Chapter 9, Problem 64RE

To determine

To calculate: To use the vectors $v=-2i+j$ and $w=4i-3j$ to find,

  $\Vert v+w\Vert$

Answer to Problem 64RE

The value of the vector is $2\sqrt{2}$ .

Explanation of Solution

Given information:

The vectors:

  $v=-2i+j$ and $w=4i-3j$

Formula used:

For the vectors by the additivity of $a+b$ in these components of ‘i’ added to ‘i’ and components of ‘j’ added to ‘j’,

Such as, $\left(a_{1}i+b_{1}j\right)+\left(a_{2}i+b_{2}j\right)=\left(a_{1}i+a_{2}i\right)+\left(b_{1}j+b_{2}j\right)$ .

The magnitude of $\Vert v\Vert$ is find by the help of formula:

  $\Vert v\Vert =\sqrt{a^2+b^2}$

In the given equation or the given type of situation first add the $v+w$ and then find the magnitude of this term.

Calculation:

Consider the vectors:

  $v=-2i+j$ and $w=4i-3j$

Put the value of the vectors in the given expression $\Vert v+w\Vert$ .

Now $v=-2i+j$ and $w=4i-3j$ . $v+w$

With the help of magnitude of vectors formula:

  $\Vert v\Vert =\sqrt{a^2+b^2}$

  $\Vert v+w\Vert =\Vert \left(-2i+j\right)+\left(4i-3j\right)\Vert$

Now find the magnitude of $v$ with the help of the given formula $\Vert v\Vert =\sqrt{a^2+b^2}$ .

Same as the magnitude of $v+w$ is: $\Vert v+w\Vert =\Vert \sqrt{v^2+w^2}\Vert$ .

To find the magnitude first we find $v+w$ .

So $v+w=\left(-2i+j\right)+\left(4i-3j\right)$

Now in this apply additivity: $\left(a_{1}i+b_{1}j\right)+\left(a_{2}i+b_{2}j\right)=\left(a_{1}i+a_{2}i\right)+\left(b_{1}j+b_{2}j\right)$

Rewrite the expression:

  $\begin{array}{c}v+w=\left(-2i+j\right)+\left(4i-3j\right)\\ =-2i+4i+j-3j\\ =2i-2j\end{array}$

Now find the determinant $\Vert v+w\Vert =\Vert \sqrt{v^2+w^2}\Vert$ .

  $\begin{array}{l}\Vert v+w\Vert =\sqrt{2i^2-2j^2}\\ \Vert v\Vert =\sqrt{4+4}\\ \Vert v\Vert =2\sqrt{2}\end{array}$

Thus, the value of the magnitude of vectors are $2\sqrt{2}$ .