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

Vector Operations in Three-Space

C2.1 C1.4About 70 minutes

Learning goals

I can add, subtract and multiply vectors by scalars in Cartesian form in three-space.
I can write three-space vectors using 𝐢, 𝐣 and 𝐤 and find unit vectors.
I can test vectors and points for collinearity and find missing vertices of parallelograms.
I can decide whether a vector is a linear combination of two given vectors in three-space.

Words to know

𝐢, 𝐣, 𝐤
The standard unit vectors [1, 0, 0], [0, 1, 0] and [0, 0, 1] in three-space.

collinear
Lying on one line; for vectors, one is a scalar multiple of the other.

linear combination
An expression a𝐮 + b𝐯 (+ c𝐰 …) of scalars times vectors.

midpoint
The point halfway between A and B: average the coordinates.

Same rules, one more number

Good news: every operation from two-space works in three-space with one extra component. A drone that climbs while it flies north-east, the forces in the cables of a suspension bridge, and the motion of a character in a 3-D video game are all handled this way. This lesson collects the rules, introduces 𝐤, and uses scalar multiples and linear combinations to answer geometric questions in three-space.

Operations in three-space

For 𝐮 = [u₁, u₂, u₃], 𝐯 = [v₁, v₂, v₃] and a scalar k:

𝐮 + 𝐯 = [u₁ + v₁, u₂ + v₂, u₃ + v₃]
𝐮 − 𝐯 = [u₁ − v₁, u₂ − v₂, u₃ − v₃]
k𝐮 = [ku₁, ku₂, ku₃]
|𝐮| = √(u₁² + u₂² + u₃²)

The geometric meaning is exactly the same as in two-space: tip-to-tail addition, the parallelogram law, opposites and scalar multiples all still apply. The only difference is that the arrows no longer have to lie flat on the page.

The standard unit vectors 𝐢, 𝐣 and 𝐤

In three-space, 𝐢 = [1, 0, 0], 𝐣 = [0, 1, 0] and 𝐤 = [0, 0, 1] point along the positive axes. Every vector can be written as

[a, b, c] = a𝐢 + b𝐣 + c𝐤

Left: [3, 4, 2] = 3𝐢 + 4𝐣 + 2𝐤, built from steps along the three axis directions. Right: the unit vectors 𝐢, 𝐣 and 𝐤.

Worked example

Example 1: Combining vectors in three-space

Given 𝐚 = [2, −1, 3] and 𝐛 = [−1, 4, 0], find 3𝐚 − 2𝐛 and its magnitude.

3𝐚 = [6, −3, 9] and 2𝐛 = [−2, 8, 0].
3𝐚 − 2𝐛 = [6 − (−2), −3 − 8, 9 − 0] = [8, −11, 9].
|3𝐚 − 2𝐛| = √(64 + 121 + 81) = √266 ≈ 16.31.

[8, −11, 9], magnitude √266 ≈ 16.31.

Worked example

Example 2: Switching notation

Write [4, −2, 7] using 𝐢, 𝐣, 𝐤, and write 2𝐢 − 𝐤 in component form.

[4, −2, 7] = 4𝐢 − 2𝐣 + 7𝐤.
2𝐢 − 𝐤 has no 𝐣 part, so its y-component is 0: [2, 0, −1].

4𝐢 − 2𝐣 + 7𝐤 and [2, 0, −1].

Worked example

Example 3: A unit vector

Find the unit vector in the direction of [2, −1, 2], and the vector of magnitude 12 in that direction.

|[2, −1, 2]| = √(4 + 1 + 4) = 3.
Unit vector: [2/3, −1/3, 2/3].
Magnitude 12: 12[2/3, −1/3, 2/3] = [8, −4, 8].

[2/3, −1/3, 2/3] and [8, −4, 8].

Collinear vectors and points

As before, two non-zero vectors are collinear when one is a scalar multiple of the other: 𝐯 = k𝐮. In three-space, all three components must share the same ratio k. Three points A, B, C are collinear when 𝐀𝐂 = k𝐀𝐁 for some k.

Worked example

Example 4: Are the points on one line?

Show that A(1, 2, 3), B(3, 5, 4) and C(7, 11, 6) are collinear.

𝐀𝐁 = [2, 3, 1] and 𝐀𝐂 = [6, 9, 3].
Compare ratios: 6/2 = 3, 9/3 = 3, 3/1 = 3. All the same, so 𝐀𝐂 = 3𝐀𝐁.
The vectors are parallel and share point A, so the three points lie on one line.

Collinear, and C is three times as far from A as B is.

Because 𝐀𝐂 = 3𝐀𝐁, the points A, B and C lie on one line. The vertical dotted lines drop each point to the xy-plane to show where it sits in space.

Worked example

Example 5: Completing a parallelogram

A(1, 0, 2), B(4, 1, −1) and C(6, 3, 1) are three vertices of parallelogram ABCD. Find D.

In parallelogram ABCD, 𝐀𝐃 = 𝐁𝐂.
𝐁𝐂 = [2, 2, 2], so 𝐎𝐃 = 𝐎𝐀 + 𝐁𝐂 = [1 + 2, 0 + 2, 2 + 2] = [3, 2, 4].
Check: 𝐀𝐁 = [3, 1, −3] and 𝐃𝐂 = [6 − 3, 3 − 2, 1 − 4] = [3, 1, −3] ✓

D(3, 2, 4)

Linear combinations in three-space

A linear combination a𝐮 + b𝐯 of two non-collinear vectors always lies in the plane that contains 𝐮 and 𝐯 (when they are drawn from the same point). So, unlike in two-space, not every 3-D vector is a combination of just two vectors. To test a vector, write one equation per component (three equations, two unknowns). Solve two of them and check the third. Planes built this way are a main topic of the unit Lines and Planes.

Combinations of two non-collinear vectors 𝐮 and 𝐯 fill a plane. A vector pointing out of that plane cannot be made from 𝐮 and 𝐯 alone.

Worked example

Example 6: Testing a linear combination

Can [4, 1, 5] be written as a linear combination of 𝐮 = [1, 1, 2] and 𝐯 = [2, −1, 1]? What about [4, 1, 6]?

Set a[1, 1, 2] + b[2, −1, 1] = [4, 1, 5]: a + 2b = 4, a − b = 1, 2a + b = 5.
Subtract the second equation from the first: 3b = 3, so b = 1 and a = 2.
Check the third equation: 2(2) + 1 = 5 ✓. So [4, 1, 5] = 2𝐮 + 𝐯.
For [4, 1, 6] the first two equations again give a = 2, b = 1, but then 2a + b = 5 ≠ 6. The system has no solution.

[4, 1, 5] = 2𝐮 + 𝐯, but [4, 1, 6] is not a linear combination of 𝐮 and 𝐯 (it does not lie in their plane).

Just for fun

In two-space you have 𝐢 and 𝐣. In three-space you add 𝐤. Mathematicians in higher dimensions keep going with more unit vectors, which is why they never run out of letters: they just start using subscripts.

Hands-on activity

Build a vector out of household objects

You need: three pencils or chopsticks, modelling clay or a ball of reusable adhesive, a ruler, and a small box (a tissue box or cereal box works).

Stick the three pencils into a lump of clay so they are mutually perpendicular. Use the right-hand rule to label them x, y and z with small paper flags.
Put the box against the clay corner so its edges run along the axes. Measure its length (x), width (y) and height (z) in centimetres.
The vector from the corner at the origin to the opposite corner is [length, width, height]. Calculate its magnitude.
Measure that space diagonal directly. (Tip: measure the floor diagonal of the box, then use it and the height to check, or carefully hold a ruler across the corner.)
Find the unit vector in the direction of the diagonal and check that its magnitude is 1.

What to look for: for a box of 24 cm × 12 cm × 8 cm, the diagonal vector is [24, 12, 8] with magnitude √(576 + 144 + 64) = 28 cm, and the unit vector is [6/7, 3/7, 2/7].

Your turn

𝐮 = [1, −2, 4] and 𝐯 = [3, 0, −1]. Find 𝐮 + 𝐯.

Find 𝐮 − 𝐯.

Find −2𝐮.

Find 2𝐮 + 3𝐯.

Find |𝐮 − 𝐯|.

Find the unit vector in the direction of 𝐯.

Write 5𝐢 − 𝐣 + 2𝐤 in component form.

More practice: warm-up

[1, 2, 3] + [4, 5, 6] = ?

2[0, −1, 3] = ?

[3, 3, 3] − [1, 2, 3] = ?

Write 𝐤 in component form.

Find |𝐢 + 𝐣 + 𝐤|.

More practice: on your own

Are A(2, −1, 0), B(5, 1, 4) and C(11, 5, 12) collinear?

Find m and n so that [m, 6, −2] is collinear with [2, n, 1].

Solve for 𝐱: 3𝐱 − [2, 1, −4] = 𝐱 + [6, −3, 2].

Find the midpoint M of A(3, −2, 5) and B(−1, 4, 1), and check that 𝐀𝐌 = 𝐌𝐁.

Find the vector of magnitude 6 in the direction of [1, 2, −2].

Challenge

Is [1, 3, −2] a linear combination of [1, 1, 0] and [0, 1, −1]? Is [1, 2, 3]?

Write [2, −1, 5] as a linear combination of 𝐢 + 𝐣, 𝐣 + 𝐤 and 𝐢 + 𝐤.

A(1, 0, 2), B(4, 1, −1) and C(6, 3, 1) are three vertices of a parallelogram, but the order of the vertices is not given. Find all possible fourth vertices.

Check your understanding

Quiz

What is [2, −1, 4] − [5, 3, −2]?

How is 2𝐢 − 3𝐤 written in component form?

What is |2𝐢 − 𝐣 + 2𝐤|?

Which vector is collinear with [2, −4, 6]?

What is the midpoint of (2, 4, −6) and (0, −2, 8)?

What is the unit vector in the direction of [0, 3, −4]?

For parents and teachers

Big idea (C2.1, C1.4): vector arithmetic in three-space is the same as in two-space with a third component. Scalar multiples test whether vectors or points line up; linear combinations of two vectors stay in one plane.

How to help: The box activity is quick and satisfying. When your teen checks collinearity, make sure they compare all three ratios, not just two.

Common difficulties:

Dropping or mixing up the third component.
Writing 2𝐢 − 3𝐤 as [2, −3, 0] (the missing 𝐣 term means a 0 in the middle).
Checking only two of the three equations in a linear-combination problem.
Assuming every 3-D vector is a combination of two given vectors.