Question

In Problems 46–49, find the dot product $\mathit{v}\mathbf{·}\mathit{w}$ and the angle between v and w.

$v=i+j+k,w=i-j+k$

Short answer

The dot product $\mathrm{v}·\mathrm{w}$is $1$ and the angle between v and w is$70.5°.$

Step-by-step solution

Step 1. Given Information 

The given vectors are$v=i+j+kandw=i-j+k.$

Step 2. Finding the dot product 

To find the dot product, use the formula of the dot product which is $\mathrm{v}·\mathrm{w}={\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2.$

So,

$$\begin{gathered} \mathrm{v}·\mathrm{w}=\left(1\right)\left(1\right)+\left(1\right)\left(-1\right)+\left(1\right)\left(1\right) \\ \mathrm{v}·\mathrm{w}=1-1+1 \\ \mathrm{v}·\mathrm{w}=1 \end{gathered}$$

Step 3. Finding the angle between v and w 

To find the angle between vectors, we will use the formula $\cos \theta =\frac{\mathrm{u}·\mathrm{v}}{\left|\left|u\right|\right|\left|\left|v\right|\right|}.$

We have the dot product but to find the length of a vector we will use$\left|\left|u\right|\right|=\sqrt{a^2+b^2}.$

So,

$$\begin{gathered} \cos \theta =\frac{\left(i+j+k\right)·\left(i-j+k\right)}{\left(\sqrt{1^2+1^2+1^2}\right)\left(\sqrt{1^2+{\left(-1\right)}^2+1^2}\right)} \\ \cos \theta =\frac{1}{\sqrt{3}\sqrt{3}} \\ \theta ={\cos }^{-1}\left(\frac{1}{3}\right) \\ \theta \approx 70.5° \end{gathered}$$