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

Worked Example 1: Finding a Vector Equation from Two Points

Work out the vector equation of the line passing through the points with position vectors $\begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}$ and $\begin{pmatrix} 3 \\ 0 \\ -1 \end{pmatrix}$.

Solution:

First, find a vector in the direction of the line. Subtract the position vector of one point from the position vector of the other point:

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

This vector is parallel to the line and serves as the direction vector.

Now use the general form $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$. You can choose either of the given position vectors as $\mathbf{a}$:

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

Alternatively, using the other point:

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

Both equations represent the same line. Different choices of the position vector or different scalar multiples of the direction vector will give different-looking equations that still represent the same line.

Worked Example 2: Converting from Cartesian to Vector Form

The Cartesian equations of a line are $\frac{x-3}{2} = \frac{y+1}{4} = \frac{z-2}{-5}$. Give the vector equation of the line.

Solution:

From the Cartesian form $\frac{x-a_1}{b_1} = \frac{y-a_2}{b_2} = \frac{z-a_3}{b_3}$, we can identify the components directly.

The direction vector is $\mathbf{b} = \begin{pmatrix} 2 \\ 4 \\ -5 \end{pmatrix}$ (the denominators).

The position vector is $\mathbf{a} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$ (obtained from $x-3$, $y-(-1)$, and $z-2$).

Therefore, using $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$:

$\mathbf{r} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 4 \\ -5 \end{pmatrix}$

Worked Example 3: Lines in Coordinate Planes

Give the Cartesian equations of the line with vector equation $\mathbf{r} = \mathbf{i} + 2\mathbf{j} - \mathbf{k} + s(3\mathbf{i} - \mathbf{k})$.

Solution:

First, write the equation in column vector form:

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

The direction vector is $\begin{pmatrix} 3 \\ 0 \\ -1 \end{pmatrix}$ and the position vector is $\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$.

Notice that the $\mathbf{j}$-component of the direction vector is zero. This means the line lies entirely on the plane $y = 2$.

The Cartesian equations are:

$\frac{x-1}{3} = \frac{z+1}{-1}, \quad y = 2$

Since we cannot write $\frac{y-2}{0}$ (division by zero is undefined), we simply state that $y = 2$ separately.

Worked Example 4: Testing if a Point Lies on a Line

Establish whether or not the point $(-5, 5, 6)$ lies on the line $\mathbf{r} = \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -2 \\ -3 \end{pmatrix}$.

Solution:

Substitute the position vector of the point into the equation in place of $\mathbf{r}$:

$\begin{pmatrix} -5 \\ 5 \\ 6 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -2 \\ -3 \end{pmatrix}$

Now examine each component separately to find the value of $\lambda$:

i-component: $-5 = 3 + 4\lambda \Rightarrow \lambda = -2$

j-component: $5 = 1 - 2\lambda \Rightarrow \lambda = -2$

k-component: $6 = 0 - 3\lambda \Rightarrow \lambda = -2$

Since you get the same value of $\lambda$ for each component, the point $(-5, 5, 6)$ does lie on the line.

Worked Example 5: Identifying the Same Line

Decide which of these equations represent the same line:

$L_1: \mathbf{r} = \begin{pmatrix} 9 \\ 1 \\ -2 \end{pmatrix} + \lambda \begin{pmatrix} -2 \\ -1 \\ 1 \end{pmatrix}$

$L_2: \mathbf{r} = \begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix}$

$L_3: \frac{x-1}{2} = \frac{y+3}{1} = \frac{z-2}{-1}$

Solution:

Step 1: Check the direction vectors.

For $L_1$, the direction vector is $\begin{pmatrix} -2 \\ -1 \\ 1 \end{pmatrix}$.

For $L_2$, the direction vector is $\begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix}$.

For $L_3$, the direction vector is $\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$.

Notice that $\begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix} = 2 \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} -2 \\ -1 \\ 1 \end{pmatrix} = -1 \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$.

All three lines have the same direction since their direction vectors are scalar multiples of each other. This means the lines are parallel (or potentially the same line).

Step 2: Check if they share a common point.

Take the point $(1, -3, 2)$ which lies on line $L_3$ (when the numerators are all zero).

Check if this point lies on $L_1$:

$\begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} = \begin{pmatrix} 9 \\ 1 \\ -2 \end{pmatrix} + \lambda \begin{pmatrix} -2 \\ -1 \\ 1 \end{pmatrix}$

• i-component: $1 = 9 - 2\lambda \Rightarrow \lambda = 4$
• j-component: $-3 = 1 - \lambda \Rightarrow \lambda = 4$
• k-component: $2 = -2 + \lambda \Rightarrow \lambda = 4$

Therefore, $(1, -3, 2)$ lies on line $L_1$ (with $\lambda = 4$).

Step 3: Since $L_1$ and $L_3$ have the same direction and share a common point, they represent the same line.

Now check if this point lies on $L_2$:

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

• i-component: $1 = 3 + 4\lambda \Rightarrow \lambda = -\frac{1}{2}$
• j-component: $-3 = 0 + 2\lambda \Rightarrow \lambda = -\frac{3}{2}$
• k-component: $2 = 4 - 2\lambda \Rightarrow \lambda = 1$

The values of $\lambda$ are inconsistent, so $(1, -3, 2)$ does not lie on $L_2$.

Conclusion: $L_2$ is parallel to the other lines but does not have a point in common, so it does not represent the same line as $L_1$ and $L_3$.

Worked Example 6: Determining Line Relationships

For each pair of lines, decide whether they are parallel, intersecting, or skew. If they intersect, find their point of intersection.

Part a:

$L_1: \mathbf{r} = 2\mathbf{i} - \mathbf{j} + \lambda(3\mathbf{i} + \mathbf{j} - 2\mathbf{k})$

$L_2: 3\mathbf{i} + 2\mathbf{j} - \mathbf{k} + \mu(2\mathbf{i} - \mathbf{k})$

Solution:

Write both equations in column vector form for clarity:

$L_1: \mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$

$L_2: \mathbf{r} = \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}$

Step 1: Check if the direction vectors are multiples of each other.

$\begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}$ are not multiples of each other, so the lines are not parallel.

Step 2: Attempt to solve simultaneously. Equate the expressions for $\mathbf{r}$:

$\begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}$

This gives three equations (one for each component):

• i: $2 + 3\lambda = 3 + 2\mu \Rightarrow 3\lambda - 2\mu = 1$
• j: $-1 + \lambda = 2 \Rightarrow \lambda = 3$
• k: $-2\lambda = -1 - \mu \Rightarrow -2\lambda + \mu = -1$

From the j-component, $\lambda = 3$.

Substitute $\lambda = 3$ into the i-component equation: $3(3) - 2\mu = 1 \Rightarrow \mu = \frac{9-1}{2} = 4$

Now check with the k-component: $-2(3) + \mu = -6 + \mu$

For this to equal $-1$, we need $\mu = 5$.

But from the i-component, we found $\mu = 4$. This is an inconsistency.

Step 3: Since the equations are inconsistent ($\mu$ cannot be both 4 and 5), there is no point of intersection. Therefore, the lines are skew.

Part b:

$L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ -8 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -2 \\ -3 \end{pmatrix}$

$L_2: \mathbf{r} = \begin{pmatrix} -4 \\ 7 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -1 \\ -5 \end{pmatrix}$

Solution:

Step 1: The direction vectors $\begin{pmatrix} 2 \\ -2 \\ -3 \end{pmatrix}$ and $\begin{pmatrix} 3 \\ -1 \\ -5 \end{pmatrix}$ are not multiples of each other, so the lines are not parallel.

Step 2: Equate the expressions:

$\begin{pmatrix} 1 \\ 0 \\ -8 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -2 \\ -3 \end{pmatrix} = \begin{pmatrix} -4 \\ 7 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -1 \\ -5 \end{pmatrix}$

This gives:

• i: $1 + 2\lambda = -4 + 3\mu$
• j: $-2\lambda = 7 - \mu$
• k: $-8 - 3\lambda = -1 - 5\mu$

From the j-component: $\mu = 7 + 2\lambda$

Adding the i and j component equations: $1 = -4 + 3\mu + 7 - \mu \Rightarrow 1 = 3 + 2\mu \Rightarrow \mu = -1$

Therefore $\lambda = \frac{\mu - 7}{2} = \frac{-1-7}{2} = -4$

Check with the k-component: $-8 - 3(-4) = -8 + 12 = 4$ $-1 - 5(-1) = -1 + 5 = 4$ ✓

The equations are consistent with $\lambda = -4$ and $\mu = -1$.

Step 3: The lines intersect. To find the point of intersection, substitute $\mu = -1$ into $L_2$ (or $\lambda = -4$ into $L_1$):

$\begin{pmatrix} -4 \\ 7 \\ -1 \end{pmatrix} + (-1) \begin{pmatrix} 3 \\ -1 \\ -5 \end{pmatrix} = \begin{pmatrix} -7 \\ 8 \\ 4 \end{pmatrix}$

Therefore, the point of intersection is $(-7, 8, 4)$.

Worked Example 7: Transforming a Line

The line with equation $\mathbf{r} = \mathbf{i} - \mathbf{j} + \lambda(\mathbf{i} + 3\mathbf{j})$ is transformed by the matrix $\mathbf{T} = \begin{pmatrix} 1 & -2 \\ 2 & 0 \end{pmatrix}$. Work out the equation of the image of the line.

Solution:

Step 1: Express a general point on the line as a single column vector.

From the line equation, the x-component is $1 + \lambda$ and the y-component is $-1 + 3\lambda$:

$\begin{pmatrix} 1 + \lambda \\ -1 + 3\lambda \end{pmatrix}$

Step 2: Pre-multiply by the transformation matrix:

$\begin{pmatrix} 1 & -2 \\ 2 & 0 \end{pmatrix} \begin{pmatrix} 1 + \lambda \\ -1 + 3\lambda \end{pmatrix} = \begin{pmatrix} (1+\lambda) - 2(-1+3\lambda) \\ 2(1+\lambda) + 0 \end{pmatrix}$

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

Step 3: Write in the form $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$:

$\mathbf{r} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -5 \\ 2 \end{pmatrix}$

Alternatively, you can write this as $\mathbf{r} = 3\mathbf{i} + 2\mathbf{j} + \lambda(-5\mathbf{i} + 2\mathbf{j})$.