Question

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

$v=4i-j+2k,w=i-2j-3k$

Short answer

The dot product $\mathrm{v}·\mathrm{w}$is $0$and the angle between v and w is $90°.$

Step-by-step solution

Step 1. Given Information 

The given vectors are$v=4i-j+2kandw=i-2j-3k.$

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,

$$\begin{gathered} \mathrm{v}·\mathrm{w}=\left(4\right)\left(1\right)+\left(-1\right)\left(-2\right)+\left(2\right)\left(-3\right) \\ \mathrm{v}·\mathrm{w}=4+2-6 \\ \mathrm{v}·\mathrm{w}=0 \end{gathered}$$

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,

$$\begin{gathered} \cos \theta =\frac{\left(4i-j+2k\right)·\left(i-2j-3k\right)}{\left(\sqrt{4^2+{\left(-1\right)}^2+2^2}\right)\left(\sqrt{1^2+{\left(-2\right)}^2+{\left(-3\right)}^2}\right)} \\ \cos \theta =\frac{0}{\left(\sqrt{4^2+{\left(-1\right)}^2+2^2}\right)\left(\sqrt{1^2+{\left(-2\right)}^2+{\left(-3\right)}^2}\right)} \\ \theta ={\cos }^{-1}\left(0\right) \\ \theta =90° \end{gathered}$$