Is it possible to add three vectors of equal magnitudes and get zero?

#### Solution

No, it is not possible to obtain zero by adding two vectors of unequal magnitudes.
Example: Let us add two vectors  $\vec{A}$ and  $\vec{B}$ of unequal magnitudes acting in opposite directions. The resultant vector is given by

$R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$

If two vectors are exactly opposite to each other, then

$\theta = 180^\circ, \cos180^\circ= - 1$

$R = \sqrt{A^2 + B^2 - 2AB}$

$\Rightarrow R = \sqrt{\left( A - B \right)^2}$

$\Rightarrow R = \left( A - B \right) \text { or } \left( B - A \right)$

From the above equation, we can say that the resultant vector is zero (R = 0) when the magnitudes of the vectors  $\vec{A}$ and $\vec{B}$ are equal (A = B) and both are acting in the opposite directions.
Yes, it is possible to add three vectors of equal magnitudes and get zero.
Lets take three vectors of equal magnitudes
$\vec{A,} \vec{B} \text { and } \vec{C}$ ,given these three vectors make an angle of $120^\circ$ with each other. Consider the figure below:
Lets examine the components of the three vectors.

$A_x = A$

$A_y = 0$

$B_x = - B \cos 60^\circ$

$B_y = B \sin 60^\circ$

$C_x = - C \cos 60^\circ$

$C_y = - C \sin 60^\circ$

$\text { Here, A = B = C }$

So, along the x - axis , we have:

$A - (2A \cos 60^\circ) = 0, as \cos 60^\circ = \frac{1}{2}$

$\Rightarrow B \sin 60^\circ - C \sin 60^\circ = 0$
Hence, proved.

Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 2 Physics and Mathematics
Short Answers | Q 2 | Page 27