4.5. Co-ordinate Geometry Exercises

Answer the following exercises based on the content from this chapter. The solutions can be found in the appendices.

Exercise 4.1

Given the following position vectors in \(\mathbb{R}^3\)

\[\begin{split} \begin{align*} \vec{a} &= \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}, & \vec{b} &= \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, & \vec{c} &= \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}, & \vec{d} &= \begin{pmatrix} 5 \\ 2 \\ 6 \end{pmatrix}. \end{align*} \end{split}\]

find:

(a)   the equation of the line that passes through \(\vec{a}\) and \(\vec{b}\);

(b)   the equation of the line that passes through \(\vec{c}\) and \(\vec{d}\);

(c)   the equation of the plane which passes through \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) lie;

(d)   the equation of the plane upon which passes through \(\vec{b}\), \(\vec{c}\) and \(\vec{d}\).

Exercise 4.2

Find the equation of the line that passes through the point with position vector \((3, 2, 1)^\mathsf{T}\) which is parallel to \(2 \vec{i} + \vec{j} + 3 \vec{k}\).

Exercise 4.3

Find the equation of the plane that passes through the point with position vector \((3, 2, 5)^\mathsf{T}\) which has a normal vector \(\vec{n} = (2, 1, 3)^\mathsf{T}\).

Exercise 4.4

A plane has the equation \(3x - 2y + z = 10\). Identify the normal to the plane and find the co-ordinates of 2 points on the plane having \(z = 2\).

Exercise 4.5

Two lines in \(\mathbb{R}^3\) are defined by \(\ell_1: (1 + 2t, -t, 1 + 3t)^\mathsf{T}\) and \(\ell_2: (1 + 2t, 4, 7 - t)^\mathsf{T}\) respectively.

(a)   find the intersection of the lines or show they are skew;

(b)   find the distance between the point with position vector \(\vec{p} = (0, -1, 3)^\mathsf{T}\) and \(\ell_1\);

(c)   find the shortest distance between the lines.

Exercise 4.6

Find the point where the line \(\ell:(1 + 2t, 2 + t, -1 + 4t)^\mathsf{T}\) meets the plane \(6x - y - 4z = 3\).

Exercise 4.7

Consider the diagram below that shows a plane that passes through the point \(\vec{p}\) and has normal vector \(\vec{n}\) and the point with position \(\vec{q}\) not on the plane.

Using the geometric definition of a dot product derive an expression for calculating the shortest distance between a point and a plane. Use your expression to find the shortest distance from the point with position vector \((2, 4, -3)^\mathsf{T}\) to the plane \(6x - y - 4z = 3\).