Chapter 9, Problem 48RE

To determine

To find: The dot product of $v·w$ and the angle between them.

Answer to Problem 48RE

The dot product of $v·w$ and the angle between them are $1$ and $70.5$ respectively.

Explanation of Solution

Given information:

The first vector is

  $v=i+j+k$

The second vector is

  $w=i-j+k$

Calculation:

The dot product of the two vectors is

  $v·w=a_1{a}_2+b_1{b}_2+c_1{c}_2$

Here, the values $a_1=1$ , $b_1=1$ , $c_1=1$ , $a_2=1$ , $b_2=-1$ and $c_2=1$ .

Substitute the values in the formula of dot product.

  $\begin{array}{c}v·w=1\cdot 1+1\cdot \left(-1\right)+1\cdot 1\\ =1-1+1\\ =1\end{array}$

Thus, the dot product is $1$ .

Write the formula for the angle $\theta$ .

  $\begin{array}{c}\cos \theta =\frac{v·w}{\Vert v\Vert \Vert w\Vert }\\ =\frac{1}{\sqrt{1^2+1^2+1^2}\cdot \sqrt{1^2+{\left(-1\right)}^2+1^2}}\\ =\frac{1}{\sqrt{3}\cdot \sqrt{3}}\\ =\frac{1}{3}\end{array}$

  $\begin{array}{c}\cos \theta =\frac{1}{3}\\ \theta ={\cos }^{-1}\left(\frac{1}{3}\right)\\ \theta \approx 70.5\end{array}$

Therefore, the dot product of $v·w$ and the angle between them are $1$ and $70.5$ respectively.