Addition of Vectors and Multiplication by a Scalar

Adding Vectors: Two Equivalent Laws

Triangle law

A vector is a displacement. If a girl walks from $A$ to $B$ and then from $B$ to $C$, her net displacement is from $A$ straight to $C$:

$$\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$$

This is the triangle law of vector addition: to add $\vec{a}$ and $\vec{b}$, slide $\vec{b}$ (a free vector!) so its initial point sits on the terminal point of $\vec{a}$; the sum runs from the tail of $\vec{a}$ to the tip of $\vec{b}$ — the third side of the triangle, taken from start to finish.

Two immediate consequences:

1. Sides of a triangle taken in order sum to zero: $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{AC} + \overrightarrow{CA} = \vec{0}$ — the walk returns to its starting point.
2. Subtraction is addition of the negative: $\vec{a} - \vec{b} = \vec{a} + \left(-\vec{b}\right)$, built by reversing $\vec{b}$ and then applying the triangle law.

Parallelogram law

If $\vec{a}$ and $\vec{b}$ are drawn from a common initial point as two adjacent sides of a parallelogram, their sum $\vec{a} + \vec{b}$ is the diagonal through that common point — think of a boat crossing a river: engine velocity one side, stream velocity the other, actual motion along the diagonal.

Key Point: the two laws are the same fact drawn differently. In parallelogram $ABCD$: $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AB} + \overrightarrow{AD}$ (since $\overrightarrow{BC} = \overrightarrow{AD}$) — triangle law on one half of the parallelogram is the parallelogram law.

Properties of vector addition

1. Commutative: $\vec{a} + \vec{b} = \vec{b} + \vec{a}$ (walk the parallelogram's two triangle-halves).
2. Associative: $\left(\vec{a} + \vec{b}\right) + \vec{c} = \vec{a} + \left(\vec{b} + \vec{c}\right)$ — so $\vec{a} + \vec{b} + \vec{c}$ needs no brackets.
3. Additive identity: $\vec{a} + \vec{0} = \vec{a}$ — the zero vector changes nothing.
4. Additive inverse: $\vec{a} + \left(-\vec{a}\right) = \vec{0}$.

Multiplying a Vector by a Scalar

For a vector $\vec{a}$ and a scalar $\lambda$, the product $\lambda\vec{a}$ is a vector collinear with $\vec{a}$:

• Direction: same as $\vec{a}$ when $\lambda > 0$; opposite when $\lambda < 0$.
• Magnitude: $\left|\lambda\vec{a}\right| = |\lambda|\,|\vec{a}|$ — note the absolute value of $\lambda$: $\left|-2\vec{a}\right| = 2|\vec{a}|$, never $-2|\vec{a}|$.

Special values of $\lambda$ tell the whole story:

1. $\lambda = -1$ gives $-\vec{a}$, the additive inverse: same length, opposite direction.
2. $\lambda = \dfrac{1}{|\vec{a}|}$ (for $\vec{a} \neq \vec{0}$) gives a vector of magnitude $1$ in the direction of $\vec{a}$ — the unit vector

$$\hat{a} = \frac{\vec{a}}{|\vec{a}|}$$

3. For any scalar $k$, $k\vec{0} = \vec{0}$.

Key Point (the workhorse recipe): to produce a vector of magnitude $M$ in the direction of $\vec{a}$, compute $M\hat{a} = \dfrac{M}{|\vec{a}|}\vec{a}$. Normalise first, then scale — this two-step is behind half the 2-mark questions in this chapter.

How big can a sum be?

From the triangle picture, the resultant's length is bounded by the triangle inequality:

$$\bigl|\,|\vec{a}| - |\vec{b}|\,\bigr| \;\leq\; \left|\vec{a} + \vec{b}\right| \;\leq\; |\vec{a}| + |\vec{b}|$$

Equality on the right holds when $\vec{a}$ and $\vec{b}$ point the same way; on the left when they point opposite ways. With $|\vec{a}| = 3$ and $|\vec{b}| = 5$, the sum's magnitude can be anything in $[2, 8]$ — and nothing outside it.

Solved Examples

Example 1: Net displacement

A girl walks $4$ km due west and then $3$ km due north. Find the magnitude of her net displacement.

Solution:

1. Set up the triangle: the two legs are perpendicular, with the net displacement as the third side (tail of the first to tip of the second).
2. Pythagoras: $\left|\text{net}\right| = \sqrt{4^2 + 3^2} = \sqrt{25} = 5$ km.

Answer: $5$ km, pointing north of west. Note that the distance walked is $7$ km — displacement (vector) and distance (scalar) part ways exactly here.

Example 2: Magnitude under scaling

If $|\vec{a}| = 3$, find $\left|-2\vec{a}\right|$ and describe the direction of $-2\vec{a}$.

Solution:

1. Magnitude rule: $\left|-2\vec{a}\right| = |-2| \cdot |\vec{a}| = 2 \times 3 = 6$.
2. Direction rule: $\lambda = -2 < 0$, so the direction is opposite to $\vec{a}$.

Answer: magnitude $6$, direction opposite to $\vec{a}$. A magnitude is never negative — $|\lambda\vec{a}| = |\lambda||\vec{a}|$ has the absolute value built in for exactly this reason.

Example 3: Building a vector of prescribed length

Given $|\vec{a}| = 4$, write a vector of magnitude $10$ in the direction of $\vec{a}$, in terms of $\vec{a}$.

Solution:

1. Normalise: $\hat{a} = \dfrac{\vec{a}}{4}$.
2. Scale: the required vector is $10\hat{a} = \dfrac{10}{4}\vec{a} = \dfrac{5}{2}\vec{a}$.

Answer: $\dfrac{5}{2}\vec{a}$. Check: $\left|\dfrac{5}{2}\vec{a}\right| = \dfrac{5}{2} \times 4 = 10$. ✓

Example 4: A closed walk

Prove that $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec{0}$ for any triangle $ABC$, and state the generalisation for a polygon.

Solution:

1. Triangle law twice: $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$.
2. Add the last side: $\overrightarrow{AC} + \overrightarrow{CA} = \overrightarrow{AA} = \vec{0}$.

Answer: the sum is $\vec{0}$ — and the same telescoping shows the directed sides of any closed polygon, taken in order, sum to the zero vector: every closed walk ends where it began.

Example 5: Diagonals of a parallelogram

In parallelogram $ABCD$, let $\overrightarrow{AB} = \vec{a}$ and $\overrightarrow{AD} = \vec{b}$. Express both diagonals in terms of $\vec{a}$ and $\vec{b}$.

Solution:

1. Diagonal from $A$: $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \vec{a} + \vec{b}$ (since $\overrightarrow{BC} = \overrightarrow{AD} = \vec{b}$).
2. Diagonal from $B$: $\overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD} = -\vec{a} + \vec{b} = \vec{b} - \vec{a}$.

Answer: $\overrightarrow{AC} = \vec{a} + \vec{b}$, $\overrightarrow{BD} = \vec{b} - \vec{a}$ — the two diagonals are the sum and the difference. This picture is worth memorising: it turns many geometry proofs into two-line vector computations.

Example 6: The boat and the stream

A boat's engine drives it at $8$ km/h straight across a river whose current flows at $6$ km/h parallel to the banks. Find the boat's actual speed.

Solution:

1. Parallelogram (rectangle) law: the two velocity vectors are perpendicular adjacent sides; the actual velocity is the diagonal.
2. Magnitude: $\sqrt{8^2 + 6^2} = \sqrt{100} = 10$ km/h.
3. Direction: downstream of straight-across by the angle $\theta$ with $\tan\theta = \dfrac{6}{8} = \dfrac{3}{4}$.

Answer: $10$ km/h, tilted downstream at $\tan^{-1}\dfrac{3}{4}$ from the crossing direction — the classic physical picture behind the parallelogram law.