Question

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

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

Short answer

The dot product of the vectors $v=i-j;w=i+j+k$is $0$and the angle between them is $\frac{\mathrm{\pi }}{2}$

Step-by-step solution

Step 1. Given

The vectors:

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

To find the dot product and the angle between them.

Step 2. Dot product of the vectors.

The dot product is given by,

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

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

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

Step 4. Find the length of the vectors.

Now,

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

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

Step 5. find the angle between given vectors

Angle between the vectors is given by,

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