Chapter 9, Problem 104RE

To determine

To calculate: The decomposition a vector $\text{v}$ into two vectors where one is parallel to $\text{w}\text{}$ and other is orthogonal to $\text{w}\text{}$ and the vectors are $\text{v}=-\text{i}+2\text{j}\text{,}\text{w = 3i}\text{}-\text{j}\text{}\text{}$ .

Answer to Problem 104RE

The decomposition of vector $\text{v}$ are ${\text{v}}_1=\frac{\left(\text{-3i+j}\right)}{2}$ and ${\text{v}}_{\text{2}}=\frac{\text{i+3j}}{2}$ .

Explanation of Solution

Given information:

The vectors are given as:

  $\text{v}=-\text{i}+2\text{j}\text{,}\text{w = 3i}\text{}-\text{j}\text{}\text{}$

Formula used:

let $\text{v}={\text{a}}_{\text{1}}\text{i}{\text{+ b}}_{\text{1}}{\text{j+ c}}_{\text{1}}\text{k}$ and ${\text{w = a}}_{\text{2}}\text{i}\text{}\text{+}\text{}{\text{b}}_{\text{2}}\text{j}\text{}\text{}\text{+}{\text{c}}_{\text{3}}\text{k}$ are two vectors,

then,

dot product: $\text{v}\cdot \text{w}\text{=}{\text{a}}_1{\text{a}}_{\text{2}}\text{}{\text{+b}}_{\text{1}}{\text{b}}_{\text{2}}{\text{+c}}_{\text{1}}{\text{c}}_{\text{2}}$

The vector $\text{v}$ are decompose into two vectors ${\text{v}}_1\text{and}{\text{v}}_{\text{2}}$ as,

where ${\text{v}}_1$ is parallel to $\text{w}\text{}$ , ${\text{v}}_{\text{1}}=\frac{\text{v}\cdot \text{w}}{{\Vert \text{w}\Vert }^{\text{2}}}\text{w}$ and ${\text{v}}_{\text{2}}$ is orthogonal to $\text{w}$ , ${\text{v}}_{\text{2}}={\text{v-v}}_{\text{1}}$

  $\Vert \text{w}\Vert =\sqrt{{\text{a}}_2^{\text{2}}{\text{+b}}_2^{\text{2}}{\text{+c}}_2^{\text{2}}}$

Calculation:

Consider the vectors,

  $\text{v}=-\text{i}+2\text{j}\text{,}\text{w = 3i}\text{}-\text{j}\text{}\text{}$

The vector $\text{v}$ decompose into ${\text{v}}_1\text{and}{\text{v}}_{\text{2}}$ vectors where ${\text{v}}_1$ is parallel to $\text{w}\text{}$ , ${\text{v}}_{\text{1}}=\frac{\text{v}\cdot \text{w}}{{\Vert \text{w}\Vert }^{\text{2}}}\text{w}$ and ${\text{v}}_{\text{2}}$ is orthogonal to $\text{w}$ , ${\text{v}}_{\text{2}}={\text{v-v}}_{\text{1}}$

The vector ${\text{v}}_1$ is parallel to $\text{w}\text{}$ .

Therefore,

  ${\text{v}}_{\text{1}}=\frac{\text{v}\cdot \text{w}}{{\Vert \text{w}\Vert }^{\text{2}}}\text{w}$

Recall that for the dot product of vectors,

  $\text{v}\cdot \text{w}\text{=}{\text{a}}_1{\text{a}}_{\text{2}}\text{}{\text{+b}}_{\text{1}}{\text{b}}_{\text{2}}{\text{+c}}_{\text{1}}{\text{c}}_{\text{2}}$

The value of ${\text{a}}_{\text{1}}{\text{=-1,b}}_{\text{1}}{\text{=2,c}}_{\text{1}}{\text{=0,a}}_{\text{2}}{\text{=3,b}}_{\text{2}}\text{=-1}\text{and}\text{}{\text{c}}_{\text{2}}\text{=0}$

Therefore, the dot product is given by,

  $\begin{array}{c}\text{v}\cdot \text{w}\text{=}{\text{a}}_1{\text{a}}_{\text{2}}\text{}{\text{+b}}_{\text{1}}{\text{b}}_{\text{2}}{\text{+c}}_{\text{1}}{\text{c}}_{\text{2}}\\ =-1\cdot 3+2\cdot -1+0\cdot 0\\ =-3-2\\ =-5\end{array}$

Furthermore,

  $\begin{array}{l}{\text{v}}_{\text{1}}=\frac{\text{v}\cdot \text{w}}{{\Vert \text{w}\Vert }^{\text{2}}}\text{w}\\ {\text{v}}_1=\frac{-5}{{\left(3\right)}^2+{\left(-1\right)}^2}\left(\text{3i-j}\right)\\ {\text{v}}_1=\frac{\left(\text{-3i+j}\right)}{2}\end{array}$

The vector ${\text{v}}_{\text{2}}$ is orthogonal to $\text{w}$ , ${\text{v}}_{\text{2}}={\text{v-v}}_{\text{1}}$

Therefor,

  $\begin{array}{l}{\text{v}}_{\text{2}}=\left(\text{-i+2j}\right)-\left(\frac{\text{-3i+j}}{2}\right)\\ {\text{v}}_{\text{2}}=\frac{\text{i+3j}}{2}\end{array}$

Thus, The decomposition of vector $\text{v}$ are ${\text{v}}_1=\frac{\left(\text{-3i+j}\right)}{2}$ and ${\text{v}}_{\text{2}}=\frac{\text{i+3j}}{2}$ .