Q. 58 - Calculus

Question

Show that \(u\cdot v=\left\|u \right\|\left\|v \right\|cos\theta\) when \(u\) and \(v\) are nonzero vectors such that \(u = kv\) with \(k < 0\).

Step-by-Step Solution

Answer

It is shown that \(u\cdot v=\left\|u \right\|\left\|v \right\|cos\theta\) when \(u\) and \(v\) are nonzero vectors such that \(u = kv\) with \(k < 0\).

Step 1. Show

It is given that \(u=kv\), so \(u\cdot v=kv\cdot v\).

Now, 

\(u\cdot v=kv\cdot v\)

\(u\cdot v =k\left\|v \right\|^{2}\) 

\(u\cdot v =k\left\|v \right\|\left\|v \right\|\)

\(u\cdot v=\left|k \right|\left\|v \right\|\left\|v \right\|\) .......(a)

Since \(u=kv\) 

Thus,

\(\left\|u \right\|=\left\|kv \right\|\)

\(\left\|u \right\|=\left|k \right|\left\|v \right\|\)

By using equation (a), we get

\(u\cdot v=\left\|u \right\|\left\|v \right\|\) ........(b)

Let's use the formula of the law of cosines,

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

So,

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

\(\left\|u \right\|\left\|v \right\|cos\theta=\frac{1}{2}\left ( \left\|u \right\|^{2}+\left\|v \right\|^{2}-\left\|u-v \right\|^{2} \right )\) .......(c)

Step 2. Show

We will let \(u=\left<u_{1},u_{2},u_{3} \right>\) and \(v=\left<v_{1},v_{2},v_{3} \right>\).

So, \(u-v=\left<u_{1}-v_{1},u_{2}-v_{2},u_{3}-v_{3} \right>\) 

Now,\(u=\left<u_{1},u_{2},u_{3}\right>\)

\(\left\|u \right\|=\sqrt{u^{2}_{1}+u^{2}_{2}+u^{2}_{3}}\)

And \(v=\left<v_{1},v_{2},v_{3}\right>\)

\(\left\|v \right\|=\sqrt{v^{2}_{1}+v^{2}_{2}+v^{2}_{3}}\)   

So,

\(\left\|u-v \right\|=\sqrt{\left ( u_{1}-v_{1} \right )^{2}+\left ( u_{2}-v_{2} \right )^{2}+\left ( u_{3}-v_{3} \right )^{2}}\). 

Now, put all the values in equation (c) 

\(\left\|u \right\|\left\|v \right\|cos\theta=\frac{1}{2}\left ( \left\|u \right\|^{2}+\left\|v \right\|^{2}-\left\|u-v \right\|^{2} \right )\)

\(\left\|u \right\|\left\|v \right\|cos\theta =\frac{1}{2}\left ( \left ( \sqrt{u^{2}_{1}+u^{2}_{2}+u^{2}_{3}} \right )^{2}+\left ( \sqrt{v^{2}_{1}+v^{2}_{2}+v^{2}_{3}} \right )^{2}-\left ( \sqrt{\left ( u_{1}-v_{1} \right )^{2}+\left ( u_{2}-v_{2} \right )^{2}+\left ( u_{3}-v_{3} \right )^{2}} \right )^{2} \right )\)

\(\left\|u \right\|\left\|v \right\|cos\theta =\frac{1}{2} \left ( 2u_{1}v_{1}+2u_{2}v_{2}+2u_{3}v_{3} \right)\)

\(\left\|u \right\|\left\|v \right\|cos\theta = \left ( u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3} \right )\)

By using (b) we get,

\(\left\|u \right\|\left\|v \right\|cos\theta =u\cdot v\)