Question

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

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

Short answer

The dot product of the vectors

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

is 0 and the angle between them is $\frac{\mathrm{\pi }}{2}$

Step-by-step solution

Step 1. Given

The vectors:

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

To find the dot product and the angle between them.

Step 2. Dot product of the vectors.

The dot product is given by,

$$\begin{gathered} v.w=(i+3j+2k).(i-j+k) \\ =\left(1\right)\left(1\right)+\left(3\right)(-1)+\left(2\right)\left(1\right) \\ =0 \end{gathered}$$

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 angle between given vectors

Angle between the vectors is given by,

$$\begin{gathered} \cos \theta =\frac{v.w}{||v||||w||} \\ =\frac{0}{||v||||w||} \\ =0 \\ \theta =\frac{\mathrm{\pi }}{2} \end{gathered}$$