Chapter 11.2, Problem 64E

To determine

To find: the component form of $v$ .

Answer to Problem 64E

The component forms of vector $v$ are $\left(-5,0,5\right)$ or $\left(5,0,5\right)$ .

Explanation of Solution

Given information:

Vector $v$ lies in the $xz$ plane, has magnitude 10, and makes an angle of $60°$ with the positive $z$ axis.

Calculation:

Component form of any vector $v$ is the representation of vector $v$ which is directed by line segment from point $P\left(p_1,p_2,p_3\right)$ and $Q\left(q_1,q_2,q_3\right)$ . Component form of $v$ is evaluated by subtracting initial coordinates $P\left(p_1,p_2,p_3\right)$ from final coordinates $Q\left(q_1,q_2,q_3\right)$ . Mathematically

  $\begin{array}{c}v=\langle v_1,v_2,v_3\rangle \\ =\overline{PQ}\\ =\langle \left(q_1-p_1\right),\left(q_2-p_2\right),\left(q_3-p_3\right)\rangle \end{array}$

As per the statement the vector $v$ lies on the $xz$ plane. It means $y$ coordinate is zero. Vector $v$ makes $60°$ angle with positive $z$ axis. It means $z$ coordinate is positive. Vector $v$ makes a triangle on $xz$ plane with $x$ axis whose one vertex is at origin $O\left(0,0,0\right)$ , one at $z$ axis $P\left(0,0,z_1\right)$ and one at its destination point that is $A\left(x_1,0,z_1\right)$ .

Hence component form $v$ will be $\langle x_1,0,z_1\rangle$ .

In $\Delta OAP$

  $\begin{array}{c}\angle AOP=60°\\ \angle OAP=90°\end{array}$

And

  $\begin{array}{c}\tan 60°=\frac{PA}{OP}\\ \sqrt{3}=\frac{PA}{OP}\\ PA=\sqrt{3}OP\end{array}$

The magnitude of vector $OA$ is 4.

It means

  $\left|OA\right|=10$

Now according to Pythagoras theorem;

  $\begin{array}{c}{\text{diagonal}}^2={\text{base}}^2+{\text{hypotenuse}}^2\\ {\left|OA\right|}^2={\left|OP\right|}^2+{\left|PA\right|}^2\\ {10}^2={\left|OP\right|}^2+{\left(\sqrt{3}\left|OP\right|\right)}^2\\ 100=4{\left|OP\right|}^2\\ \left|OP\right|=5\end{array}$

  $OP$ is calculated as

  $\begin{array}{c}OP=\sqrt{{\left(z_1-0\right)}^2}\\ =z_1\end{array}$

Hence $z$ coordinate $z_1$ is $5$ .

Now $x$ coordinate can be positive as well as negative with same value of $z_1$ as the $v$ can make $60°$ angle on either side of $z$ axis. Therefore $x$ coordinate $x_1$ is either 5 or $-5$ .

Hence the component forms of vector $v$ are $\left(-5,0,5\right)$ or $\left(5,0,5\right)$ .