Game plan 3 · Unit 3

Vectors and 3D geometry

Name the object, check what you have been given, and work out what is missing.

Reach for it when

You see $\boldsymbol{a}\times\boldsymbol{b}$, column vectors, points in three dimensions, equations of lines, planes or spheres, “perpendicular”, “parallel”, “normal”, or “use vectors to prove”.

Part 1

What am I building?

Every question in this topic builds one of three objects. Each one needs two things before you can write its equation.

A line

a point $\boldsymbol{a}$

and a direction $\boldsymbol{d}$

A plane

a point $\boldsymbol{a}$

and a normal $\boldsymbol{n}$

A sphere

a centre $C$

and a radius $r$

Game plan 3.1

The object decision

  1. 1Name the object. A line needs a point and a direction, a plane needs a point and a normal, and a sphere needs a centre and a radius.
  2. 2Check what you have been given against that list. Whatever is missing is what you need to find. Three points but no normal means a cross product. An expanded sphere with no centre means completing the square.
  3. 3Change between forms when the question asks. For a line, you can read the point and direction straight off the vector, parametric or Cartesian form. For a plane, the normal’s components are the coefficients $a$, $b$ and $c$ in $ax+by+cz+d=0$.
C(h, k, l)P(x, y, z)r
The expanded form of a sphere hides the centre. Complete the square in $x$, $y$ and $z$ to get it back, and the constant that is left is $r^2$.

In the Topic Guides

Part 2

Equations of a line

The QCAA asks you to convert between the three forms of a line every year.

Game plan 3.2

Writing the equation of a line

  1. 1Get the two ingredients, a point $\boldsymbol{a}$ and a direction $\boldsymbol{d}$. Given two points $A$ and $B$, take $\boldsymbol{a}=\overrightarrow{OA}$ and $\boldsymbol{d}=\overrightarrow{AB}$.
  2. 2Vector form: $\boldsymbol{r}=\boldsymbol{a}+t\boldsymbol{d}$, where $t\in\mathbb R$.
  3. 3Parametric form: read the components off the vector form, one equation for each axis.
  4. 4Cartesian form: solve each parametric equation for $t$ and set them equal. If a component of $\boldsymbol{d}$ is zero, that variable is constant. State it separately, and don’t divide by zero.
  5. 5To test a point, substitute it and check that the same $t$ works in every component. For a line segment, restrict $t$ to an interval and state it.
VectorParametricCartesian
$\boldsymbol{r}=\boldsymbol{a}+t\boldsymbol{d}$$x=a_1+td_1$
$y=a_2+td_2$
$z=a_3+td_3$		$\dfrac{x-a_1}{d_1}=\dfrac{y-a_2}{d_2}=\dfrac{z-a_3}{d_3}$

Tip

Every form carries the same two facts. Read $\boldsymbol{a}$ off the constants and $\boldsymbol{d}$ off the coefficients of $t$. In Cartesian form they are the numerators and the denominators.

In the Topic Guides

Part 3

Angles, projections and perpendicularity

Game plan 3.3

Testing and measuring

  1. 1Perpendicular: test $\boldsymbol{a}\cdot\boldsymbol{b}=0$. Parallel: test $\boldsymbol{a}=k\boldsymbol{b}$.
  2. 2Angle between two vectors: rearrange $\boldsymbol{a}\cdot\boldsymbol{b}=|\boldsymbol{a}||\boldsymbol{b}|\cos\theta$.
  3. 3Use the scalar projection for a length and the vector projection for a vector. Both are on the formula sheet.
abvector projection of a on b
The scalar projection is a length. The vector projection is that length pointed along $\hat{\boldsymbol{b}}$. The only decision is which one the question asked for.

Try it

Project $\boldsymbol a$ onto $\boldsymbol b$

abθ

The angle is acute, so the scalar projection is positive and the arrow points along $\boldsymbol b$.

At $90^\circ$, $\boldsymbol a\cdot\boldsymbol b=0$, so both projections are zero.

The angle is obtuse, so the scalar projection is negative and the arrow points against $\boldsymbol b$.

In the Topic Guides

Part 4

Planes

Before you write an equation, ask which coordinate the plane holds fixed. A variable missing from the Cartesian form tells you the plane is parallel to that axis.

Game plan 3.4

Building a plane from three points

  1. 1Form two direction vectors in the plane, such as $\overrightarrow{AB}$ and $\overrightarrow{AC}$.
  2. 2Cross them for the normal: $\boldsymbol{n}=\overrightarrow{AB}\times\overrightarrow{AC}$.
  3. 3Write $\boldsymbol{r}\cdot\boldsymbol{n}=\boldsymbol{a}\cdot\boldsymbol{n}$, using any of the three points as $\boldsymbol{a}$. Expand it for the Cartesian form if the question asks.
  4. 4Check by substituting a point you didn’t use.

The order matters

$\boldsymbol{b}\times\boldsymbol{a}$ points the opposite way to $\boldsymbol{a}\times\boldsymbol{b}$. Either normal gives the same plane, but if the question states a direction, use it. $|\boldsymbol{a}\times\boldsymbol{b}|$ is also the area of the parallelogram the two vectors span.

Game plan 3.5

Applying a plane you are given

  1. 1A given plane almost always arrives as $ax+by+cz+d=0$. Read the normal straight off the coefficients: $\boldsymbol{n}=(a,b,c)$.
  2. 2To use a point, substitute it. The point lies on the plane exactly when the equation balances.
  3. 3That one fact answers three kinds of question: whether a point is on the plane, an unknown coordinate, and an axis intercept, where you set the other two coordinates to zero.

Try it

Build a plane from three points

xyz ABCn

The shaded patch is only part of the plane. A plane carries on forever in every direction, and the dashed triangle just marks the three points. Swap the order and the normal flips. Every coefficient changes sign, but it is still the same plane. The arrow shows the direction of $\boldsymbol n$, not its length.

In the Topic Guides

Part 5

Intersections

Game plan 3.6

Finding where things meet

  1. 1Line and plane: substitute the line’s parametric equations into the plane’s Cartesian equation, solve for $t$, then substitute back for the point. No solution means the line is parallel to the plane and off it. An identity such as $0=0$ means the line lies in the plane.
  2. 2Line and line: set $\boldsymbol{r}_1(t_1)=\boldsymbol{r}_2(t_2)$, with a different parameter for each line. Solve two of the component equations for $t_1$ and $t_2$, then check the third.
  3. 3If the third component agrees, the lines intersect. If it fails and the directions aren’t parallel, the lines are skew. If the directions are parallel, the lines are parallel or the same line.
Intersectall three components agree Paralleldirections are multiples Skewthird component fails
Testing the third component is what separates intersecting lines from skew lines. Skipping it is where most students lose the mark.

Try it

Intersect, parallel or skew?

xyz r1r2

All three components agree, so the lines meet at $(3,1,2)$.

The directions are multiples, and a point on $\boldsymbol r_2$ is not on $\boldsymbol r_1$, so the lines are parallel and separate.

The third component fails and the directions are not multiples, so the lines are skew. Turn the view: from some angles they look as if they cross.

In the Topic Guides

Part 6

Vector geometric proofs

Game plan 3.7

Proving with vectors

  1. 1Give each key point a position vector, and write every segment in the diagram as a difference of position vectors.
  2. 2Translate what you are proving into vectors. Perpendicular means a dot product of zero, parallel or collinear means a scalar multiple, and a midpoint is an average.
  3. 3Do the calculation, then finish with a sentence saying what has been shown.

In the Topic Guides

Formulas

Formula panel

Each formula is labelled with how to treat it in the exam.

Magnitude, unit vector, the dot product in both forms, and scalar and vector projection
On the sheet
The cross product in both forms
On the sheet
A line in vector, parametric and Cartesian form
On the sheet
A plane: $\boldsymbol{r}\cdot\boldsymbol{n}=\boldsymbol{a}\cdot\boldsymbol{n}$ and $ax+by+cz+d=0$
On the sheet
A sphere: $(x-h)^2+(y-k)^2+(z-l)^2=r^2$
On the sheet
Perpendicular test $\boldsymbol{a}\cdot\boldsymbol{b}=0$, parallel test $\boldsymbol{a}=k\boldsymbol{b}$
Memorise
Parallelogram area $|\boldsymbol{a}\times\boldsymbol{b}|$, triangle area $\frac12|\boldsymbol{a}\times\boldsymbol{b}|$
Memorise
Midpoint $\overrightarrow{OM}=\frac12(\boldsymbol{a}+\boldsymbol{b})$. A point dividing $AB$ in the ratio $m:n$ is $\frac{n\boldsymbol{a}+m\boldsymbol{b}}{m+n}$.
Memorise
Angle between vectors: $\cos\theta=\dfrac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{a}||\boldsymbol{b}|}$, from the dot product on the sheet
Derive

Watch for

There is no perpendicular-distance formula in this course. If a question asks for a distance, use a diagram and reasoning, not a formula.

QCAA language

What the question says, and what it means

Question typeWhat you’ll seeWhat it means
Building an object“Determine the equation of the plane”, line or sphereGame plan 3.1. Check the point, direction or normal
Lines“The vector, parametric or Cartesian equation of the line”, two named pointsGame plan 3.2. A point plus a direction, then convert
Perpendicular or parallel“Show … is perpendicular to …”A dot product of zero, or a scalar multiple
A plane from pointsThree named pointsGame plan 3.4. Cross for the normal
Intersection“Where does the line meet the plane”, “do the lines intersect”Game plan 3.6
Proof“Use vector methods to prove”Game plan 3.7. Position vectors first

Tip

The most useful skill in this topic is reading the point and the direction, or the normal, straight off any form of the equation.

Back to the External Game plan or the Topic Guides.

Wording quoted from past QCAA Specialist Mathematics external assessments is © State of Queensland (QCAA), licensed under CC BY 4.0. Tangent Tuition is not affiliated with the QCAA.

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