Vectors 2 (AQA A-Level Further Maths): Revision Notes

Worked Example: Intersection with parametric plane

Question: Find the point of intersection between the line with equation $\mathbf{r} = -4\mathbf{j} + 2\mathbf{k} + \lambda(3\mathbf{i} + \mathbf{j} + \mathbf{k})$ and the plane with equation $\mathbf{r} = \mathbf{i} - 2\mathbf{j} + s(\mathbf{j} + 3\mathbf{k}) + t(2\mathbf{i} + 2\mathbf{j} + \mathbf{k})$.

Solution:

At the point of intersection, we can equate the two expressions:

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

Now consider each component separately:

i-component: $3\lambda = 1 + 2t$

j-component: $-4 + \lambda = -2 + s + 2t$, which simplifies to $\lambda = 2 + s + 2t$

Subtracting the first equation from the second gives:

$2\lambda = -1 - s$

This can be rearranged to: $s = -1 - 2\lambda$

k-component: $2 + \lambda = 3s + t$

Substitute $s = -1 - 2\lambda$:

$2 + \lambda = 3(-1 - 2\lambda) + t$

$t = 5 + 7\lambda$

Now solve together with $3\lambda = 1 + 2t$:

Substituting $t = 5 + 7\lambda$:

$3\lambda = 1 + 2(5 + 7\lambda)$

$3\lambda = 1 + 10 + 14\lambda$

$-11\lambda = 11$

$\lambda = -1$

Therefore $t = 5 + 7(-1) = -2$ and $s = -1 - 2(-1) = 1$.

Substitute $\lambda = -1$ into the equation of the line:

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

$\mathbf{r} = -3\mathbf{i} - 5\mathbf{j} + \mathbf{k}$

Answer: The point of intersection is (-3, -5, 1).

Worked Example: Intersection with Cartesian plane

Question: Find the point of intersection between the line with equation $\mathbf{r} = \mathbf{i} - \mathbf{k} + t(\mathbf{i} - 2\mathbf{j} + 3\mathbf{k})$ and the plane with equation $2x - y - 3z = 20$.

Solution:

From the line equation, we can write the coordinates in terms of $t$:

• $x = 1 + t$
• $y = -2t$
• $z = -1 + 3t$

Substitute these into the plane equation:

$2(1 + t) - (-2t) - 3(-1 + 3t) = 20$

$2 + 2t + 2t + 3 - 9t = 20$

$5 - 5t = 20$

$-5t = 15$

$t = -3$

Now substitute $t = -3$ back into the line equation:

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

$\mathbf{r} = \mathbf{i} - \mathbf{k} - 3\mathbf{i} + 6\mathbf{j} - 9\mathbf{k}$

$\mathbf{r} = -2\mathbf{i} + 6\mathbf{j} - 10\mathbf{k}$

Answer: The point of intersection is (-2, 6, -10).

Worked Example: Distance from point to plane

Question: Find the shortest distance from the plane $\Pi$ with equation $\mathbf{r} \cdot \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix} = -4$ to the point $P(4, 0, -7)$.

Solution:

The normal vector to the plane is $\mathbf{n} = \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix}$.

The equation of the line through $P$ perpendicular to the plane is:

$\mathbf{r} = \begin{pmatrix} 4 \\ 0 \\ -7 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix}$

At point $Q$ where the line meets the plane:

$\begin{pmatrix} 4 + \lambda \\ 3\lambda \\ -7 - 2\lambda \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix} = -4$

Expanding the dot product:

$(4 + \lambda)(1) + (3\lambda)(3) + (-7 - 2\lambda)(-2) = -4$

$4 + \lambda + 9\lambda + 14 + 4\lambda = -4$

$18 + 14\lambda = -4$

$14\lambda = -22$

Wait, let me recalculate this more carefully. We have:

$4 + \lambda + 2(3\lambda) - 2(-7 - 2\lambda) = -4$

$4 + \lambda + 6\lambda + 14 + 4\lambda = -4$

$11\lambda = -22$

$\lambda = -2$

The position vector of $Q$ is:

$\overrightarrow{OQ} = \begin{pmatrix} 4 \\ 0 \\ -7 \end{pmatrix} - 2\begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 4 - 2 \\ 0 - 6 \\ -7 + 4 \end{pmatrix} = \begin{pmatrix} 2 \\ -6 \\ -3 \end{pmatrix}$

The vector $\overrightarrow{PQ}$ is:

$\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \begin{pmatrix} 2 \\ -6 \\ -3 \end{pmatrix} - \begin{pmatrix} 4 \\ 0 \\ -7 \end{pmatrix} = \begin{pmatrix} -2 \\ -6 \\ 4 \end{pmatrix}$

This is just $\lambda\mathbf{n}$ where $\lambda = -2$.

The magnitude is:

$|\overrightarrow{PQ}| = \sqrt{(-2)^2 + (-6)^2 + 4^2} = \sqrt{4 + 36 + 16} = \sqrt{56} = 2\sqrt{14}$

Answer: The shortest distance from point P to plane Π is $2\sqrt{14}$ units.

Key observation: Notice that the distance was given by $|\lambda\mathbf{n}|$ where $\lambda$ is the parameter value at the point of intersection.

Worked Example: Distance between skew lines

Question: Find the shortest distance between the lines $l_1$ and $l_2$ with equations $l_1: \mathbf{r} = 2\mathbf{i} - \mathbf{j} + s(\mathbf{i} + \mathbf{j} - 3\mathbf{k})$ and $l_2: \mathbf{r} = 4\mathbf{i} + \mathbf{j} - \mathbf{k} + t(2\mathbf{i} - \mathbf{k})$.

Solution:

Step 1: Find the normal vector using the vector product:

$\mathbf{n} = (\mathbf{i} + \mathbf{j} - 3\mathbf{k}) \times (2\mathbf{i} - \mathbf{k})$

$\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -3 \\ 2 & 0 & -1 \end{vmatrix}$

$\mathbf{n} = \mathbf{i}(1 \times (-1) - (-3) \times 0) - \mathbf{j}(1 \times (-1) - (-3) \times 2) + \mathbf{k}(1 \times 0 - 1 \times 2)$

$\mathbf{n} = \mathbf{i}(-1) - \mathbf{j}(-1 + 6) + \mathbf{k}(-2)$

$\mathbf{n} = -\mathbf{i} - 5\mathbf{j} - 2\mathbf{k}$

Step 2: Find the equation of the plane containing $l_1$:

$\Pi_1: \mathbf{r} \cdot (-\mathbf{i} - 5\mathbf{j} - 2\mathbf{k}) = (2\mathbf{i} - \mathbf{j}) \cdot (-\mathbf{i} - 5\mathbf{j} - 2\mathbf{k})$

$\mathbf{r} \cdot (-\mathbf{i} - 5\mathbf{j} - 2\mathbf{k}) = -2 + 5 = 3$

The distance of $\Pi_1$ from the origin is:

$\frac{|3|}{\sqrt{1^2 + 5^2 + 2^2}} = \frac{3}{\sqrt{30}} = \frac{3\sqrt{30}}{30} = \frac{\sqrt{30}}{10}$

Step 3: Find the equation of the plane containing $l_2$:

$\Pi_2: \mathbf{r} \cdot (-\mathbf{i} - 5\mathbf{j} - 2\mathbf{k}) = (4\mathbf{i} + \mathbf{j} - \mathbf{k}) \cdot (-\mathbf{i} - 5\mathbf{j} - 2\mathbf{k})$

$\mathbf{r} \cdot (-\mathbf{i} - 5\mathbf{j} - 2\mathbf{k}) = -4 - 5 + 2 = -7$

The distance of $\Pi_2$ from the origin is:

$\frac{|-7|}{\sqrt{30}} = \frac{7\sqrt{30}}{30}$

Step 4: Since the planes are on different sides of the origin (one has $p = 3$, the other $p = -7$), we add the distances:

$\text{Distance between planes} = \frac{\sqrt{30}}{10} + \frac{7\sqrt{30}}{30} = \frac{3\sqrt{30}}{30} + \frac{7\sqrt{30}}{30} = \frac{10\sqrt{30}}{30} = \frac{\sqrt{30}}{3}$

Answer: The shortest distance between the lines is $\frac{\sqrt{30}}{3}$ units.

Worked Example: Reflection of a point

Question: Find the image of the point $A(1, 0, 2)$ in the plane $\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} + \mu\begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$.

Solution:

Step 1: Find the normal vector using the vector product:

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

$\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 3 & 2 \\ 0 & 2 & 1 \end{vmatrix}$

$\mathbf{n} = \mathbf{i}(3 \times 1 - 2 \times 2) - \mathbf{j}((-1) \times 1 - 2 \times 0) + \mathbf{k}((-1) \times 2 - 3 \times 0)$

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

$\mathbf{n} = -\mathbf{i} + \mathbf{j} - 2\mathbf{k}$

Step 2: The equation of the line through $A$ perpendicular to the plane is:

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

Step 3: At the intersection point, this line equation equals the plane equation:

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

Comparing components:

i-component: $1 - t = 1 - \mu$ gives $t = \mu$ ... (1)

j-component: $t = 3\mu + 2\lambda$ ... (2)

k-component: $2 - 2t = -2 + 2\mu + \lambda$ ... (3)

From equation (1), $t = \mu$. Substituting into equation (2):

$t = 3t + 2\lambda$

$-2t = 2\lambda$

$\lambda = -t$

Substituting $t = \mu$ and $\lambda = -t$ into equation (3):

$2 - 2t = -2 + 2t + (-t)$

$2 - 2t = -2 + t$

$4 = 3t$

$t = \frac{4}{3}$

Step 4: The image $A'$ is at parameter $2t = \frac{8}{3}$:

$\overrightarrow{OA'} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \frac{8}{3}\begin{pmatrix} -1 \\ 1 \\ -2 \end{pmatrix}$

$\overrightarrow{OA'} = \begin{pmatrix} 1 - \frac{8}{3} \\ 0 + \frac{8}{3} \\ 2 - \frac{16}{3} \end{pmatrix} = \begin{pmatrix} -\frac{5}{3} \\ \frac{8}{3} \\ -\frac{10}{3} \end{pmatrix}$

Answer: The coordinates of the image $A'$ are $\left(-\frac{5}{3}, \frac{8}{3}, -\frac{10}{3}\right)$.

Key insight: Remember to double the parameter value $t$ to find the reflection. The intersection point is the midpoint between the original point and its image.