Question

Plotting Points and Finding Distance in Three Dimensions Two points$P$and$Q$are given. (a) Plot $P$ and $Q$ (b) Find the distance between $P$ and $Q$

$P(3,1,0),Q(-1,2,-5)$

Short answer

(a). The points $P(3,1,0)$and $Q(-1,2,-5)$are plotted and shown as,

(b). The distance between the two points $P(3,1,0)$and $Q(-1,2,-5)$ is $\sqrt{42}$ .

Step-by-step solution

Step 1.Given information.

The given points are $P(3,1,0)$ and $Q(-1,2,-5)$ .

Step 2. We have to plot the given points in the xyz  plane.

Plot the points in the xyz plane.We get,

b.Step 1.Given information.

The given points are $P(3,1,0)$ and $Q(-1,2,-5)$.

Step 2. find the distance between the points.

Substitute the given points in the distance formula in three-dimensional.

The distance between the two points in three-dimension is given by,

$d(P,Q)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Substitute the points in the equation, we get

$\begin{array}{l}d(P,Q)=\sqrt{{(-1-3)}^2+{(2-1)}^2+{(-5-0)}^2}\\ =\sqrt{{(-4)}^2+1^2+{(-5)}^2}\\ =\sqrt{16+1+25}\\ \therefore d(P,Q)=\sqrt{42}\end{array}$