Chapter 11, Problem 8CT

To determine

To find: the angle $\theta$ between the vectors $u$ and $v$ .

Answer to Problem 8CT

  ${\cos }^{-1}\left(\frac{42}{\sqrt{38}\sqrt{97}}\right)$

Explanation of Solution

Given points : $u$ and $v$ are the vectors from A(8,-2,5) to B(6,4,-1) and from A to C(-4,3,0) respectively.

Calculation:

We know that if $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ are initial and terminal points of a vector $a$ respectively, then component form of vector

  $a$ is

  $a=\langle x_2-x_1,y_2-y_1,z_2-z_1\rangle$

Therefore,

  $\begin{array}{l}u=\langle 6-8,4+2,-1-5\rangle \Rightarrow u=\langle -2,6,-6\rangle \\ v=\langle -4-8,3+2,0-5\rangle \Rightarrow v=\langle -12,5,-5\rangle \end{array}$

Now,

  $\begin{array}{l}u.v=\langle -2,6,-6\rangle .\langle -12,5,-5\rangle \\ u.v=\left(-2\right)\left(-12\right)+6\left(5\right)+\left(-6\right)\left(-5\right)=24+30+30=84\end{array}$

The angle $\theta$ between the vectors $u$ and $v$ is

  $\begin{array}{l}\cos \theta =\frac{u.v}{\left|u\right|\left|v\right|}\\ \cos \theta =\frac{84}{\sqrt{{\left(-2\right)}^2+{\left(6\right)}^2+{\left(-6\right)}^2}\sqrt{{\left(-12\right)}^2+{\left(5\right)}^2+{\left(-5\right)}^2}}\\ \cos \theta =\frac{84}{\sqrt{76}\sqrt{194}}\\ \cos \theta =\frac{42}{\sqrt{38}\sqrt{97}}\\ \theta ={\cos }^{-1}\left(\frac{42}{\sqrt{38}\sqrt{97}}\right)\end{array}$