Chapter 9.5, Problem 44AYU

To determine

Create an application required dot product in different form.

Answer to Problem 44AYU

  $a^2=b^2+c^2+2\left|\left|\overrightarrow{b}\right|\right|.\left|\left|\overrightarrow{c}\right|\right|\cos \alpha$

Explanation of Solution

Given information:

A dot product application.

Calculation:

Here, we can prove the law of cosines by using dot product.

Now, the law of cosines relates the length of the sides of a triangle to the cosines of one of its angle:

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos \alpha \\ b^2=a^2+c^2-2ac\cos \beta \\ c^2=a^2+b^2-2ab\cos \gamma \end{array}$

Now, let us suppose:

  $\begin{array}{l}\left|\left|\overrightarrow{a}\right|\right|=a\\ \left|\left|\overrightarrow{b}\right|\right|=b\\ \left|\left|\overrightarrow{c}\right|\right|=c\end{array}$

Now, we will apply triangle law of vector addition:

  $\begin{array}{l}\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}\\ \overrightarrow{a}.\overrightarrow{a}=\left(\overrightarrow{b}+\overrightarrow{c}\right).\left(\overrightarrow{b}+\overrightarrow{c}\right)\\ a^2=\overrightarrow{b.}\left(\overrightarrow{b}+\overrightarrow{c}\right)+\overrightarrow{c}.\left(\overrightarrow{b}+\overrightarrow{c}\right)\\ a^2=\overrightarrow{b.}\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{b}+\overrightarrow{c}.\overrightarrow{c}\\ a^2=b^2+2\overrightarrow{b}.\overrightarrow{c}+c^2\\ a^2=b^2+c^2+2\left|\left|\overrightarrow{b}\right|\right|.\left|\left|\overrightarrow{c}\right|\right|\cos \alpha \end{array}$

Now, the dot product of two vectors $v$ and $u$ with including angle $\theta$ ,

  $\begin{array}{l}vu=\left|\left|v\right|\right|\left|\left|u\right|\right|\cos \theta \\ =b^2+c^2+2bc\cos \alpha \end{array}$

Now, similarly we can prove:

  $\begin{array}{l}{b}^2=a^2+c^2-2ac\cos \beta \\ c^2=a^2+b^2-2ab\cos \gamma \end{array}$