Question

In Problems 51–58, find the dot product v.w and the angle between v and w.$$\begin{gathered} v=3i-4j+k; \\ w=6i-8j+2k \end{gathered}$$

Short answer

The dot product of the vectors

$$\begin{gathered} v=3i-4j+k; \\ w=6i-8j+2k \end{gathered}$$

is 52 and the angle between them is 0.

Step-by-step solution

Step 1. Given

The vectors:

$$\begin{gathered} v=3i-4j+k; \\ w=6i-8j+2k \end{gathered}$$

To find the dot product and the angle between them.

Step 2. Dot product of the vectors.

The dot product is given by,

$$\begin{gathered} v.w=(3i-4j+k).(6i-8j+2k) \\ =\left(3\right)\left(6\right)+(-4)(-8)+\left(1\right)\left(2\right) \\ =52 \end{gathered}$$

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,

$$\begin{gathered} ||v||=||3i-4j+k|| \\ =\sqrt{3^2+{(-4)}^2+{\left(1\right)}^2} \\ =\sqrt{26} \end{gathered}$$

$$\begin{gathered} ||w||=||6i-8j+2k|| \\ =\sqrt{6^2+{(-8)}^2+2^2} \\ =\sqrt{102} \end{gathered}$$

Step 5. find the angle between given vectors

Angle between the vectors is given by,

$$\begin{gathered} \cos \theta =\frac{v.w}{||v||||w||} \\ =\frac{52}{\sqrt{26}\sqrt{102}} \\ =\frac{52}{52} \\ \theta ={\cos }^{-1}\left(1\right) \\ \simeq 0 \end{gathered}$$