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Vectors in Three Dimensions

Q: What does k represent?

The unit vector along the z-axis, perpendicular to both i and j. The three together form the standard basis for three-dimensional space.

Adding a third axis adds a third component and changes remarkably little else. Addition, subtraction, scalar multiples, the parallel test and position vectors all carry over verbatim; magnitude gains one more square under the root. The single thing that does not survive is describing direction with one angle.

30 min Vectors

The big picture

Three dimensions is where vectors stop being a convenience and become a necessity. In two dimensions you can usually get by with coordinate geometry — gradients, distances, the equation of a line. In three, gradient is meaningless (a line can slope in infinitely many ways), and there is no single equation of a line in $x$ , $y$ and $z$ . Vector methods keep working without modification, which is why every field that deals with real space — mechanics, engineering, computer graphics, machine learning — describes it with vectors rather than coordinates. The generality you get here is the point of having learned the machinery.

What you'll be able to do

Use $i$ , $j$ , $k$ and three-component column vectors
Find magnitudes and distances in three dimensions
Apply position vectors, midpoints and the parallel test unchanged
Explain why direction in 3D needs more than a single angle

A third axis and a third unit vector

The $z$ -axis is added perpendicular to both $x$ and $y$ , with unit vector $k$ along it. A general vector becomes $a_{1} i + a_{2} j + a_{3} k$ , or a column of three.

Every rule from two dimensions applies componentwise as before: add by adding components, scale by scaling all three, and two vectors are parallel exactly when one is a scalar multiple of the other. Nothing about those rules mentioned how many components there were, which is why they transfer without comment.

a = a_{1} i + a_{2} j + a_{3} k = a_{1} a_{2} a_{3}

Three components, same idea.

1Double

q

- 2 i + 10 j + 2 k

2Add componentwise:

(2 - 2) i + (- 1 + 10) j + (4 + 2) k

Answer

9 j + 6 k

Tip — A zero component is still a component. Writing $9 j + 6 k$ is fine, but in column form it must be $096$ — the slot cannot be left out.

Magnitude: Pythagoras used twice

The three-dimensional magnitude formula looks like the two-dimensional one with an extra term, and it is — but it is worth seeing where the extra term comes from.

Take the vector as the diagonal of a rectangular box with edges $a_{1}$ , $a_{2}$ , $a_{3}$ . The diagonal of the base is $a_{1}^{2} + a_{2}^{2}$ by Pythagoras. That base diagonal and the vertical edge $a_{3}$ form a second right-angled triangle whose hypotenuse is the vector itself. Applying Pythagoras again gives the result.

∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2}

Pythagoras applied twice.

1Squares:

4 + 9 + 36 = 49

2Root:

49

Answer

7

Why this works

Notice the second example needed no new formula at all — $A B = b - a$ and "distance is the magnitude" both transferred untouched. That is the pattern for this whole lesson.

What carries over: position vectors, midpoints, collinearity

Every result from the position-vectors lesson holds in three dimensions with a third component appended. $A B = b - a$ , the midpoint is $\frac{1}{2} (a + b)$ , a ratio point is start plus a fraction of the journey, and collinearity is still parallel displacements sharing a point.

Vector proof also transfers, with one caveat: in three dimensions a base needs $three$ non-coplanar vectors to describe every point, not two. For problems confined to a single plane within 3D space, two are still enough.

A B = 2 i + 3 j + 4 k

B C = 4 i + 6 j + 8 k

B C = 2 A B

, and both involve the point

B

AnswerParallel with a common point

B

, so the three points are collinear.

Tip — Check all three components when testing for a scalar multiple. Two matching and one not means the vectors are $not$ parallel — a mismatch it is easy to miss by stopping early.

Direction: the one thing that changes

In two dimensions a single angle fixes a direction. In three it cannot: infinitely many directions make the same angle with the $x$ -axis, sweeping out a cone around it.

A-Level handles this by asking for the angle a vector makes with a $specified$ axis, one at a time, using the component along that axis over the magnitude. The three angles together — the direction cosines — do pin the direction down, and they satisfy a neat identity.

The practical consequence is that the unit vector, not an angle, is the natural way to record a direction in three dimensions. $\hat{a} = a /∣ a ∣$ still works exactly as before.

cos θ_{x} = \frac{a _{1}}{∣ a ∣}

Angle with the

x

-axis; similarly for

y

and

z

cos^{2} θ_{x} + cos^{2} θ_{y} + cos^{2} θ_{z} = 1

The three direction cosines are not independent.

1Magnitude:

4 + 9 + 36 = 7

cos θ_{x} = \frac{2}{7}

θ_{x} = arccos (\frac{2}{7})

Answer

θ_{x} \approx 73. 4^{\circ}

Why this works

The identity is just the magnitude formula in disguise: dividing $a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = ∣ a ∣^{2}$ through by $∣ a ∣^{2}$ gives it immediately. It also confirms that only two of the three angles are free.

Think like an examiner

Include zero components explicitly in column vectors — an omitted row is a different-sized vector.
Check every component when testing for parallel vectors; two out of three is not enough.
When asked for an angle in 3D, state which axis it is measured from — "the angle" alone is ambiguous.
Distances and magnitudes stay exact as surds unless a decimal is asked for.

Common misconceptions

Using

a_{1}^{2} + a_{2}^{2}

for a 3D magnitude.

All three components are squared:

a_{1}^{2} + a_{2}^{2} + a_{3}^{2}

Declaring vectors parallel after checking two components.

The same scalar must work for all three, or they are not parallel.

Trying to describe a 3D direction with one angle.

One angle leaves a whole cone of directions. Use a unit vector, or give the angle to a named axis.

Talking about the gradient of a line in 3D.

Gradient is a 2D idea. In three dimensions direction is carried by a vector.

3D vector results

a = a_{1} i + a_{2} j + a_{3} k

Three components.

∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2}

Magnitude.

A B = b - a

Unchanged from 2D.

cos θ_{x} = \frac{a _{1}}{∣ a ∣}

Angle to an axis.

Stretch yourself

Points $A (2, 1, 0)$ , $B (4, 4, 6)$ and $C$ are such that $B$ divides $A C$ in the ratio $1 : 2$ . Find the coordinates of $C$ and the exact length $A C$ .

Hint — $B$ being $\frac{1}{3}$ of the way from $A$ to $C$ means $A B = \frac{1}{3} A C$ .

Questions students ask

Which results from 2D do not transfer to 3D?

Only direction-by-one-angle. Addition, scalar multiples, magnitude, the parallel test, position vectors, midpoints, ratios and collinearity all carry over with a third component added.

Why is there no gradient in three dimensions?

Gradient measures rise over run in a plane. In space a line can rise in infinitely many directions for the same horizontal run, so a single number cannot capture it — which is why direction is given as a vector.

Can three points in 3D always be made collinear?

No. Three arbitrary points generally form a triangle. Collinearity still has to be proved, by showing one displacement is a scalar multiple of another with a shared point.

What does

k

represent?

The unit vector along the

z

-axis, perpendicular to both

i

and

j

. The three together form the standard basis for three-dimensional space.

Key takeaways

A third axis brings a third unit vector $k$ and a third component.
Addition, scalar multiples and the parallel test work componentwise, unchanged.
Magnitude is Pythagoras applied twice: $a_{1}^{2} + a_{2}^{2} + a_{3}^{2}$ .
Position vectors, midpoints, ratios and collinearity all transfer directly.
Direction needs a unit vector or an angle to a named axis — one angle is not enough.

How this fits the course

Build on

Solving Geometric Problems

Mechanics: forces as vectorssoon

Leads to

Mechanicssoon

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Vectors in Three Dimensions

A third axis and a third unit vector

Magnitude: Pythagoras used twice

What carries over: position vectors, midpoints, collinearity

Direction: the one thing that changes

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Related topics

Vectors in Three Dimensions

A third axis and a third unit vector

Magnitude: Pythagoras used twice

What carries over: position vectors, midpoints, collinearity

Direction: the one thing that changes

Practise this topic

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