Chapter 10, Problem 9CT

To determine

To find: The angle between the vectors $\text{u}$ and $\text{v}$ .

Answer to Problem 9CT

The angle between the vectors $\text{u}$ and $\text{v}$ is $75.82°$ .

Explanation of Solution

Given information:

Let $u$ and $v$ be the vectors from $A\left(8,-2,5\right)$ to $B\left(4,4,-1\right)$ and from $A$ to $C\left(-4,3,0\right)$ respectively.

Calculation:

The component form of a vector $\text{v}$ with initial point $P\left(p_1,p_2,p_3\right)$ and terminal point $Q\left(q_1,q_2,q_3\right)$ is $\text{v}=\langle q_1-p_1,q_2-p_2,q_3-p_3\rangle$ .

Substitute $8$ , $-2$ , $5$ for $p_1$ , $p_2$ , $p_3$ and $4$ , $4$ , $-1$ for $q_1$ , $q_2$ , $q_3$ in the formula.

  $\begin{array}{c}u=\langle q_1-p_1,q_2-p_2,q_3-p_3\rangle \\ =\langle 4-8,4-\left(-2\right),-1-5\rangle \\ =\langle -4,6,-6\rangle \end{array}$

So, the component form of the vector $u$ is $\langle -4,6,-6\rangle$ .

Substitute $8$ , $-2$ , $5$ for $p_1$ , $p_2$ , $p_3$ and $-4$ , $3$ , $0$ for $q_1$ , $q_2$ , $q_3$ in the formula.

  $\begin{array}{c}\text{v}=\langle q_1-p_1,q_2-p_2,q_3-p_3\rangle \\ =\langle -4-8,3-\left(-2\right),0-5\rangle \\ =\langle -12,5,-5\rangle \end{array}$

So, the component form of the vector $\text{v}$ is $\langle -12,5,-5\rangle$ .

The formula for angle between the vectors $\text{u}$ and $\text{v}$ .

  $\cos \theta =\frac{\text{u}\cdot \text{v}}{\Vert \text{u}\Vert \Vert \text{v}\Vert }$

The dot product of vectors $\text{u}=\langle -4,6,-6\rangle$ and $\text{v}=\langle -12,5,-5\rangle$ .

  $\begin{array}{c}u\cdot v=\langle -4,6,-6\rangle \cdot \langle -5,3,1\rangle \\ =-4\times \left(-5\right)+6\times 3+\left(-6\right)\times 1\\ =20+18-6\\ =32\end{array}$

Calculate the magnitude of vector $\text{u}=\langle -4,6,-6\rangle$ .

  $\begin{array}{c}\Vert \text{u}\Vert =\sqrt{{\left(-4\right)}^2+{\left(6\right)}^2+{\left(-6\right)}^2}\\ =\sqrt{16+36+36}\\ =2\sqrt{22}\end{array}$

Calculate the magnitude of vector $\text{v}=\langle -12,5,-5\rangle$ .

  $\begin{array}{c}\Vert \text{v}\Vert =\sqrt{v_1^2+v_2^2+v_3^2}\\ =\sqrt{{\left(-12\right)}^2+{\left(5\right)}^2+{\left(-5\right)}^2}\\ =\sqrt{144+25+25}\\ =\sqrt{194}\end{array}$

Substitute the values in the formula for angle.

  $\begin{array}{c}\cos \theta =\frac{\text{u}\cdot \text{v}}{\Vert \text{u}\Vert \Vert \text{v}\Vert }\\ =\frac{32}{2\sqrt{22}\times \sqrt{194}}\\ =\frac{8\sqrt{1067}}{1067}\end{array}$

The solution of the equation $\cos \theta =\frac{8\sqrt{1067}}{1067}$ is the angle $\theta =75.82°$ .

Therefore, the angle between the vectors $\text{u}$ and $\text{v}$ is $75.82°$ .