Chapter 9, Problem 102RE

To determine

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

Answer to Problem 102RE

The decomposition of vector $\text{v}$ are $v_1=\frac{\left(8i-4j\right)}{5}$ and $v_2=\frac{-23i+14j}{5}$ .

Explanation of Solution

Given information:

The vectors are given as:

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

Formula used:

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

then,

dot product: $v\cdot w=a_1{a}_2\text{}+b_1{b}_2+c_1{c}_2$

The vector $v$ are decompose into two vectors $v_1\text{and}{v}_2$ as,

where $v_1$ is parallel to $w\text{}$ , $v_1=\frac{v\times w}{{\Vert w\Vert }^2}w$ and $v_2$ is orthogonal to $\text{w}$ , $v_2=v-v_1$

  $\Vert w\Vert =\sqrt{a_2^2+b_2^2+c_2^2}$

Calculation:

Consider the vectors,

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

The vector $v$ decompose into $v_1\text{and}{v}_2$ vectors where $v_1$ is parallel to $w\text{}$ , $v_1=\frac{v\times w}{{\Vert w\Vert }^2}w$ and $v_2$ is orthogonal to $w\text{}$ , $v_2=v-v_1$

The vector $v_1$ is parallel to $w\text{}$ .

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 $a_1=-3,b_1=2,c_1=0,a_2=-2,b_2=1\text{and}\text{}{c}_2=0$

Therefore, the dot product is given by,

  $\begin{array}{c}v\cdot w=a_1{a}_2\text{}+b_1{b}_2+c_1{c}_2\\ =-3\times -2+2\times 1+0\times 0\\ =-6+2\\ =-4\end{array}$

Furthermore,

  $\begin{array}{l}{v}_1=\frac{v\cdot w}{{\Vert w\Vert }^2}w\\ v_1=\frac{-4}{{\left(-2\right)}^2+{\left(1\right)}^2}\left(-2i+j\right)\\ v_1=\frac{\left(8i-4j\right)}{5}\end{array}$

The vector $v_2$ is orthogonal to $w$ , $v_2=v-v_1$

Therefor,

  $\begin{array}{l}{v}_2=\left(-3i+2j\right)-\left(\frac{8i-4j}{5}\right)\\ v_2=\frac{-23i+14j}{5}\end{array}$

Thus, The decomposition of vector $v$ are $v_1=\frac{\left(8i-4j\right)}{5}$ and $v_2=\frac{-23i+14j}{5}$ .