Chapter 9, Problem 30RE

To determine

To find: The position vector of the two given vectors and value of $\Vert v\Vert$ .

Answer to Problem 30RE

The position vector of the two given vectors and value of $\Vert v\Vert$ are $\left(-1,3\right)$ and $\sqrt{10}$ .

Explanation of Solution

Given information:

The given vectors are

  $P=\left(0,-2\right)$ and $Q=\left(-1,1\right)$

Calculation:

Consider the two points of vectors as

The initial point of the vector is,

  $P_1=\left(x_1,y_1\right)$

The final point of the vector is,

  $P_2=\left(x_2,y_2\right)$

Calculate the position vector.

  $\begin{array}{c}v=\overrightarrow{P_1{P}_2}\\ =\left(x_2-x_1,y_2-y_1\right)\\ =\left(\left(-1\right)-0,1-\left(-2\right)\right)\\ \left(-1,3\right)\end{array}$

If $\left(a,b\right)$ denote the position vector of a vector $v$ , then the vector $v$ is expressed as,

  $\begin{array}{c}\left(a,b\right)=a\left(-1,0\right)+b\left(0,1\right)\\ =ai+bj\end{array}$

Calculate the vector $v$ .

  $\begin{array}{c}\left(-1,3\right)=a\left(-1,0\right)+b\left(0,3\right)\\ =-i+3j\end{array}$

Thus, the position vector of $v=-i+3j$ is $\left(-1,3\right)$ .

Compare $v=-i+3j$ with the standard form $v=a_{1}i+b_{1}j$ .

  $\begin{array}{c}{a}_1=-1\\ b_1=3\end{array}$

Calculate the magnitude of the vector.

  $\begin{array}{c}\Vert v\Vert =\sqrt{a_1^2+b_1^2}\\ =\sqrt{{\left(-1\right)}^2+3^2}\\ =\sqrt{1+9}\\ =\sqrt{10}\end{array}$

Therefore, the position vector of the two given vectors and value of $\Vert v\Vert$ are $\left(-1,3\right)$ and $\sqrt{10}$ .