Question

Let \(u\) and \(v\) be two nonzero vectors in \(\mathbb{R}^{2}\). Prove that \(u\cdot v=\left\|u \right\|\left\|v \right\|cos\theta\) where \(θ\) is the angle between \(u\) and \(v\).

Short answer

It is proven that \(u\cdot v=\left\|u \right\| \left\|v \right\|cos\theta\).

Step-by-step solution

Step 1. Prove

To prove that \(u\cdot v=\left\|u \right\|\left\|v \right\|cos\theta\) where \(θ\) is the angle between \(u\) and \(v\) we will use the law of cosines and the properties of dot-product.

So, the law of cosines for the vectors \(u\) and \(v\) is

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

Step 2. Prove

Now, by using the dot-product we get,

\(\left\|u-v \right\|^{2}=\left ( u-v \right )\cdot \left ( u-v \right )\)

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

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

\(\left\|u-v \right\|^{2}=\left\|u \right\|^{2}-2u\cdot v+\left\|v \right\|^{2}\) .......(b)

By using equations (a) and (b)

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

\(-2u\cdot v=-2\left\|u \right\| \left\|v \right\|cos\theta\)

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

Hence proved.