C. Geometry and Algebra of Vectors · Vectors in Two-Space and Three-Space

Subtracting Vectors and Multiplying by a Scalar

C2.1About 70 minutes

Learning goals

I can subtract vectors geometrically, using 𝐮 − 𝐯 = 𝐮 + (−𝐯) or the tail-to-tail method.
I can multiply a vector by a scalar and describe the result's magnitude and direction.
I can use scalar multiples to recognize collinear vectors and points, and find unit vectors.
I can write vectors in a figure in terms of two given vectors.

Words to know

opposite vector
−𝐯: the same magnitude as 𝐯, opposite direction.

scalar multiple
The vector k𝐮, with magnitude |k||𝐮|, in the direction of 𝐮 if k > 0 and opposite if k < 0.

collinear vectors
Vectors that are parallel (same or opposite direction); one is a scalar multiple of the other.

unit vector
A vector of magnitude 1. The unit vector in the direction of 𝐮 is (1/|𝐮|)𝐮.

change in velocity
Δ𝐯 = 𝐯₂ − 𝐯₁, the vector that must be added to the first velocity to get the second.

Changes and multiples

A goalie's glove stops a puck moving at 30 m/s and sends it back at 5 m/s. How much did its velocity change? A drone doubles its speed without turning. How do we write its new velocity? Questions like these need two more operations: subtracting vectors and multiplying a vector by a scalar (a real number). Both are easy to draw, and together with addition they let you write almost any vector in terms of a few others.

Subtraction

The opposite of 𝐯 is −𝐯: same magnitude, opposite direction. Subtracting means adding the opposite:

𝐮 − 𝐯 = 𝐮 + (−𝐯)

There is a quicker picture. Put 𝐮 and 𝐯 tail to tail. Then 𝐮 − 𝐯 is the arrow from the head of 𝐯 to the head of 𝐮. Check: going along 𝐯 and then along 𝐮 − 𝐯 brings you to the head of 𝐮, so 𝐯 + (𝐮 − 𝐯) = 𝐮 ✓.

With position vectors this gives a rule you already know as "head minus tail":

𝐀𝐁 = 𝐎𝐁 − 𝐎𝐀

In the parallelogram formed by 𝐮 and 𝐯, the two diagonals are 𝐮 + 𝐯 and 𝐮 − 𝐯. If θ is the angle between 𝐮 and 𝐯 (tail to tail), the triangle for 𝐮 − 𝐯 contains θ itself, so

|𝐮 − 𝐯|² = |𝐮|² + |𝐯|² − 2|𝐮||𝐯| cos θ

Two ways to draw 𝐮 − 𝐯. Both give the same dashed arrow. Tail to tail, 𝐮 − 𝐯 points from the head of 𝐯 to the head of 𝐮.

Worked example

Example 1: Sum and difference

|𝐮| = 6, |𝐯| = 4 and the angle between them is 50°. Find |𝐮 − 𝐯| and |𝐮 + 𝐯|.

For 𝐮 − 𝐯 (tail to tail) the angle in the triangle is 50°: |𝐮 − 𝐯|² = 36 + 16 − 48 cos 50° ≈ 52 − 30.85 = 21.15, so |𝐮 − 𝐯| ≈ 4.60.
For 𝐮 + 𝐯 (tip to tail) the angle in the triangle is 130°: |𝐮 + 𝐯|² = 36 + 16 − 48 cos 130° ≈ 52 + 30.85 = 82.85, so |𝐮 + 𝐯| ≈ 9.10.
Reasonable: the vectors point roughly the same way, so the sum is long and the difference is short.

|𝐮 − 𝐯| ≈ 4.60 and |𝐮 + 𝐯| ≈ 9.10.

Worked example

Example 2: Change in velocity

A car rounds a corner, changing its velocity from 20 m/s [N] to 20 m/s [E]. Find the change in velocity, Δ𝐯 = 𝐯₂ − 𝐯₁.

Δ𝐯 = 𝐯₂ + (−𝐯₁) = 20 m/s [E] + 20 m/s [S].
These are perpendicular: |Δ𝐯| = √(20² + 20²) = 20√2 ≈ 28.3 m/s.
Equal east and south parts: the direction is exactly halfway between east and south, S 45° E (bearing 135°).

Δ𝐯 = 20√2 ≈ 28.3 m/s [S 45° E], even though the speed never changed.

Scalar multiplication

For a scalar k and a vector 𝐮, the vector k𝐮 has

magnitude |k𝐮| = |k| |𝐮|, and
the same direction as 𝐮 if k > 0, the opposite direction if k < 0.

Also 0𝐮 = 𝟎 and (−1)𝐮 = −𝐮. For example, if 𝐮 = 8 km/h [N 30° E], then 3𝐮 = 24 km/h [N 30° E] and −½𝐮 = 4 km/h [S 30° W].

Two non-zero vectors are collinear (parallel) exactly when one is a scalar multiple of the other: 𝐯 = k𝐮. This gives a test for points on a line: A, B and C are collinear if 𝐀𝐁 = k𝐀𝐂 for some scalar k.

Dividing a non-zero vector by its own magnitude gives a unit vector (magnitude 1) in the same direction: (1/|𝐮|)𝐮. Unit vectors are used to describe pure direction.

Scalar multiples of 𝐮 are all parallel to 𝐮. Multiplying by 2 doubles the length; by ½ halves it; a negative scalar also reverses the direction.

Worked example

Example 3: Describing scalar multiples

𝐮 is a force of 15 N [W]. Describe 4𝐮, −𝐮/3 and the unit vector in the direction of 𝐮.

4𝐮: magnitude 4 × 15 = 60 N, same direction: 60 N [W].
−𝐮/3 = (−1/3)𝐮: magnitude 15/3 = 5 N, opposite direction: 5 N [E].
Unit vector: (1/15)𝐮 has magnitude 1 and points west. (A unit vector has no unit of force; it is just a direction.)

60 N [W], 5 N [E], and a unit vector pointing west.

Worked example

Example 4: Writing vectors in terms of two others

In parallelogram ABCD, let 𝐚 = 𝐀𝐁 and 𝐛 = 𝐀𝐃. M is the midpoint of DC. Write 𝐀𝐂, 𝐁𝐃, 𝐀𝐌 and 𝐌𝐁 in terms of 𝐚 and 𝐛.

𝐀𝐂 = 𝐀𝐁 + 𝐁𝐂 = 𝐚 + 𝐛 (because 𝐁𝐂 = 𝐀𝐃 = 𝐛).
𝐁𝐃 = 𝐁𝐀 + 𝐀𝐃 = −𝐚 + 𝐛 = 𝐛 − 𝐚.
𝐀𝐌 = 𝐀𝐃 + 𝐃𝐌 = 𝐛 + ½𝐃𝐂 = 𝐛 + ½𝐚 (𝐃𝐂 = 𝐚, and M is halfway along it).
𝐌𝐁 = 𝐌𝐂 + 𝐂𝐁 = ½𝐚 + (−𝐛) = ½𝐚 − 𝐛.

𝐀𝐂 = 𝐚 + 𝐛, 𝐁𝐃 = 𝐛 − 𝐚, 𝐀𝐌 = 𝐛 + ½𝐚, 𝐌𝐁 = ½𝐚 − 𝐛.

Any path from one point to another gives the same vector. To find 𝐀𝐌, go up 𝐛 to D, then half of 𝐚 to M.

Find a path

To write a vector like 𝐌𝐁 in terms of given vectors, find any path from M to B along vectors you know, then add them, flipping the sign of any you travel backwards. Different paths always give the same answer, which is a great way to check.

Just for fun

Multiplying a vector by −1 is the math version of a U-turn: same speed, opposite direction, and no one gets a ticket.

Hands-on activity

Velocity change on a bike path (or a hallway)

You need: a sheet of paper, a ruler, a protractor, a pencil, and (optional) a phone stopwatch.

Walk at a steady pace along a hallway or path, then turn a corner of about 90° and keep walking at the same pace. Estimate your speed (count steps over 10 s and measure a step).
On paper, draw your first velocity 𝐯₁ and your second velocity 𝐯₂ tail to tail, to scale (for example 1 cm : 0.2 m/s).
Draw Δ𝐯 = 𝐯₂ − 𝐯₁ from the head of 𝐯₁ to the head of 𝐯₂. Measure its length and direction.
Check with a calculation: for a 90° turn at speed s, |Δ𝐯| = s√2.
Repeat on paper for a U-turn (180°). What is |Δ𝐯| now?

What to look for: for a 90° turn at about 1.3 m/s, |Δ𝐯| ≈ 1.8 m/s, pointing halfway between your new direction and the direction opposite your old one. For a U-turn, |Δ𝐯| = 2s, the biggest possible change at that speed.

Your turn

Simplify 𝐎𝐁 − 𝐎𝐀.

𝐮 and 𝐯 are perpendicular with |𝐮| = 5 and |𝐯| = 12. Find |𝐮 − 𝐯|.

|𝐮| = 8, |𝐯| = 5 and the angle between them is 60°. Find |𝐮 − 𝐯|.

𝐮 = 3 m/s [E]. Describe −2𝐮.

|𝐮| = 4 and k = −2.5. Find |k𝐮|.

Are 𝐮 and −3𝐮 collinear? Explain.

In parallelogram ABCD with 𝐚 = 𝐀𝐁 and 𝐛 = 𝐀𝐃, write 𝐂𝐀 in terms of 𝐚 and 𝐛.

More practice: warm-up

Simplify 𝐮 − 𝐮.

What is 0𝐮?

Simplify −(−𝐮).

|𝐮| = 2. Find |−3𝐮|.

Fill in the blank: 𝐮 − 𝐯 = 𝐮 + ___.

More practice: on your own

Simplify 𝐀𝐁 − 𝐀𝐂.

A ball moving at 15 m/s [E] bounces off a wall and moves at 12 m/s [W]. Find its change in velocity.

|𝐮| = 10, |𝐯| = 6 and |𝐮 − 𝐯| = 8. Find the angle between 𝐮 and 𝐯.

A car's velocity changes from 60 km/h [N] to 80 km/h [E]. Find Δ𝐯.

𝐀𝐂 = 3𝐀𝐁. What can you say about the points A, B and C? Find the ratio AB : BC.

Challenge

Prove that the diagonals of the parallelogram formed by 𝐮 and 𝐯 have equal length if and only if the parallelogram is a rectangle.

𝐮 and 𝐯 are non-zero and not collinear, and 2𝐮 + 3𝐯 = a𝐮 + (b − 1)𝐯. Find a and b, and explain why "not collinear" matters.

ABCDEF is a regular hexagon with 𝐚 = 𝐀𝐁 and 𝐛 = 𝐁𝐂. Show that 𝐂𝐃 = 𝐛 − 𝐚 and that 𝐀𝐃 = 2𝐛.

Check your understanding

Quiz

𝐮 and 𝐯 are drawn tail to tail. Which arrow represents 𝐮 − 𝐯?

𝐮 = 6 N [N 20° E]. What is −½𝐮?

What is 𝐎𝐁 − 𝐎𝐀?

|𝐮| = 3 and |𝐯| = 4, and the angle between them is 90°. What is |𝐮 − 𝐯|?

Points satisfy 𝐏𝐑 = −2𝐏𝐐. What must be true?

Which vector is a unit vector in the direction of 𝐮 if |𝐮| = 5?

For parents and teachers

Big idea (C2.1): subtracting a vector means adding its opposite, and multiplying by a scalar stretches, shrinks or reverses a vector. Together with addition, these let students describe any point in a figure using just two vectors.

How to help: Ask your teen to explain Example 2 aloud: how can a car's velocity change if its speed stays at 20 m/s? (Because direction is part of velocity.) For figure problems, ask them to trace a path with a finger from the start point to the end point.

Common difficulties:

Drawing 𝐮 − 𝐯 from the head of 𝐮 to the head of 𝐯 (that is 𝐯 − 𝐮).
Writing a negative magnitude, such as "−3 N [N]", instead of "3 N [S]".
Forgetting to change the sign of a vector travelled backwards along a path.
Using the cosine-law angle for a sum when finding a difference (for 𝐮 − 𝐯 the angle in the triangle is θ itself).