Chapter 11, Problem 41RE

To determine

Find whether given points are collinear using vectors.

Answer to Problem 41RE

The points $P\left(6,3,–1\right),Q\left(5,8,3\right),R\left(7,–2,–5\right)$ are collinear

Explanation of Solution

Given:

The points

  $P\left(6,3,–1\right),Q\left(5,8,3\right),R\left(7,–2,–5\right)$

If there are two points in space $P\left(x_1,y_1,z_1\right)$ and $Q\left(x_2,y_2,z_2\right)$ then it can written in component form as $\overline{PQ}=\langle \left(x_2-x_1\right),\left(y_2-y_1\right),\left(z_2-z_1\right)\rangle$

Three points are said to be collinear if the component vector $\overline{PQ}$ and $\overline{PR}$ of these points is in the form $\overline{PQ}=c\overline{PR}$

Consider the following points

  $P\left(6,3,–1\right),Q\left(5,8,3\right),R\left(7,–2,–5\right)$

First compute the component forms of $\overline{PQ}$ and $\overline{PR}$ that is,

  $\begin{array}{l}\overline{PQ}=\langle 5–6,8–3,3+1\rangle \hfill \\ =\langle –1,5,4\rangle \hfill \end{array}$

  $\begin{array}{l}\overline{PR}=\langle 7–6,–2–3,–5+1\rangle \hfill \\ =\langle 1,–5,–4\rangle \hfill \end{array}$

Since, $\overline{PQ}=\left(-1\right)\overline{PR}$ where $\left(-1\right)$ is some constant.

Therefore, the points $P\left(6,3,–1\right),Q\left(5,8,3\right),R\left(7,–2,–5\right)$ are collinear.