Chapter 9, Problem 73RE

To determine

To calculate:The distance from $p_1=\left(1,3,-2\right)$ to $p_2=\left(4,-2,1\right)$ .

Answer to Problem 73RE

The value of the vectorsis $\sqrt{43}$ .

Explanation of Solution

Given information:

The points:

  $p_1=\left(1,3,-2\right)$ to $p_2=\left(4,-2,1\right)$ .

Formula used:

The distance formula:

  $\left|p_1{p}_2\right|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Calculation:

Consider thepoints:

  $p_1=\left(1,3,-2\right)$ to $p_2=\left(4,-2,1\right)$ .

Now to calculate first find the values of $x_1,x_2,y_1,y_2,z_1,z_2$ .

So here the values are:

  $\begin{array}{c}{x}_1=1,x_2=4\\ y_1=3,y_2=-2\\ z_1=-2,z_2=1\end{array}$

Now apply these values in the formula:

  $\begin{array}{l}\left|p_1{p}_2\right|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}\\ \left|p_1{p}_2\right|=\sqrt{{\left(4-1\right)}^2+{\left(3+2\right)}^2+{\left(1+2\right)}^2}\\ \left|p_1{p}_2\right|=\sqrt{{\left(3\right)}^2+{\left(5\right)}^2+{\left(3\right)}^2}\\ \left|p_1{p}_2\right|=\sqrt{9+25+9}\end{array}$

Now the distance is:

  $\begin{array}{l}\left|p_1{p}_2\right|=\sqrt{9+25+9}\\ \left|p_1{p}_2\right|=\sqrt{43}\end{array}$

Thus, the distance of the pointsare: $\sqrt{43}$ .