Question

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

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

Short answer

The dot product of the vectors

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

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

Step-by-step solution

Step 1. Given

The vectors:

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

To find the dot product and the angle between them.

Step 2. Dot product of the vectors.

The dot product is given by,

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

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,localid="1646807346417" $\cos \theta =\frac{u.v}{||u||||v||}$

Step 4. Find the length of the vectors.

Now,

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

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

Step 5. find the angle between given vectors

Angle between the vectors is given by,

$$\begin{gathered} \cos \theta =\frac{v.w}{||v||||w||} \\ =\frac{0}{\sqrt{14}\sqrt{3}} \\ =\frac{0}{\sqrt{3}\sqrt{14}} \\ \theta ={\cos }^{-1}\left(\frac{0}{3\sqrt{14}}\right) \\ \simeq \frac{\mathrm{\pi }}{2} \end{gathered}$$