Unit P4: Vectors

1. Vector Foundations & Operations

Vectors are represented in 2D and 3D using column notation or unit vector notation ($i$, $j$, $k$).

Position Vectors (7.4)

The vector joining point A to point B is given by:

$$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\mathbf{b}-\mathbf{a}$$

Magnitude & Unit Vectors (7.2)

The magnitude of a vector $\mathbf{a}$ is its length. A unit vector in the direction of $\mathbf{a}$ has a magnitude of 1.

$$|\mathbf{a}|=\sqrt{x^2+y^2+z^2}$$

2. Distance Between Two Points (7.5)

To find the distance $d$ between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$:

$$d^2=(x_1-x_2{)}^2+(y_1-y_2{)}^2+(z_1-z_2{)}^2$$

3. Vector Equations of Lines (7.6)

A line passing through point $\mathbf{a}$ with direction vector $\mathbf{b}$ is given by:

$$\mathbf{r}=\mathbf{a}+t\mathbf{b}$$

Or through two points $\mathbf{c}$ and $\mathbf{d}$:

$$\mathbf{r}=\mathbf{c}+t(\mathbf{d}-\mathbf{c})$$

Exam Tip

You must be able to determine if two lines are parallel, intersecting, or skew (neither parallel nor intersecting in 3D).

4. The Scalar Product (7.7)

The scalar (dot) product is used to find the angle between two vectors.

$$\mathbf{a}\cdot \mathbf{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3$$

To find the angle $\theta$:

$$\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$$

Perpendicular Condition: If $\mathbf{a}\cdot \mathbf{b}=0$ and the vectors are non-zero, then the vectors are perpendicular.

---

Problem 1: Angle between two lines

Question: Find the acute angle θ between the lines $L_1$ and $L_2$ defined by:

$L_1:\mathbf{r}=(\mathbf{i}+2\mathbf{j}-\mathbf{k})+\lambda (2\mathbf{i}+\mathbf{j}+2\mathbf{k})$
$L_2:\mathbf{r}=(3\mathbf{i}+\mathbf{k})+\mu (3\mathbf{i}-4\mathbf{j})$

Solution:

Step 1: Identify Direction Vectors
We only care about the vectors attached to the scalars $\lambda$ and $\mu$.

${\mathbf{d}}_1=\left(\begin{array}{c}2\\ 1\\ 2\end{array}\right),{\mathbf{d}}_2=\left(\begin{array}{c}3\\ -4\\ 0\end{array}\right)$

Step 2: Calculate Scalar Product and Magnitudes

${\mathbf{d}}_1\cdot {\mathbf{d}}_2=(2\times 3)+(1\times -4)+(2\times 0)=2$
$\left|{\mathbf{d}}_1\right|=\sqrt{2^2+1^2+2^2}=3$
$\left|{\mathbf{d}}_2\right|=\sqrt{3^2+{\left(-4\right)}^2+0^2}=5$

Step 3: Find θ

$\cos \theta =\frac{\left|2\right|}{3\times 5}=\frac{2}{15}$

Result: $\theta \approx 82.3°$

---

Problem 2: Proving Lines are Skew

The Logic: To prove lines are skew, you must show they are not parallel and that they do not intersect.

Part A: Check Parallelism
Check if the direction vectors are multiples of each other. If ${\mathbf{d}}_1\ne k{\mathbf{d}}_2$, they are not parallel.

Part B: Check Intersection
Set the components of $L_1$ equal to $L_2$:

1. Solve for $\lambda$ and $\mu$ using the $x$ and $y$ equations.
2. Substitute these values into the $z$ equation.
3. If the $z$ equation is not consistent (e.g., $5=12$), the lines do not intersect.

Conclusion: Not parallel + No intersection = Skew Lines.