Co-ordinate Geometry Exercises Solutions

Solution to Exercise 4.1

(a)   Using the vector equation of a line we have \(\vec{r} = \vec{a} + t\vec{d}_1\)

\[\begin{split} \begin{align*} \vec{d}_1 &= \vec{b} - \vec{a} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix}, \\ \therefore \vec{r} &= \vec{a} + t\vec{d}_1 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + t \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 - t \\ 1 \\ 0 \end{pmatrix} \end{align*} \end{split}\]

(b)   Using the vector equation of a line we have \(\vec{r} = \vec{c} + t\vec{d}_2\)

\[\begin{split} \begin{align*} \vec{d}_2 &= \vec{d} - \vec{c} = \begin{pmatrix} 5 \\ 2 \\ 6 \end{pmatrix} - \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 2 \end{pmatrix}, \\ \therefore \vec{r} &= \vec{c} + t\vec{d}_2 = \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 2 \\ 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 + 2 t \\ -1 + 3 t \\ 4 + 2 t \end{pmatrix} \end{align*} \end{split}\]

(c)   Calculate the normal vector to the plane

\[\begin{split} \begin{align*} \vec{b} - \vec{a} &= \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix}, \\ \vec{c} - \vec{a} &= \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix}, \\ \therefore \vec{n} &= (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -1 & 0 & 0 \\ 1 & -2 & 4 \end{vmatrix} = \begin{pmatrix} 0 \\ 4 \\ 2 \end{pmatrix}, \end{align*} \end{split}\]

Using the point normal definition of a plane

\[\begin{split} \begin{align*} n_x(x - a_x) + n_y(y - a_y) + n_z(z - a_z) &= 0 \\ 0(x - 2) + 4(y - 1) + 2(z - 0) &= 0 \\ 4 y + 2 z - 4 &= 0. \end{align*} \end{split}\]

(d)   Calculate the normal vector to the plane

\[\begin{split} \begin{align*} \vec{c} - \vec{b} &= \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}, \\ \vec{d} - \vec{b} &= \begin{pmatrix} 5 \\ 2 \\ 6 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 6 \end{pmatrix}, \\ \vec{n} &= (\vec{c} - \vec{b}) \times (\vec{d} - \vec{b}) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & -2 & 4 \\ 4 & 1 & 6 \end{vmatrix} = \begin{pmatrix} -16 \\ 4 \\ 10 \end{pmatrix}, \end{align*} \end{split}\]

Using the point normal definition of a plane

\[\begin{split} \begin{align*} n_x(x - b_x) + n_y(y - b_y) + n_z(z - b_z) &= 0 \\ -16(x - 1) + 4(y - 1) + 10(z - 0) &= 0\\ -16 x + 4 y + 10 z + 12 &= 0. \end{align*} \end{split}\]

Solution to Exercise 4.2

\[\begin{split} \vec{r} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 + 2 t \\ 2 + t \\ 1 + 3 t \end{pmatrix} \end{split}\]

Solution to Exercise 4.3

Using the point normal definition of a plane

\[\begin{split} \begin{align*} \vec{n} \cdot \begin{pmatrix} x - x_0 \\ y - y_0 \\ z - z_0 \end{pmatrix} &= 0 \\ \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} x - 3 \\ y - 2 \\ z - 5 \end{pmatrix} \\ 2(x - 3) + (y - 2) + 3(z - 5) &= 0 \\ 2 x + y + 3 z - 23 &= 0 \end{align*} \end{split}\]

Solution to Exercise 4.4

\(\vec{n} = (3, -2, 1)\).

Let \(x=0\) then \(3(0) - 2 y + 2 = 10\) so \(y = -4\) and a point on the plane has co-ordinates \((0, -4, 2)\).

Let \(x = 2\) then \(3(2) - 2 y + 2 = 10\) so \(y = -1\) and a point on the plane has co-ordinates \((2, -1, 2)\).

Solution to Exercise 4.5

(a)   Equating \(\ell_1\) and \(\ell_2\) and attempting to solve for \(t\)

\[\begin{split} \begin{align*} 1 + 2t_1 &= 1 + 2t_2 \\ -t_1 &= 4 \\ 1 + 3t_1 &= 7 - t_2 \end{align*} \end{split}\]

From the second equation \(t_1 = -4\) which when substituted into the third equation gives \(t_2 = 18\). Substituting these into the first equation gives \(-7 = 37\) which is a contradiction so \(\ell_1\) and \(\ell_2\) do not intersect.

We also need to show that they are not parallel, i.e., there is no value \(k\) such that \(\vec{d}_1 = k \vec{d}_2\). The direction vectors for \(\ell_1\) and \(\ell_2\) are \(\vec{d}_1 = (2, -1, 3)\) and \(\vec{d}_2 = (2, 0, -1)\) so

\[\begin{split} \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} &= k \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}, \end{split}\]

which gives the system

\[\begin{split} \begin{align*} 2 &= 2k ,\\ -1 &= 0, \\ 3 &= -k. \end{align*} \end{split}\]

The second equation is a contradiction so \(\ell_1\) and \(\ell_2\) are not parallel, and since they do not intersect then they must be skew.

(b)   Using the shortest distance between a point and a line

\[\begin{split} \begin{align*} t &= \frac{(\vec{p} - \vec{p}_1)\cdot \vec{d}_1}{\vec{d}_1 \cdot \vec{d}_1} = \frac{ \left( \begin{pmatrix} 0 \\ -1 \\ 3 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \right) \cdot \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}}{ \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} } = \frac{5}{14}, \\ \therefore \vec{r} &= \vec{p}_1 + t\vec{d}_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + \frac{5}{14} \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} \frac{12}{7} \\ - \frac{5}{14} \\ \frac{29}{14} \end{pmatrix}, \\ \overrightarrow{\vec{rp}} &= \begin{pmatrix} 0 \\ -1 \\ 3 \end{pmatrix} - \begin{pmatrix} \frac{12}{7} \\ -\frac{5}{14} \\ \frac{29}{14} \end{pmatrix} = \begin{pmatrix} -\frac{12}{7} \\ -\frac{9}{14} \\ \frac{13}{14} \end{pmatrix}, \\ \therefore d &= |\overrightarrow{\vec{rp}}| = \sqrt{\left(-\frac{12}{7}\right)^2 + \left(-\frac{9}{14}\right)^2 + \left(\frac{13}{14}\right)^2} = \frac{\sqrt{826}}{14}. \end{align*} \end{split}\]

(c)   The direction vectors for lines \(\ell_1\) and \(\ell_2\) are \(\vec{d}_1 = (2, -1, 3)^\mathsf{T}\) and \(\vec{d}_2 = (2, -, -1)^\mathsf{T}\) respectively. Calculating a vector perpendicular to both \(\vec{d}_1\) and \(\vec{d}_2\)

\[\begin{split} \begin{align*} \vec{n} &= \vec{d}_1 \times \vec{d_1} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \times \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & -1 & 3 \\ 2 & 0 & -1 \end{vmatrix} = \begin{pmatrix} 1 \\ 8 \\ 2 \end{pmatrix}, \\ \end{align*} \end{split}\]

and normalising gives

\[\begin{split} \begin{align*} \hat{\vec{n}} = \frac{\vec{n}}{|\vec{n}|} = \frac{1}{\sqrt{69}} \begin{pmatrix} 1 \\ 8 \\ 2 \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{69}}{69} \\ \frac{8\sqrt{69}}{69} \\ \frac{2\sqrt{69}}{69} \end{pmatrix}. \end{align*} \end{split}\]

Using the distance between two lines

\[\begin{split} \begin{align*} d &= (\vec{p}_2 - \vec{p}_1) \cdot \hat{\vec{n}} \\ &= \left( \begin{pmatrix} 1 \\ 4 \\ 7 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \right) \cdot \begin{pmatrix} \frac{\sqrt{69}}{69} \\ \frac{8\sqrt{69}}{69} \\ \frac{2\sqrt{69}}{69} \end{pmatrix} \\ &= \begin{pmatrix} 0 \\ 4 \\ 6 \end{pmatrix} \cdot \begin{pmatrix} \frac{\sqrt{69}}{69} \\ \frac{8\sqrt{69}}{69} \\ \frac{2\sqrt{69}}{69} \end{pmatrix} \\ &= \frac{44\sqrt{69}}{69}. \end{align*} \end{split}\]

Solution to Exercise 4.6

First we need to find the position vector of a point, \(\vec{r}\) say, that lies on the plane. Let \(x=0\) and \(y=1\) then \(z=-1\) so we know that \(\vec{r} = (0, 1, -1)^\mathsf{T}\) lies on the plane. Using the point normal definition of a plane

\[\begin{split} \begin{align*} \vec{n} \cdot \begin{pmatrix} x - x_0 \\ y - y_0 \\ z - z_0 \end{pmatrix} &= 0 \\ \begin{pmatrix} 6 \\ -1 \\ -4 \end{pmatrix} \cdot \begin{pmatrix} 0 - (1 + 2 t) \\ 1 - (2 + t) \\ -1 - (-1 + 4 t) \end{pmatrix} &= 0 \\ \begin{pmatrix} 6 \\ -1 \\ -4 \end{pmatrix} \cdot \begin{pmatrix} - 1 - 2 t \\ -1 - t \\ -4 t \end{pmatrix} &= 0 \\ -6 - 12 t + 1 + t + 16 t &= 0 \\ 5 t - 5 &= 0 \\ \therefore t = 1. \end{align*} \end{split}\]

So the line intersects with the plane at

\[\begin{split} \begin{align*} \vec{p} + t \vec{d} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \begin{pmatrix} 2 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix}. \end{align*} \end{split}\]

Solution to Exercise 4.7

Using the geometric definition of a dot product

\[ \begin{align*} (\vec{q} - \vec{p})\cdot \vec{n} &= |\vec{n}||\vec{q} - \vec{p}| \cos(\theta). \end{align*} \]

Since \(d\) is the length of the adjacent side of the right-angled triangle then

\[ \begin{align*} \cos(\theta) &= \frac{d}{|\vec{q} - \vec{p}|}, \end{align*} \]

so

\[\begin{split} \begin{align*} (\vec{q} - \vec{p}) \cdot \vec{n} &= |\vec{n}| |\vec{q} - \vec{p}| \frac{d}{|\vec{q} - \vec{p}|} \\ \therefore d &= (\vec{q} - \vec{p})\cdot \frac{\vec{n}}{|\vec{n}|}. \end{align*} \end{split}\]

The equation of the plane is \(6 x-y-4 z=3\) so letting \(x=0\) and \(y=1\) then \(z = 1\) so we know that \(\vec{p} = (0, 1, -1)\) lies on the plane. Since \(\vec{q} = (2, 4, -3)^\mathsf{T}\) and \(\vec{n} = (6, -1, -4)^\mathsf{T}\) then applying the above formula gives

\[\begin{split} \begin{align*} d &= \left( \begin{pmatrix} 2 \\ 4 \\ -3 \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} \right) \cdot \begin{pmatrix} 6 \\ -1 \\ -4 \end{pmatrix} / \left| \begin{pmatrix} 6 \\ -1 \\ -4 \end{pmatrix} \right| \\ &= \begin{pmatrix} 2 \\ 3 \\ -2 \end{pmatrix} \cdot \frac{1}{\sqrt{53}} \begin{pmatrix} 6 \\ -1 \\ -4 \end{pmatrix} \\ &= \frac{12 - 3 + 8}{\sqrt{53}} = \frac{17}{\sqrt{53}} \end{align*} \end{split}\]