$c^2\=a^2\+b^2-2 a b \cos \left(\mathrm{\θ}\right)$

to the triangle formed by the vectors A (in red), B (in green), and $\mathbf{B}-\mathbf{A}$ (in black) in Figure 1.3.3(a). The angle $\mathrm{\θ}$ is formed by the vectors A and B.

$c^2$

$\= a^2\+b^2-2 a b \cos \left(\mathrm{\theta }\right)$

${∥\mathbf{B}-\mathbf{A}∥}^2$

$\= {∥\mathbf{A}∥}^2\+$ ${∥\mathbf{B}∥}^2$ $-2∥\mathbf{A}∥ ∥\mathbf{B}∥ \cos \(\mathrm{\θ}\)$ 

$\left(\mathbf{B}-\mathbf{A}\right)·\left(\mathbf{B}-\mathbf{A}\right)$

$\= \mathbf{ }\mathbf{A}·\mathbf{A}\+\mathbf{B}·\mathbf{B}-2∥\mathbf{A}∥ ∥\mathbf{B}∥ \cos \(\mathrm{\θ}\)$

$\mathbf{B}·\mathbf{B}-2 \mathbf{A}·\mathbf{B}\+\mathbf{A}·\mathbf{A}$

$\= \mathbf{A}·\mathbf{A}\+\mathbf{B}·\mathbf{B}-2∥\mathbf{A}∥ ∥\mathbf{B}∥ \cos \(\mathrm{\θ}\)$

$-2 \mathbf{A}·\mathbf{B}$

$\= -2∥\mathbf{A}∥ ∥\mathbf{B}∥ \cos \(\mathrm{\θ}\)$

$\mathbf{A}·\mathbf{B}$

$\= ∥\mathbf{A}∥ ∥\mathbf{B}∥ \cos \(\mathrm{\θ}\)$

Initialize

Loading Student:-MultivariateCalculus

$\mathbf{A}\=⟨a_1\,a_2\,a_3⟩$$\stackrel{\text{assign}}{\to }$

$\mathbf{B}\=⟨b_1\,b_2\,b_3⟩$$\stackrel{\text{assign}}{\to }$

Obtain the dot product of A and B

$\mathbf{A}·\mathbf{B}$ = $a_1{b}_1\+a_2{b}_2\+a_3{b}_3$$$

Write the law of cosines in the form $a b \cos \left(\mathrm{\θ}\right)\=\left(a^2\+b^2-c^2\right)\/2$.

Apply this to the vectors A, B, and $\mathbf{B}-\mathbf{A}$ in the triangle in Figure 1.3.3(a).

${∥ \mathbf{A} ∥}^2\+{\∥\mathbf{ }\mathbf{B}\mathbf{ }\∥}^2-{∥\mathbf{B}-\mathbf{A}∥}^2$

$a_1^2\+a_2^2\+a_3^2\+b_1^2\+b_2^2\+b_3^2-{\left(b_1-a_1\right)}^2-{\left(b_2-a_2\right)}^2-{\left(b_3-a_3\right)}^2$

$\stackrel{\text{expand}}{\=}$

$\frac{2}{}$

$a_1{b}_1\+a_2{b}_2\+a_3{b}_3$

Initialize

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{MultivariateCalculus}\right)\:$$$

$\mathbf{A}\,\mathbf{B}≔⟨a_1\,a_2\,a_3⟩\,⟨b_1\,b_2\,b_3⟩\:$

Compute the dot product

$\mathrm{DotProduct}\left(\mathbf{A}\,\mathbf{B}\right)$

Obtain ${∥\mathbf{A}∥}^2$, ${∥\mathbf{B}∥}^2$, and ${∥\mathbf{B}-\mathbf{A}∥}^2$ 

$\mathrm{Norm}{\left(\mathbf{A}\right)}^2$

$\mathrm{Norm}{\left(\mathbf{B}\right)}^2$

$\mathrm{Norm}{\left(\mathbf{B}-\mathbf{A}\right)}^2$

Combine to obtain $\({\∥ \mathbf{A} \∥}^2\+{∥\mathbf{ }\mathbf{B}\mathbf{ }∥}^2-{\∥\mathbf{B}-\mathbf{A}\∥}^2\)\/2$

$\mathrm{expand}\left(\+-\right)\/2$