Chapter 6, Problem 11PS

To determine

To find:The magnitude of sum is greater in figure (a) or (b).

Answer to Problem 11PS

  ${\left|u+v\right|}_b\le {\left|u+v\right|}_a$

Explanation of Solution

Given information:

Two vectors $u$ and $v$ are of same magnitude.

If two vectors $u$ and $v$ are given and the angle between them is $\theta$ , then the magnitude of their sum is :

  $\left|u+v\right|=\sqrt{{\left|u\right|}^2+{\left|v\right|}^2+2\left|u\right|\left|v\right|\cos \left(\theta \right)}$

From the given information,

  $\left|u\right|=\left|v\right|$

Let $\left|u\right|=\left|v\right|=a$ and angle between the vectors is $\theta$

  $\begin{array}{l}\left|u+v\right|=\sqrt{{\left|u\right|}^2+{\left|v\right|}^2+2\left|u\right|\left|v\right|\cos \left(\theta \right)}\\ \Rightarrow \left|u+v\right|=\sqrt{a^2+a^2+2\cdot a\cdot a\cos \left(\theta \right)}\\ \Rightarrow \left|u+v\right|=\sqrt{2a^2+2a^2\cos \left(\theta \right)}\\ \Rightarrow \left|u+v\right|=a\sqrt{2\left(1+\cos \left(\theta \right)\right)}\\ \Rightarrow \left|u+v\right|=a\sqrt{2\left(2{\cos }^2\left(\frac{\theta }{2}\right)\right)}\\ \Rightarrow \left|u+v\right|=a\left(2\cos \left(\frac{\theta }{2}\right)\right)\\ \Rightarrow \left|u+v\right|=2a\cos \left(\frac{\theta }{2}\right)\\ \Rightarrow \left|u+v\right|=2\left|u\right|\cos \left(\frac{\theta }{2}\right)\end{array}$

In figure (a),

The angle between two vectors ( $\theta$ ) is an acute angle.

Then $\left(\frac{\theta }{2}\right)$ is also an acute angle.

In figure (b),

The angle between two vectors is an obtuse angle.

Then $\left(\frac{\theta }{2}\right)$ is an acute angle.

Since the function $\cos \left(x\right)$ is a decreasing function for the values of $0\le x\le \frac{\pi }{2}$ (i.e. acute angle)

If $x_1\ge x_2$ for both $0\le x_1,x_2\le \frac{\pi }{2}$

then $\cos \left(x_1\right)\le \cos \left(x_2\right)$

Since the angle $\left(\frac{\theta }{2}\right)$ in figure (b) is greater than the angle $\left(\frac{\theta }{2}\right)$ in figure (a).

Therefore, $\cos {\left(\frac{\theta }{2}\right)}_b\le \cos {\left(\frac{\theta }{2}\right)}_a$

Therefore, the magnitude of sum of vectors in figure (b) is less than magnitude of sum of vectors in figure(a).

Therefore, ${\left|u+v\right|}_b\le {\left|u+v\right|}_a$