Vectors for Intermediate Students

A vector is a quantity with both size and direction, like a velocity of 20 m/s due north or a force of 50 N pulling up a slope. Vectors let you describe space with algebra, which is why they appear in intermediate mathematics and run straight through physics and engineering. This guide explains the key ideas, from components to the scalar triple product, and works through the calculations you will use again and again.

Scalars and Vectors

A scalar has only magnitude: mass, time, temperature, energy. A vector has magnitude and direction: displacement, velocity, acceleration, force. In print, vectors are shown in bold (\( \mathbf{a} \)) or with an arrow (\( \vec{a} \)). In your notebook, underline them or draw an arrow over them, because losing track of which quantities are vectors is the most common source of errors.

Components in Two and Three Dimensions

Using the unit vectors \( \mathbf{i} \), \( \mathbf{j} \) and \( \mathbf{k} \) along the \( x \), \( y \) and \( z \) axes, any vector can be written as

$$ \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} $$

Its magnitude is

$$ |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} $$

and the unit vector in its direction is \( \hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|} \).

The vector from point \( A \) to point \( B \) is \( \overrightarrow{AB} = \text{(position vector of } B) - \text{(position vector of } A) \). Remember: "end minus start".

Direction Cosines

If \( \mathbf{a} \) makes angles \( \alpha, \beta, \gamma \) with the axes, then

$$ \cos\alpha = \frac{a_1}{|\mathbf{a}|}, \quad \cos\beta = \frac{a_2}{|\mathbf{a}|}, \quad \cos\gamma = \frac{a_3}{|\mathbf{a}|} $$

and \( \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \).

Basic Operations

• Addition: add matching components. Geometrically, place the vectors head to tail.
• Scalar multiplication: \( k\mathbf{a} \) scales every component by \( k \). Negative \( k \) reverses direction.
• Parallel vectors: \( \mathbf{a} \) and \( \mathbf{b} \) are parallel if \( \mathbf{a} = k\mathbf{b} \) for some scalar \( k \).

The Dot (Scalar) Product

The dot product of two vectors is a scalar:

$$ \mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta = a_1b_1 + a_2b_2 + a_3b_3 $$

where \( \theta \) is the angle between them. Key facts:

• If \( \mathbf{a}\cdot\mathbf{b} = 0 \) (and neither is zero), the vectors are perpendicular.
• \( \mathbf{i}\cdot\mathbf{i} = \mathbf{j}\cdot\mathbf{j} = \mathbf{k}\cdot\mathbf{k} = 1 \), and \( \mathbf{i}\cdot\mathbf{j} = 0 \).
• The projection of \( \mathbf{a} \) on \( \mathbf{b} \) is \( \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|} \).
• Work done by a constant force \( \mathbf{F} \) through displacement \( \mathbf{d} \) is \( W = \mathbf{F}\cdot\mathbf{d} \).

The Cross (Vector) Product

The cross product is a vector perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \):

$$ \mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} $$

Its magnitude is \( |\mathbf{a}||\mathbf{b}|\sin\theta \), which equals the area of the parallelogram with sides \( \mathbf{a} \) and \( \mathbf{b} \). Half of this is the area of the triangle. The cross product is not commutative: \( \mathbf{b}\times\mathbf{a} = -(\mathbf{a}\times\mathbf{b}) \). In physics, torque is \( \boldsymbol{\tau} = \mathbf{r}\times\mathbf{F} \).

Worked Example 1: Dot Product and the Angle Between Vectors

Let \( \mathbf{a} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k} \) and \( \mathbf{b} = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k} \). Find the angle between them.

Step 1. Dot product:

$$ \mathbf{a}\cdot\mathbf{b} = 2(1) + (-1)(2) + 3(-2) = 2 - 2 - 6 = -6 $$

Step 2. Magnitudes:

$$ |\mathbf{a}| = \sqrt{4 + 1 + 9} = \sqrt{14}, \qquad |\mathbf{b}| = \sqrt{1 + 4 + 4} = 3 $$

Step 3. Cosine of the angle:

$$ \cos\theta = \frac{-6}{3\sqrt{14}} = \frac{-2}{\sqrt{14}} \approx -0.5345 $$

Step 4. \( \theta = \cos^{-1}(-0.5345) \approx 122.3^\circ \).

Sanity check. The dot product is negative, so the angle must be obtuse (between \( 90^\circ \) and \( 180^\circ \)). It is.

Worked Example 2: Cross Product and Area

Using the same vectors, find \( \mathbf{a}\times\mathbf{b} \) and the area of the triangle with sides \( \mathbf{a} \) and \( \mathbf{b} \).

Step 1. Set up the determinant:

$$ \mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ 1 & 2 & -2 \end{vmatrix} $$

Step 2. Expand:

• \( \mathbf{i} \): \( (-1)(-2) - (3)(2) = 2 - 6 = -4 \)
• \( \mathbf{j} \): \( -\big[(2)(-2) - (3)(1)\big] = -(-4 - 3) = 7 \)
• \( \mathbf{k} \): \( (2)(2) - (-1)(1) = 4 + 1 = 5 \)

So \( \mathbf{a}\times\mathbf{b} = -4\mathbf{i} + 7\mathbf{j} + 5\mathbf{k} \).

Step 3. Check perpendicularity. With \( \mathbf{a} \): \( -8 - 7 + 15 = 0 \). With \( \mathbf{b} \): \( -4 + 14 - 10 = 0 \). Both zero, so the result is perpendicular to both vectors.

Step 4. Area of parallelogram:

$$ |\mathbf{a}\times\mathbf{b}| = \sqrt{16 + 49 + 25} = \sqrt{90} = 3\sqrt{10} \approx 9.49 $$

The triangle's area is half of this: \( \tfrac{3\sqrt{10}}{2} \approx 4.74 \) square units.

Consistency check. \( \sin\theta \) for \( \theta \approx 122.3^\circ \) is about 0.845, and \( |\mathbf{a}||\mathbf{b}|\sin\theta \approx 3.742 \times 3 \times 0.845 \approx 9.49 \). It matches.

Worked Example 3: Work Done by a Force

A force \( \mathbf{F} = 3\mathbf{i} + 2\mathbf{j} - \mathbf{k} \) newtons moves a particle from \( A(1, 1, 1) \) to \( B(4, 3, 2) \), with distances in metres. Find the work done.

Step 1. Displacement: \( \mathbf{d} = \overrightarrow{AB} = (4 - 1)\mathbf{i} + (3 - 1)\mathbf{j} + (2 - 1)\mathbf{k} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} \) m.

Step 2. Work:

$$ W = \mathbf{F}\cdot\mathbf{d} = 3(3) + 2(2) + (-1)(1) = 9 + 4 - 1 = 12 \text{ J} $$

The work done is 12 joules.

The Scalar Triple Product

The scalar triple product \( \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) \) equals the determinant of the three vectors' components. Its absolute value is the volume of the parallelepiped formed by the three vectors. If it equals zero, the three vectors are coplanar. For example, with \( \mathbf{c} = \mathbf{k} \) and the vectors above, \( \mathbf{c}\cdot(\mathbf{a}\times\mathbf{b}) = 5 \), so the parallelepiped on \( \mathbf{a} \), \( \mathbf{b} \) and \( \mathbf{k} \) has volume 5 cubic units.

Did You Know?

• The \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) notation comes from William Rowan Hamilton's quaternions, which he famously carved into a bridge in Dublin in 1843.
• Modern vector analysis was shaped independently by Josiah Willard Gibbs and Oliver Heaviside in the 1880s, largely to make electromagnetism easier to write down.
• GPS receivers and flight simulators solve vector equations continuously to work out position and direction.

Key Takeaways

• A vector from \( A \) to \( B \) is "end minus start".
• The dot product gives a scalar and finds angles, projections and work; zero means perpendicular.
• The cross product gives a perpendicular vector whose magnitude is a parallelogram's area.
• The scalar triple product gives volume and tests whether three vectors are coplanar.
• Always check a cross product by dotting it with both original vectors.

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