Question

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

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

Short answer

The dot product$v·w$is$-11$and the angle betweenv andwis$\theta \approx 169°.$

Step-by-step solution

Step 1. Given Information

The given vectors are$v=-2i+j,w=4i-3j.$

Step 2. Finding the dot product

To find the dot product, use the formula of the dot product which is$v·w=a_1{a}_2+b_1{b}_2$

So,

$$\begin{gathered} v·w=\left(-2\right)\left(4\right)+\left(1\right)\left(-3\right) \\ v·w=-11 \end{gathered}$$

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,

$$\begin{gathered} \cos \theta =\frac{\left(-2i+j\right)·\left(4\mathrm{i}-3\mathrm{j}\right)}{\left(\sqrt{{\left(-2\right)}^2+1^2}\right)\left(\sqrt{{\left(4\right)}^2+{\left(-3\right)}^2}\right)} \\ \cos \theta =\frac{-11}{5\sqrt{5}} \\ \theta ={\cos }^{-1}\left(\frac{-11}{5\sqrt{5}}\right) \\ \theta \approx 169° \end{gathered}$$