4.5. Co-ordinate Geometry Exercises

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 \(6x - y - 4z = 3\).