Geometric Proof with Vectors (Edexcel IGCSE Further Maths)

Vector Paths

Spec point — 7A

How do I use vector paths?

• A vector path is a path of vectors taking you from a start point to an end point

• On the diagram shown

  • F to B has the path F to A to B

    • $\overset{\rightarrow}{FB}=\overset{\rightarrow}{FA}+\overset{\rightarrow}{AB}$

  • There are other possible paths too

    • F to G to B

• You can write vector paths in terms of vectors given (a, b, ...)

  • The rule $\overset{\rightarrow}{XY}=-\overset{\rightarrow}{YX}$ helps

    • $\overset{\rightarrow}{FB}=\overset{\rightarrow}{FA}+\overset{\rightarrow}{AB}=-\overset{\rightarrow}{AF}+\overset{\rightarrow}{AB}=-b+a$

• Vectors given can be used to build bigger paths

  • $\overset{\rightarrow}{FM}=\overset{\rightarrow}{FG}+\overset{\rightarrow}{GH}+\overset{\rightarrow}{HM}=a+a+b=2a+b$

    • Different correct paths simplify to the same final answer

How do I use ratios in vector paths?

• Convert ratios into fractions

• In the example shown, if $AX:XB=3:5$ then

  • $\overset{\rightarrow}{AX}=\frac{3}{8}\overset{\rightarrow}{AB}$

  • $\overset{\rightarrow}{XB}=\frac{5}{8}\overset{\rightarrow}{AB}$

    • The ratio 3:5 has 3 + 5 = 8 parts

• Always check which ratio you are being asked for

  • $\overset{\rightarrow}{AX}=\frac{3}{5}\overset{\rightarrow}{XB}$
  • $\overset{\rightarrow}{XB}=\frac{5}{3}\overset{\rightarrow}{AX}$

Exam Hint

• Mark schemes will accept different correct paths, as long as the final answer is fully simplified.
• Check for symmetries in the diagram to see if the vectors given can be used anywhere else.

Worked Example The following diagram shows a grid formed of identical parallelograms.

Vectors $a$* *and $b$ are given by $\overset{\rightarrow}{AB}$ and $\overset{\rightarrow}{AF}$ respectively.

Find the following vectors in terms of $a$ and $b$, fully simplifying your answer.

(a) $\overset{\rightarrow}{EK}$

*There are many ways to get from E to *K * One option is E to O (b twice) then O to K( -a four times). *

$\begin{gathered} \vec{EK} & = \vec{EJ} + \vec{JO} + \vec{ON} + \vec{NM} + \vec{ML} + \vec{LK} \\ = b + b - a - a - a - a \end{gathered}$

• *$\overset{\rightarrow}{EK}=2b-4a$

-4a + 2b also accepted

(b) $\overset{\rightarrow}{AX}$, where $X$is the point on the line $AT$such that $AX:XT=1:3$.

Start by imagining the vector $\begin{gathered} \vec{AT} \end{gathered}$ *It is easy to write this vector in terms of a and b *

$\begin{gathered} \vec{AT} \end{gathered} = 4 a + 3 b$

To find $\begin{gathered} \vec{AX} \end{gathered}$, use the given ratio to write it as a fraction of $\begin{gathered} \vec{AT} \end{gathered}$ There are 1 + 3 = 4 parts

$\begin{gathered} \vec{AX} & = 1 \frac{1}{4} \vec{AT} \end{gathered}$

Substitute in the known expression for $\begin{gathered} \vec{AT} \end{gathered}$ Expand and simplify to get the final answer

$\begin{gathered} \vec{AX} & = 1 \frac{1}{4} ( 4 a + 3 b ) \\ = a + \frac{3}{4} b \end{gathered}$

| A X -vec | = a + 3 / 4 b | |

Parallel Vectors

Spec point — 7A

How do I know if two vectors are parallel?

• Two vectors are parallel if one is a scalar multiple of the other

  • This means if b is parallel to a, then b = *ka

    • where k is a constant number(scalar)

• For example, $a = ( \begin{gathered} 1 \\ 3 \end{gathered} )$ and $b = ( \begin{gathered} 2 \\ 6 \end{gathered} )$

  • $( \begin{gathered} 2 \\ 6 \end{gathered} ) = 2 \times ( \begin{gathered} 1 \\ 3 \end{gathered} )$ so $b=2a$
  • b is a scalar multiple of a, so b is parallel to a

• If the scalar multiple is negative, then the vectors are parallel and in opposite directions

  • $c = ( \begin{gathered} - 3 \\ - 9 \end{gathered} ) = - 3 a$

    • c is parallel to a and in the opposite direction

How do I use factorisation to show two vectors are parallel?

• If two vectors factorise with a common bracket, then they are parallel

  • They can be written as scalar multiples

• For example

  • 9a + 6b factorises to 3(3a + 2b)

  • 12a + 8b factorises to 4(3a + 2b)

  • This means $12 a + 8 b = \frac{4}{3} ( 9 a + 6 b )$

    • so they are scalar multiples of each other
    • and therefore parallel

Exam Hint

• If a question asks you to show that two vectors are parallel

  • Don't stop after showing two vectors are scalar multiples of each other
  • You must conclude with "therefore a and b are parallel" to get all the marks!

Worked Example Show that the vectors $a=2p-4q$ and $b=6q--3p$ are parallel.

It helps to write b in the same order as a

$b=-3p+6q$

Factorise both vectors to form a common bracket

$\begin{gathered} a = 2 ( p - 2 q ) \\ b = - 3 ( p - 2 q ) \end{gathered}$

Write b in terms of a It can help to make the brackets the subject of the first equation, i.e. $( p - 2 q ) = \frac{a}{2}$ *Then substitute that into the equation for *b

$b=-3\times\frac{a}{2}=-\frac{3}{2}a$

Write a conclusion showing the scalar multiple State that this shows they are parallel

$b=-\frac{3}{2}a$ therefore b is parallel to a

You could also use$a=-\frac{2}{3}b$

Collinearity

Spec point — 7F

What does collinear mean?

• The points A, B and C are collinear if they all lie along the same straight line

• In the image on the left in the diagram below, the following vectors are all parallel:

  • $\overset{\rightarrow}{AB},\overset{\rightarrow}{BC},\overset{\rightarrow}{AC},\overset{\rightarrow}{BA},\overset{\rightarrow}{CA},\overset{\rightarrow}{CB}$

How do I use vectors to show that three points are collinear?

• To show the points A, B and C are collinear

  • prove that two line segments are parallel

  • and show that there is at least one point that lies on both segments

    • This makes them parallel and connected(not side-by-side)

• For example, if you show that $\overset{\rightarrow}{BC}=2\overset{\rightarrow}{AB}$ then

  • the line segments AB and BC are parallel

  • and they have a common point, B

    • So A, B and C must be collinear

• Similarly, $\overset{\rightarrow}{AC}=3\overset{\rightarrow}{AB}$ means AC and AB are parallel

  • and they have a common point, A

    • so A, B and C must be collinear

How can collinearity be used when extend lines?

• Extending the line segment OA means continuing the straight line beyond A*

  • If the extended line passes through the point X

    • then O , A and X are collinear

• To test whether another point, Y, is on OA extended*

  • test whether O , A and Y are collinear

Exam Hint

• Even though it's 'obvious' that $\overset{\rightarrow}{AB}$ and $\overset{\rightarrow}{AC}$ have a common point at A, you should still write this fact in your answer to complete a collinearity proof.

Worked Example The diagram shows a quadrilateral, $OABC$, where

$\overset{\rightarrow}{OA}=2a$

$\overset{\rightarrow}{OC}=c$

It is also know that

$\overset{\rightarrow}{OB}=2a+3c$

(a) Juan believe that the point $X$, which is halfway along $AC$, lies on the line $OB$.

Show that $O$, $X$ and $B$ are not collinear.

O , X and B are collinear if two line segments are parallel and share a common point Consider the vectors $\overset{\rightarrow}{OX}$ and $\overset{\rightarrow}{OB}$ *They have a common point, *O Check if they are parallel by first finding $\overset{\rightarrow}{OX}$

$\begin{gathered} \vec{OX} & = \vec{OA} + 1 \frac{1}{2} \vec{AC} \\ = 2 a + 1 \frac{1}{2} ( \vec{AO} + \vec{OC} ) \\ = 2 a + 1 \frac{1}{2} ( - 2 a + c ) \end{gathered}$

Expand and simplify

$\begin{gathered} \vec{OX} & = 2 a - a + 1 \frac{1}{2} c \\ = a + 1 \frac{1}{2} c \end{gathered}$

Now compare it to $\overset{\rightarrow}{OB}=2a+3c$ (given in the question) Try factorisation to write $\overset{\rightarrow}{OB}$ in terms of $\overset{\rightarrow}{OX}$

$\begin{gathered} \vec{OB} & = 2 a + 3 c \\ = 2 ( a + \frac{3}{2} c ) \end{gathered}$

The $a$ inside the brackets matches the $a$ in $\overset{\rightarrow}{OX}=a+\frac{1}{2}c$ But the $+\frac{3}{2}c$ inside the brackets does not match the $+\frac{1}{2}c$ in $\overset{\rightarrow}{OX}=a+\frac{1}{2}c$ Therefore you cannot write $\overset{\rightarrow}{OB}$ and $\overset{\rightarrow}{OX}$ as scalar multiples of each other

Final answer: $\overset{\rightarrow}{OX}$is not a scalar multiple of$\overset{\rightarrow}{OB}$ Therefore OX and OB are not parallel O, X and B cannot be collinear

(b) Maria believes that the point $Y$, which is three quarters along $AC$ from $A$, lies on the line $OB$.

Show that $O$, $Y$ and $B$ are collinear.* *

Similarly to part (a), find $\overset{\rightarrow}{OY}$ and see if it is parallel to $\overset{\rightarrow}{OB}$* *

$\begin{gathered} \vec{OY} & = \vec{OA} + \frac{3}{4} \vec{AC} \\ = 2 a + \frac{3}{4} ( - 2 a + c ) \\ = 1 \frac{1}{2} a + \frac{3}{4} c \end{gathered}$

Use factorisation to write $\overset{\rightarrow}{OB}$ in terms of $\overset{\rightarrow}{OY}$* *

$\vec{OB} = 2 a + 3 c = 4 ( 1 \frac{1}{2} a + \frac{3}{4} c ) = 4 \vec{OY}$

*So OB is parallel to OY * They also have a common point, O

Final answer: $\overset{\rightarrow}{OB}=4\overset{\rightarrow}{OY}$ Therefore OY and OB are parallel They also have a common point, O Therefore O, Y and B are collinear

Equating Coefficients of Vectors

Spec point — 7B

How do I equate coefficients of vectors?

• In the vector $10a-3b$

  • the coefficient of a is $10$
  • the coefficient of b is $-3$

• Two vectors can only be equal if all their coefficients are equal

• So if the vectors $3a+kb$ and $ma+5b$ are both equal

  • then $3=m$ (by equating coefficients of a)
  • and $k=5$ (by equating coefficients of b)

How do I use equating coefficients to find points of intersection?

• Some diagrams have vectors that meet at a point

  • These are called concurrent vectors

• To find this point, create two different vector paths that approach the point from different directions

  • Then set them equal and equate coefficients

• For example, if the lines AB and CD intersect at the point P,* *find $\overset{\rightarrow}{AP}$

  • One path is along the line AB

    • A , P and B are collinear
    • so $\overset{\rightarrow}{AP}=k\overset{\rightarrow}{AB}$ (they are parallel)

  • Another path is along the line CD

    • For example, $\overset{\rightarrow}{AP}=\overset{\rightarrow}{AC}+\overset{\rightarrow}{CP}$
    • C , P and D are also collinear so $\overset{\rightarrow}{CP}=λ\overset{\rightarrow}{CD}$
    • $λ$ is a different scalar to $k$

  • Set both paths equal and equate coefficients

    • This forms simultaneous equations in $k$ and $λ$
    • Solve these and substitute their values back in to find $\overset{\rightarrow}{AP}$

Exam Hint

• The method of equating two different paths is really useful for other hard situations.

  • For example, extending lines out of a diagram, or continuing lines within a diagram.

Worked Example The diagram shows the diagonals of quadrilateral $OABC$, where

$\overset{\rightarrow}{OA}=2a$

$\overset{\rightarrow}{OC}=c$

You are also given that

$\overset{\rightarrow}{AB}=3\overset{\rightarrow}{OC}$

The diagonals $AC$ and $OB$ intersect at the point $P$.

Find the vector $\overset{\rightarrow}{OP}$, giving your answer fully simplified in terms of $a$ and $c$.

Find two different paths from O to P that come in along different diagonals The first path is along OB itself *Although the length OP is unknown, *$\overset{\rightarrow}{OP}$ is definitely a scalar multiple of $\overset{\rightarrow}{OB}$ (they are parallel)

$\overset{\rightarrow}{OP}=k\overset{\rightarrow}{OB}$

Use $\begin{array}{rcl} \vec{OB} & = \vec{OA} + \vec{AB} = \vec{OA} + 3 \vec{OC} \end{array}$* to write this first path in terms of a and c*

$\vec{OP} = k ( 2 a + 3 c )$

The second path needs to come in along the AC diagonal

$\begin{gathered} \vec{OP} & = \vec{OA} + \vec{AP} \end{gathered}$* *

*Although the length AP is unknown, *$\overset{\rightarrow}{AP}$ is definitely a scalar multiple of $\overset{\rightarrow}{AC}$ (they are parallel) Give the scalar a letter different to $k$ used above

• *$\begin{gathered} \vec{OP} & = \vec{OA} + \\lambda \vec{AC} \end{gathered}$

Use $\overset{\rightarrow}{AC}=\overset{\rightarrow}{AO}+\overset{\rightarrow}{OC}$* to write this second path in terms of a and c *

$\vec{OP} = 2 a + \\lambda ( - 2 a + c )$* *

Set the first path equal to the second path

$k ( 2 a + 3 c ) = 2 a + \\lambda ( - 2 a + c )$

*Collect the terms into their a and c components on each side *

$2 k a + 3 k c = ( 2 - 2 \\lambda ) a + \\lambda c$

*Equate (and simplify) the a coefficients *

$\begin{gathered} 2 k & = 2 - 2 \\lambda \\ k = 1 - \\lambda \end{gathered}$

*Equate the c coefficients *

$3k=λ$

This gives two simultaneous equations in $k$ and $λ$ Solve these simultaneously

$\begin{gathered} k = 1 - 3 k \\ 4 k & = 1 \\ k = 1 \frac{1}{4} \end{gathered}$ *and ** *$λ=\frac{3}{4}$

• *

Go back to either of the two paths for $\overset{\rightarrow}{OP}$* * Substitute in $k$ or $λ$ Simplify to get the final answer

$\begin{gathered} \vec{OP} & = k ( 2 a + 3 c ) \\ = 1 \frac{1}{4} ( 2 a + 3 c ) \\ = 1 \frac{1}{2} a + \frac{3}{4} c \end{gathered}$

| O P -vec | = 1 / 2 a + 3 / 4 c | |