Question

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

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

Short answer

The dot product of the vectors $v=2i+j-3k;w=i+2j+2k$
is $-2$and the angle between them is $100.3°$

Step-by-step solution

Step 1. Given

The vectors:

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

To find the dot product and the angle between them.

Step 2. Dot product of the vectors.

The dot product is given by,

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

Step 3. Angle between the vectors.

If u and v are two nonzero vectors, the angle $\theta ,0\le \theta \le \pi$, between u and v is determined by the formula,

$\cos \theta =\frac{u.v}{||u||||v||}$

Step 4. Find the length of the vectors.

Now,

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

$$\begin{gathered} ||w||=||i+2j+2k|| \\ =\sqrt{1^2+2^2+2^2} \\ =\sqrt{9} \end{gathered}$$

Step 5. find the angle between given vectors

Angle between the vectors is given by,

$$\begin{gathered} \cos \theta =\frac{v.w}{||v||||w||} \\ =\frac{-2}{\sqrt{14}\sqrt{9}} \\ =\frac{-2}{3\sqrt{14}} \\ \theta ={\cos }^{-1}(-\frac{2}{3\sqrt{14}}) \\ \simeq 100.3° \end{gathered}$$