Chapter 11.2, Problem 65E

To determine

To explain $\mathrm{w}$ can be represented by the arrow starting from end terminal of v to the end terminal of u.

Explanation of Solution

Given:

The $\mathrm{u}=\langle {\mathrm{u}}_1,{\mathrm{u}}_2\rangle$ and $\mathrm{v}=\langle {\mathrm{v}}_1,{\mathrm{v}}_2\rangle$ are the vectors in the plane.

Calculation:

The diagram shows, by vector addition rule, that

  $\begin{array}{l}\mathrm{v}+\mathrm{w}=\mathrm{u}\\ \mathrm{w}=\mathrm{u}-\mathrm{v}\end{array}$

So, $\mathrm{w}$ can be represented by the arrow starting from end terminal of v to the end terminal of u.

To determine

To prove why ${\left|\mathrm{w}\right|}^2={\left|\mathrm{u}\right|}^2+{\left|\mathrm{v}\right|}^2-2\left|\mathrm{u}\right|\left|\mathrm{v}\right|\cos \mathrm{\theta }$

Explanation of Solution

Given:

The $\mathrm{u}=\langle {\mathrm{u}}_1,{\mathrm{u}}_2\rangle$ and $\mathrm{v}=\langle {\mathrm{v}}_1,{\mathrm{v}}_2\rangle$ are the vectors in the plane.

Calculation:

This is just the law of cosines applied to the triangle, the sides of which are the magnitudes of the vectors. We state this as

  ${\left|\mathrm{w}\right|}^2={\left|\mathrm{u}\right|}^2+{\left|\mathrm{v}\right|}^2-2\left|\mathrm{u}\right|\left|\mathrm{v}\right|\cos \mathrm{\theta }$

To determine

To find the component form of $\mathrm{w}$ and use it to prove ${\left|\mathrm{u}\right|}^2+{\left|\mathrm{v}\right|}^2-{\left|\mathrm{w}\right|}^2=2({\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2)$

Answer to Problem 65E

  $\mathrm{w}=\langle {\mathrm{u}}_1-{\mathrm{v}}_1,{\mathrm{u}}_2-{\mathrm{v}}_2\rangle$

Explanation of Solution

Given:

The given vectors are

  $\mathrm{w}=\mathrm{u}-\mathrm{v}$

Calculation:

The component form of the vector w is

  $\begin{array}{l}\mathrm{w}=\mathrm{u}-\mathrm{v}\\ =\langle {\mathrm{u}}_1,{\mathrm{u}}_2\rangle -\langle {\mathrm{v}}_1,{\mathrm{v}}_2\rangle \\ \mathrm{w}=\langle {\mathrm{u}}_1-{\mathrm{v}}_1,{\mathrm{u}}_2-{\mathrm{v}}_2\rangle \end{array}$

And the magnitude or absolute value of a vector $<\mathrm{a},\mathrm{b}>$ is given by

  $\left|\langle \mathrm{a},\mathrm{b}\rangle \right|=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$

So, ${\left|\mathrm{u}\right|}^2+{\left|\mathrm{v}\right|}^2-{\left|\mathrm{w}\right|}^2=2({\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2)$

To determine

To prove $\mathrm{u}\cdot \mathrm{v}=\left|\mathrm{u}\right|\left|\mathrm{v}\right|\cos \mathrm{\theta }$

Explanation of Solution

Given:

The $\mathrm{u}=\langle {\mathrm{u}}_1,{\mathrm{u}}_2\rangle$ and $\mathrm{v}=\langle {\mathrm{v}}_1,{\mathrm{v}}_2\rangle$ are the vectors in the plane.

Calculation:

From part (b), we have

  $\begin{array}{l}{\left|\mathrm{w}\right|}^2={\left|\mathrm{u}\right|}^2+{\left|\mathrm{v}\right|}^2-2\left|\mathrm{u}\right|\left|\mathrm{v}\right|\cos \mathrm{\theta }\\ {\left|\mathrm{u}\right|}^2+{\left|\mathrm{v}\right|}^2-{\left|\mathrm{w}\right|}^2=2\left|\mathrm{u}\right|\left|\mathrm{v}\right|\cos \mathrm{\theta }\end{array}$

And, from part (c), we have

  ${\left|\mathrm{u}\right|}^2+{\left|\mathrm{v}\right|}^2-{\left|\mathrm{w}\right|}^2=2({\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2)$

Substituting, we get

  $2({\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2)=2\left|\mathrm{u}\right|\left|\mathrm{v}\right|\cos \mathrm{\theta }$

So, $\mathrm{u}\cdot \mathrm{v}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2=\left|\mathrm{u}\right|\left|\mathrm{v}\right|\cos \mathrm{\theta }$