Chapter 10.2, Problem 71E

To determine

To find: the component form of $v$ .

Answer to Problem 71E

The component forms of vector $v$ are $\langle 0,2\sqrt{2},2\sqrt{2}\rangle$ or $\langle 0,2\sqrt{2},-2\sqrt{2}\rangle$ .

Explanation of Solution

Given information:

Vector $v$ lies in the $yz$ plane, has magnitude 4, and makes an angle of $45°$ with the positive $y$ 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 $yz$ plane. It means $x$ coordinate is zero. Vector $v$ makes $45°$ angle with positive $y$ axis. It means $y$ coordinate is positive. Also it can be concluded that the magnitude of $y$ and $z$ coordinates will be same irrespective of their sign as for a right angled triangle, having other angle $45°$ , has equal base and hypotenuse.

Vector $v$ makes a triangle on $yz$ plane with $y$ axis whose one vertex is at origin $O\left(0,0,0\right)$ , one at $y$ axis $P\left(0,y_1,0\right)$ and one at its destination point that is $A\left(0,y_1,z_1\right)$ .

Hence component form $v$ will be $\langle 0,y_1,z_1\rangle$ .

In $\Delta OAP$

  $\begin{array}{c}\angle AOP=45°\\ \angle OAP=90°\\ OP=PA\end{array}$

The magnitude of vector $OA$ is 4.

It means

  $\left|OA\right|=4$

Now according to Pythagoras theorem;

  $\begin{array}{c}diagona{l}^2=bas{e}^2+hypotenus{e}^2\\ {\left|OA\right|}^2={\left|OP\right|}^2+{\left|PA\right|}^2\\ 4^2=2{\left|OP\right|}^2\\ 4={\left|OP\right|}^2\\ \left|OP\right|=2\sqrt{2}\end{array}$

  $OP$ is calculated as

  $\begin{array}{c}OP=\sqrt{{\left(y_1-0\right)}^2}\\ =y_1\end{array}$

Hence $y$ coordinate $y_1$ is $2\sqrt{2}$ .

Now $z$ coordinate can be positive as well as negative with same value of $y_1$ as the $v$ can make $45°$ angle on either side of $y$ axis. Therefore $z$ coordinate $z_1$ is either $2\sqrt{2}$ or $-2\sqrt{2}$ .

Hence the component forms of vector $v$ are $\langle 0,2\sqrt{2},2\sqrt{2}\rangle$ or $\langle 0,2\sqrt{2},-2\sqrt{2}\rangle$ .