Chapter 9, Problem 49RE

To determine

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

Answer to Problem 49RE

The dot product of $v·w$ and the angle between them are $0$ and $90°$ respectively.

Explanation of Solution

Given information:

The first vector is

  $v=4i-j+2k$

The second vector is

  $w=i-2j-3k$

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 $a_1=4$ , $b_1=-1$ , $c_1=2$ , $a_2=1$ , $b_2=-2$ and $c_2=-3$ .

Substitute the values in the formula of dot product.

  $\begin{array}{c}v·w=4\cdot 1+\left(-1\right)\cdot \left(-2\right)+2\cdot \left(-3\right)\\ =4+2-6\\ =0\end{array}$

Thus, the dot product is $0$ .

Write the formula for the angle $\theta$ .

  $\begin{array}{c}\cos \theta =\frac{v·w}{\Vert v\Vert \Vert w\Vert }\\ \cos \theta =\frac{1}{\sqrt{4^2+{\left(-1\right)}^2+2^2}\cdot \sqrt{1^2+{\left(-2\right)}^2+{\left(-3\right)}^2}}\\ \theta ={\cos }^{-1}\left(0\right)\\ \theta =90°\end{array}$

Therefore, the dot product of $v·w$ and the angle between them are $0$ and $90°$ respectively.