Question

In Problems 51–58, find the dot product v.w and the angle between v and w.

$$\begin{gathered} v=2i+2j-k; \\ w=i+2j+3k \end{gathered}$$

Short answer

The dot product of the vectors

$$\begin{gathered} v=2i+2j-k; \\ w=i+2j+3k \end{gathered}$$

is 3 and the angle between them is$74.5^{\circ }$.

Step-by-step solution

Step 1. Given

The vectors:

$$\begin{gathered} v=2i+2j-k; \\ w=i+2j+3k \end{gathered}$$

To find the dot product and the angle between them.

Step 2. Dot product of the vectors.

The dot product is given by,

$$\begin{gathered} v.w=(2i+2j-k).(i+2j+3k) \\ =\left(2\right)\left(1\right)+\left(2\right)\left(2\right)+(-1)\left(3\right) \\ =3 \end{gathered}$$

Step 3. Angle between the vectors.

Ifu andv are two nonzero vectors, the angle$\theta ,0\le \theta \le \pi$betweenu andv is determined by the formula,$\cos \theta =\frac{u.v}{||u||||v||}$

Step 4. Find the length of the vectors.

Now,

$$\begin{gathered} ||v||=||2i+2j-k|| \\ =\sqrt{2^2+2^2+{(-1)}^2} \\ =\sqrt{9} \\ =3 \end{gathered}$$
$$\begin{gathered} ||w||=||i+2j+3k|| \\ =\sqrt{1^2+2^2+3^2} \\ =\sqrt{14} \end{gathered}$$

Step 5. find the angle between given vectors

Angle between the vectors is given by,

$$\begin{gathered} \cos \theta =\frac{v.w}{||v||||w||} \\ =\frac{3}{\sqrt{9}\sqrt{14}} \\ =\frac{3}{3\sqrt{14}} \\ \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{14}}\right) \\ \simeq 74.5° \end{gathered}$$