Chapter 9.4, Problem 36E

To determine

To find:The angle between the given vector.

Answer to Problem 36E

The angle between the given vector is $48.19°$ .

Explanation of Solution

Given information:

The vectoris $u=i+2j-2k,v=4i-3k$ .

Calculation:

The dot product of two vectors is,

$\begin{array}{c}u\cdot v=\langle u_1,u_2,u_3\rangle \langle v_1,v_2,v_3\rangle \\ =u_1{v}_1+u_2{v}_2+u_3{v}_3\end{array}$

Substitute the values in above equation.

$\begin{array}{c}u\cdot v=\left(i+2j-2k\right)\left(4i-3k\right)\\ =\left(1\right)\left(4\right)+\left(0\right)\left(0\right)+\left(-2\right)\left(-3\right)\\ =10\end{array}$

The magnitude of vector $u$ is,

$\begin{array}{c}\left|u\right|=\sqrt{u_1^2+u_2^2+u_3^2}\\ =\sqrt{{\left(1\right)}^2+{\left(2\right)}^2+{\left(-2\right)}^2}\\ =3\end{array}$

The magnitude of vector $v$ is,

$\begin{array}{c}\left|v\right|=\sqrt{v_1^2+v_2^2+v_3^2}\\ =\sqrt{{\left(4\right)}^2+{\left(0\right)}^2+{\left(-3\right)}^2}\\ =5\end{array}$

The angle between the vectors is,

  $\theta ={\cos }^{-1}\left(\frac{u\cdot v}{\left|u\right|\left|v\right|}\right)$

Substitute the values in above equation.

  $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{10}{\left(3\right)\left(5\right)}\right)\\ =48.19°\end{array}$