Finding Distances 1 (AQA A-Level Further Maths): Revision Notes

Worked Example 1: Point to Line Distance

Calculate the shortest distance from the point $(3, 3, 1)$ to the line with equation:

$\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ -6 \end{pmatrix} + \lambda \begin{pmatrix} -3 \\ 0 \\ -1 \end{pmatrix}$

Solution:

First, create a vector from the point $(3, 3, 1)$ to a general point on the line:

$\begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix} - \left[\begin{pmatrix} 2 \\ 0 \\ -6 \end{pmatrix} + \lambda \begin{pmatrix} -3 \\ 0 \\ -1 \end{pmatrix}\right] = \begin{pmatrix} 1 + 3\lambda \\ 3 \\ 7 + \lambda \end{pmatrix}$

Since the vectors are perpendicular, their scalar product equals zero:

$\begin{pmatrix} 1 + 3\lambda \\ 3 \\ 7 + \lambda \end{pmatrix} \cdot \begin{pmatrix} -3 \\ 0 \\ -1 \end{pmatrix} = 0$

Expanding this gives:

$-3 - 9\lambda - 7 - \lambda = 0$

$\Rightarrow \lambda = -1$

Substitute $\lambda = -1$ to find the position vector of the intersection point:

$\begin{pmatrix} 2 \\ 0 \\ -6 \end{pmatrix} - \begin{pmatrix} -3 \\ 0 \\ -1 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \\ -5 \end{pmatrix}$

Finally, calculate the distance between $(3, 3, 1)$ and $(5, 0, -5)$:

$\text{Distance} = \sqrt{(5-3)^2 + (0-3)^2 + (-5-1)^2} = \sqrt{4 + 9 + 36} = 7$

The shortest distance is 7 units.

Worked Example 2: Distance Between Parallel Lines

Lines $l_1$ and $l_2$ have equations:

$l_1: \mathbf{r} = 2\mathbf{i} - \mathbf{j} + 8\mathbf{k} + s(4\mathbf{i} + 2\mathbf{j} - \mathbf{k})$

$l_2: \mathbf{r} = 3\mathbf{i} + \mathbf{k} + t(-12\mathbf{i} - 6\mathbf{j} + 3\mathbf{k})$

Calculate the shortest distance between $l_1$ and $l_2$.

Solution:

First, verify the lines are parallel. The direction vector of $l_2$ is:

$-12\mathbf{i} - 6\mathbf{j} + 3\mathbf{k} = -3(4\mathbf{i} + 2\mathbf{j} - \mathbf{k})$

This confirms the lines are parallel since one direction vector is a scalar multiple of the other.

Choose the point $(0, 3, 1)$ on $l_2$ (when $t = 0$) and find its distance to line $l_1$.

Create a vector from this point to a general point on $l_1$:

$(3\mathbf{j} + \mathbf{k}) - [2\mathbf{i} - \mathbf{j} + 8\mathbf{k} + s(4\mathbf{i} + 2\mathbf{j} - \mathbf{k})]$

$= (-2 - 4s)\mathbf{i} + (3 + 1 - 2s)\mathbf{j} + (1 - 8 + s)\mathbf{k}$

$= (-2 - 4s)\mathbf{i} + (4 - 2s)\mathbf{j} + (-7 + s)\mathbf{k}$

Apply the perpendicular condition with the direction vector:

$[(-2 - 4s)\mathbf{i} + (4 - 2s)\mathbf{j} + (-7 + s)\mathbf{k}] \cdot [4\mathbf{i} + 2\mathbf{j} - \mathbf{k}] = 0$

$4(-2 - 4s) + 2(4 - 2s) + (-7 + s)(-1) = 0$

$-8 - 16s + 8 - 4s + 7 - s = 0$

$\Rightarrow s = \frac{7}{21} = \frac{1}{3}$

Substitute $s = \frac{1}{3}$ into the line equation:

$2\mathbf{i} - \mathbf{j} + 8\mathbf{k} + \frac{1}{3}(4\mathbf{i} + 2\mathbf{j} - \mathbf{k}) = \frac{10}{3}\mathbf{i} - \frac{1}{3}\mathbf{j} + \frac{23}{3}\mathbf{k}$

Calculate the distance:

$\text{Distance} = \sqrt{\left(\frac{10}{3} - 0\right)^2 + \left(-\frac{1}{3} - 3\right)^2 + \left(\frac{23}{3} - 1\right)^2} = \frac{20}{3}$

The shortest distance is $\frac{20}{3}$ units.

Worked Example 3: Distance Between Skew Lines

Find the minimum distance between the skew lines:

$\mathbf{r} = 2\mathbf{i} + \mathbf{j} + \lambda(-\mathbf{j} + 2\mathbf{k})$

$\mathbf{r} = \mathbf{j} - 2\mathbf{k} + \mu(\mathbf{i} + 2\mathbf{j})$

Solution:

A general point $P$ on the first line has position vector:

$\overrightarrow{OP} = \begin{pmatrix} 2 \\ 1 - \lambda \\ 2\lambda \end{pmatrix}$

A general point $Q$ on the second line has position vector:

$\overrightarrow{OQ} = \begin{pmatrix} \mu \\ 1 + 2\mu \\ -2 \end{pmatrix}$

The vector joining these points is:

$\overrightarrow{PQ} = \begin{pmatrix} \mu \\ 1 + 2\mu \\ -2 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 - \lambda \\ 2\lambda \end{pmatrix} = \begin{pmatrix} \mu - 2 \\ 2\mu + \lambda \\ -2 - 2\lambda \end{pmatrix}$

For minimum distance, $\overrightarrow{PQ}$ must be perpendicular to both direction vectors.

Perpendicular to first line's direction vector:

$\begin{pmatrix} \mu - 2 \\ 2\mu + \lambda \\ -2 - 2\lambda \end{pmatrix} \cdot \begin{pmatrix} 0 \\ -1 \\ 2 \end{pmatrix} = 0$

$-2\mu - \lambda + 2(-2 - 2\lambda) = 0$

$\Rightarrow -2\mu - 5\lambda = 4 \quad \text{...(1)}$

Perpendicular to second line's direction vector:

$\begin{pmatrix} \mu - 2 \\ 2\mu + \lambda \\ -2 - 2\lambda \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} = 0$

$(\mu - 2) + 2(2\mu + \lambda) = 0$

$\Rightarrow 5\mu + 2\lambda = 2 \quad \text{...(2)}$

Solving equations (1) and (2) simultaneously:

$\mu = \frac{6}{7}, \quad \lambda = -\frac{8}{7}$

Substitute these values into the vector $\overrightarrow{PQ}$:

$\overrightarrow{PQ} = \begin{pmatrix} \frac{6}{7} - 2 \\ 2\left(\frac{6}{7}\right) + \left(-\frac{8}{7}\right) \\ -2 - 2\left(-\frac{8}{7}\right) \end{pmatrix} = \begin{pmatrix} -\frac{8}{7} \\ \frac{4}{7} \\ \frac{2}{7} \end{pmatrix}$

Calculate the magnitude:

$PQ = \sqrt{\left(-\frac{8}{7}\right)^2 + \left(\frac{4}{7}\right)^2 + \left(\frac{2}{7}\right)^2} = \frac{2\sqrt{21}}{7} \approx 1.31 \text{ units}$

The perpendicular distance between the two lines is 1.31 units (to 3 significant figures).

Worked Example 4: Calculus Method for Point to Line

Find the minimum distance between the line $\mathbf{r} = \mathbf{i} + \mathbf{j} + t(\mathbf{j} - \mathbf{k})$ and the point $(2, 1, -2)$.

Solution:

The vector from a general point on the line to the given point is:

$\mathbf{i} + \mathbf{j} + t(\mathbf{j} - \mathbf{k}) - (2\mathbf{i} + \mathbf{j} - 2\mathbf{k})$

$= -\mathbf{i} + t\mathbf{j} + (2 - t)\mathbf{k}$

The length of this vector is:

$x = \sqrt{1 + t^2 + (2-t)^2} = \sqrt{2t^2 - 4t + 5}$

Therefore:

$x^2 = 2t^2 - 4t + 5$

Differentiate with respect to $t$:

$\frac{d(x^2)}{dt} = 4t - 4$

Set the derivative equal to zero:

$\frac{d(x^2)}{dt} = 0 \Rightarrow 4t - 4 = 0 \Rightarrow t = 1$

Substitute $t = 1$ into the expression for $x$:

$x = \sqrt{1^2 + 1^2 + (2-1)^2} = \sqrt{3}$

The minimum distance is $\sqrt{3}$ units.

Worked Example 5: Calculus Method for Parallel Lines

Lines $l_1$ and $l_2$ have equations:

$l_1: \frac{x-1}{4} = \frac{y+1}{2} = \frac{z+2}{-1}$

$l_2: \mathbf{r} = \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -2 \\ -1 \\ 0.5 \end{pmatrix}$

Use calculus to calculate the shortest distance between $l_1$ and $l_2$. Give your answer to 3 significant figures.

Solution:

First, convert $l_1$ to vector form:

$\mathbf{r} = \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix}$

Verify the lines are parallel:

$\begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix} = -2 \begin{pmatrix} -2 \\ -1 \\ 0.5 \end{pmatrix}$

The lines are parallel. Consider the shortest distance from point $(1, 3, 0)$ on $l_2$ to line $l_1$.

Create the vector from $(1, 3, 0)$ to a general point on $l_1$:

$\begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix} - \left[\begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix}\right] = \begin{pmatrix} -4\lambda \\ 4 - 2\lambda \\ 2 + \lambda \end{pmatrix}$

The distance is:

$x = \sqrt{(-4\lambda)^2 + (4 - 2\lambda)^2 + (2 + \lambda)^2} = \sqrt{21\lambda^2 - 12\lambda + 20}$

Square both sides:

$x^2 = 21\lambda^2 - 12\lambda + 20$

Differentiate:

$\frac{d(x^2)}{d\lambda} = 42\lambda - 12$

Set equal to zero:

$42\lambda - 12 = 0 \Rightarrow \lambda = \frac{2}{7}$

Substitute back:

$x = \sqrt{21\left(\frac{2}{7}\right)^2 - 12\left(\frac{2}{7}\right) + 20}$

Calculate:

$x = 4.28 \text{ units (to 3 s.f.)}$

The shortest distance is 4.28 units (to 3 significant figures).

Worked Example 6: Reflection in a Line

Find the coordinates of the image of the point $A(-1, 5, 2)$ after it is reflected in the line with equation:

$\mathbf{r} = \begin{pmatrix} 0 \\ 3 \\ -2 \end{pmatrix} + t \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$

Solution:

A general point on the line has position vector:

$\overrightarrow{OM} = \begin{pmatrix} t \\ t + 3 \\ 2 - t \end{pmatrix}$

The vector from $A$ to this general point is:

$\overrightarrow{AM} = \begin{pmatrix} t \\ t + 3 \\ 2 - t \end{pmatrix} - \begin{pmatrix} -1 \\ 5 \\ 2 \end{pmatrix} = \begin{pmatrix} t + 1 \\ t - 2 \\ -4 - t \end{pmatrix}$

Since $\overrightarrow{AM}$ is perpendicular to the line:

$\begin{pmatrix} t + 1 \\ t - 2 \\ -4 - t \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = 0$

$(t + 1) + (t - 2) - (-4 - t) = 0$

$3t = -3$

$t = -1$

Substitute $t = -1$:

$\overrightarrow{AM} = \begin{pmatrix} 0 \\ -3 \\ -3 \end{pmatrix}$

Using the doubling relationship:

$\overrightarrow{AA'} = 2\overrightarrow{AM} = 2\begin{pmatrix} 0 \\ -3 \\ -3 \end{pmatrix} = \begin{pmatrix} 0 \\ -6 \\ -6 \end{pmatrix}$

Finally:

$\overrightarrow{OA'} = \begin{pmatrix} -1 \\ 5 \\ 2 \end{pmatrix} + \begin{pmatrix} 0 \\ -6 \\ -6 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \\ -4 \end{pmatrix}$

The coordinates of the image $A'$ are $(-1, -1, -4)$.