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 $Q$.

$P(5,-4,-6),Q(8,-7,4)$

Short answer

(a). The points $P(5,-4,-6)$and $Q(8,-7,4)$are plotted and shown as

(b). The distance between the two points $P(5,-4,-6)$ and $Q(8,-7,4)$ is $\sqrt{118}$.

Step-by-step solution

a.Step 1.Given information.

The given points are $P(5,-4,-6)$ and $Q(8,-7,4)$.

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 $P(5,-4,-6)$ and $Q(8,-7,4)$

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,

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

. $\begin{array}{l}(P,Q)=\sqrt{{(8-5)}^2+{(-7-(-4\left)\right)}^2+{(4-(-6\left)\right)}^2}\\ =\sqrt{{\left(3\right)}^2+{(-3)}^2+{\left(10\right)}^2}\\ =\sqrt{9+9+100}\\ \therefore d(P,Q)=\sqrt{118}\end{array}$