Problem Solving with Vectors (Cambridge International IGCSE Mathematics)

Vector Problem Solving

Vector problem-solving

What are vector proofs?

• Vectors can be used to prove things that are true in geometrical diagrams

  • Vector proofs can be used to find additional information that can help us to solve problems

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

• 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

How do I know if three points lie on a straight line?

• You may be asked to prove that three points lie on a straight line

  • Points that lie on a straight line are collinear

• To show that 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 parallel and 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 do I use ratios in vector paths?

Example of a point dividing a line segment

• 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

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

Worked Example The diagram shows trapezium OABC.

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

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

AB is parallel to OC, with $\overset{\rightarrow}{AB}=3\overset{\rightarrow}{OC}$.

(a) Find expressions for vectors $\overset{\rightarrow}{OB}$ and $\overset{\rightarrow}{AC}$ in terms of a *and c.

Answer:

$\overset{\rightarrow}{AB}=3\overset{\rightarrow}{OC}$ and $\overset{\rightarrow}{OC}=c$ so $\overset{\rightarrow}{AB}=3c$.

$\begin{gathered} \vec{OB} & = \vec{OA} + \vec{AB} \\ = 2 a + 3 c \end{gathered}$

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

$\begin{gathered} \vec{AC} & = \vec{AO} + \vec{OC} \\ = - \vec{OA} + \vec{OC} \\ = - 2 a + c \end{gathered}$

$\overset{\rightarrow}{AC}=c-2a$

(b) Point P lies on AC such that AP: PC = 3 : 1.

Answer:

Find expressions for vectors $\overset{\rightarrow}{AP}$ and $\overset{\rightarrow}{OP}$ in terms of a and c.

*AP : PC = 3 : 1 means that *$\overset{\rightarrow}{AP}=\frac{3}{3+1}\overset{\rightarrow}{AC}=\frac{3}{4}\overset{\rightarrow}{AC}$

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

$\overset{\rightarrow}{AP}=-\frac{3}{2}a+\frac{3}{4}c$

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

$\overset{\rightarrow}{OP}=\frac{1}{2}a+\frac{3}{4}c$

(c) Hence, prove that point P lies on line OB, and determine the ratio $\overset{\rightarrow}{OP}:\overset{\rightarrow}{PB}$.

Answer:

*To show that O, P, and B are collinear (lie on the same line), note that *$\vec{OP} = 1 \frac{1}{2} a + \frac{3}{4} c = 1 \frac{1}{4} ( 2 a + 3 c )$

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

$\overset{\rightarrow}{OP}=\frac{1}{4}\overset{\rightarrow}{OB}$ therefore OP is parallel to OB ** ** and so P must lie on the line OB

If $\begin{gathered} \vec{OP} & = 1 \frac{1}{4} \vec{OB} \end{gathered}$ then $\begin{gathered} \vec{PB} & = \frac{3}{4} \vec{OB} \end{gathered}$

$\overset{\rightarrow}{OP}:\overset{\rightarrow}{PB}=1:3$