Teach Me Maths

Vectors

Question 11 mark

Write down vector d. Enter the two components as (\text{horizontal},\text{vertical}).

Write your answer as (x, y)

Hint

Count horizontal movement first, then vertical movement.

Worked solution

Write down vector d. Enter the two components as (\text{horizontal},\text{vertical}).

  1. The arrow goes 3 squares right and 2 squares up.
  2. Its column vector is \begin{pmatrix}3\\2\end{pmatrix}.

Question 21 mark

Work out \begin{pmatrix}2\\3\end{pmatrix} + \begin{pmatrix}4\\1\end{pmatrix}. Enter the two components as (\text{horizontal},\text{vertical}).

Write your answer as (x, y)

Hint

Add the two top numbers, then add the two bottom numbers.

Worked solution

Work out \begin{pmatrix}2\\3\end{pmatrix} + \begin{pmatrix}4\\1\end{pmatrix}. Enter the two components as (\text{horizontal},\text{vertical}).

  1. Add matching components: 2+4=6 and 3+1=4.
  2. The result is \begin{pmatrix}6\\4\end{pmatrix}.

Question 31 mark

Work out \begin{pmatrix}-3\\5\end{pmatrix} + \begin{pmatrix}7\\-2\end{pmatrix}. Enter the two components as (\text{horizontal},\text{vertical}).

Write your answer as (x, y)

Hint

Keep the signs with their components.

Worked solution

Work out \begin{pmatrix}-3\\5\end{pmatrix} + \begin{pmatrix}7\\-2\end{pmatrix}. Enter the two components as (\text{horizontal},\text{vertical}).

  1. The top component is -3+7=4.
  2. The bottom component is 5+(-2)=3.
  3. The result is \begin{pmatrix}4\\3\end{pmatrix}.

Question 41 mark

Work out \begin{pmatrix}8\\3\end{pmatrix} - \begin{pmatrix}2\\6\end{pmatrix}. Enter the two components as (\text{horizontal},\text{vertical}).

Write your answer as (x, y)

Hint

Subtract each component in the stated order.

Worked solution

Work out \begin{pmatrix}8\\3\end{pmatrix} - \begin{pmatrix}2\\6\end{pmatrix}. Enter the two components as (\text{horizontal},\text{vertical}).

  1. Subtract matching components: 8-2=6 and 3-6=-3.
  2. The result is \begin{pmatrix}6\\-3\end{pmatrix}.

Question 51 mark

Work out \begin{pmatrix}-1\\4\end{pmatrix} - \begin{pmatrix}-6\\-3\end{pmatrix}. Enter the two components as (\text{horizontal},\text{vertical}).

Write your answer as (x, y)

Hint

Subtracting a negative adds its positive value.

Worked solution

Work out \begin{pmatrix}-1\\4\end{pmatrix} - \begin{pmatrix}-6\\-3\end{pmatrix}. Enter the two components as (\text{horizontal},\text{vertical}).

  1. The top component is -1-(-6)=5.
  2. The bottom component is 4-(-3)=7.
  3. The result is \begin{pmatrix}5\\7\end{pmatrix}.

Question 61 mark

Work out 4\begin{pmatrix}2\\-3\end{pmatrix}. Enter the two components as (\text{horizontal},\text{vertical}).

Write your answer as (x, y)

Hint

The scalar multiplies both components.

Worked solution

Work out 4\begin{pmatrix}2\\-3\end{pmatrix}. Enter the two components as (\text{horizontal},\text{vertical}).

  1. Multiply both components by 4: 4×2=8, 4×(-3)=-12.
  2. The result is \begin{pmatrix}8\\-12\end{pmatrix}.

Question 71 mark

Work out -3\begin{pmatrix}-2\\5\end{pmatrix}. Enter the two components as (\text{horizontal},\text{vertical}).

Write your answer as (x, y)

Hint

Check the signs of both products.

Worked solution

Work out -3\begin{pmatrix}-2\\5\end{pmatrix}. Enter the two components as (\text{horizontal},\text{vertical}).

  1. Multiply both components by -3: (-3)(-2)=6, (-3)(5)=-15.
  2. The result is \begin{pmatrix}6\\-15\end{pmatrix}.

Question 81 mark

\overrightarrow{AB}=\begin{pmatrix}5\\-2\end{pmatrix}. Work out \overrightarrow{AB}+\overrightarrow{BA}. Enter the two components as (\text{horizontal},\text{vertical}).

Write your answer as (x, y)

Hint

The return vector has opposite components.

Worked solution

\overrightarrow{AB}=\begin{pmatrix}5\\-2\end{pmatrix}. Work out \overrightarrow{AB}+\overrightarrow{BA}. Enter the two components as (\text{horizontal},\text{vertical}).

  1. Reversing direction changes the sign: \overrightarrow{BA}=\begin{pmatrix}-5\\2\end{pmatrix}.
  2. Adding gives \begin{pmatrix}0\\0\end{pmatrix}. The two movements return to the start.

Question 91 mark

Write \overrightarrow{AC} in terms of a and b.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Add the directed vectors along a route from the start to the finish.

Worked solution

Write \overrightarrow{AC} in terms of a and b.

  1. Follow the route from A to B, then B to C.
  2. \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=a+b.

Question 101 mark

\overrightarrow{PQ}=3a-2b. Write \overrightarrow{QP} in terms of a and b.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Reverse the sign of every term.

Worked solution

\overrightarrow{PQ}=3a-2b. Write \overrightarrow{QP} in terms of a and b.

  1. The reverse vector is the negative of the original.
  2. \overrightarrow{QP}=-(3a-2b)=-3a+2b.

Question 112 marks

a=\begin{pmatrix}3\\-4\end{pmatrix} and b=\begin{pmatrix}-2\\5\end{pmatrix}. Work out 2a+3b. Enter the two components as (\text{horizontal},\text{vertical}).

Write your answer as (x, y)

Hint

Calculate each scalar multiple before adding.

Worked solution

a=\begin{pmatrix}3\\-4\end{pmatrix} and b=\begin{pmatrix}-2\\5\end{pmatrix}. Work out 2a+3b. Enter the two components as (\text{horizontal},\text{vertical}).

  1. 2a=\begin{pmatrix}6\\-8\end{pmatrix} and 3b=\begin{pmatrix}-6\\15\end{pmatrix}.
  2. Add matching components: 6-6=0, -8+15=7.
  3. The result is \begin{pmatrix}0\\7\end{pmatrix}.

Question 122 marks

a=\begin{pmatrix}-1\\4\end{pmatrix} and b=\begin{pmatrix}5\\-3\end{pmatrix}. Work out 3a-2b. Enter the two components as (\text{horizontal},\text{vertical}).

Write your answer as (x, y)

Hint

Find both multiples and take care when subtracting a negative.

Worked solution

a=\begin{pmatrix}-1\\4\end{pmatrix} and b=\begin{pmatrix}5\\-3\end{pmatrix}. Work out 3a-2b. Enter the two components as (\text{horizontal},\text{vertical}).

  1. 3a=\begin{pmatrix}-3\\12\end{pmatrix} and 2b=\begin{pmatrix}10\\-6\end{pmatrix}.
  2. Subtract: -3-10=-13 and 12-(-6)=18.
  3. The result is \begin{pmatrix}-13\\18\end{pmatrix}.

Question 132 marks

a+\begin{pmatrix}4\\-7\end{pmatrix} = \begin{pmatrix}9\\2\end{pmatrix}. Find a. Enter the two components as (\text{horizontal},\text{vertical}).

Write your answer as (x, y)

Hint

Undo vector addition component by component.

Worked solution

a+\begin{pmatrix}4\\-7\end{pmatrix} = \begin{pmatrix}9\\2\end{pmatrix}. Find a. Enter the two components as (\text{horizontal},\text{vertical}).

  1. Subtract \begin{pmatrix}4\\-7\end{pmatrix} from both sides: a=\begin{pmatrix}9-4\\2-(-7)\end{pmatrix}.
  2. So a=\begin{pmatrix}5\\9\end{pmatrix}.

Question 142 marks

The arrow shows vector d. A new vector 2d starts at P. Give the coordinates of its endpoint.

Write your answer as (x, y)

Hint

Work out the vector displacement before applying it from P.

Worked solution

The arrow shows vector d. A new vector 2d starts at P. Give the coordinates of its endpoint.

  1. Vector d is \begin{pmatrix}3\\-2\end{pmatrix}, so 2d=\begin{pmatrix}6\\-4\end{pmatrix}.
  2. Starting at P(-3,3), move 6 right and 4 down.
  3. The endpoint is (-3+6,3-4)=(3,-1).
  4. The completed figure is shown below.
    A directed vector on a gridA square coordinate grid with one-unit intervals. x runs from -4 to 4; y runs from -3 to 5. P at (-3, 3), Q at (3, -1). The arrow labelled 2d starts at (-3, 3) and ends at (3, -1).−4−3−2−11234−3−2−112345xyOPQ2d
    The image uses the calculated result.

Question 152 marks

ABCD is a parallelogram. \overrightarrow{AB}=a and \overrightarrow{BC}=b. Write \overrightarrow{BD} in terms of a and b.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Go from B to A, then from A to D. Which side of the parallelogram is parallel to AD and the same length?

Worked solution

ABCD is a parallelogram. \overrightarrow{AB}=a and \overrightarrow{BC}=b. Write \overrightarrow{BD} in terms of a and b.

  1. \overrightarrow{BA}=-a, because it is \overrightarrow{AB} reversed.
  2. AD is parallel to BC and the same length, so \overrightarrow{AD}=\overrightarrow{BC}=b.
  3. \overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{AD}=-a+b.

Question 16Challenge3 marks

k\begin{pmatrix}2\\-3\end{pmatrix} + \begin{pmatrix}1\\t\end{pmatrix} = \begin{pmatrix}9\\-5\end{pmatrix}.

(a)

Find k.

1 mark

(b)

Find t.

2 marks

Hint

Use the top components to find k, then use it in the bottom components.

Worked solution

k\begin{pmatrix}2\\-3\end{pmatrix} + \begin{pmatrix}1\\t\end{pmatrix} = \begin{pmatrix}9\\-5\end{pmatrix}.

Part (a)

Find k.

  1. Top components give 2k+1=9.
  2. 2k=8, so k=4.

Part (b)

Find t.

  1. Bottom components give -3k+t=-5.
  2. With k=4, -12+t=-5, so t=7.

Question 17Challenge3 marks

3a-b=\begin{pmatrix}10\\-11\end{pmatrix} and b=\begin{pmatrix}2\\5\end{pmatrix}. Find a. Enter the two components as (\text{horizontal},\text{vertical}).

Write your answer as (x, y)

Hint

Undo subtraction first, then undo the scalar multiplication.

Worked solution

3a-b=\begin{pmatrix}10\\-11\end{pmatrix} and b=\begin{pmatrix}2\\5\end{pmatrix}. Find a. Enter the two components as (\text{horizontal},\text{vertical}).

  1. Add b to obtain 3a=\begin{pmatrix}12\\-6\end{pmatrix}.
  2. Divide both components by 3: a=\begin{pmatrix}4\\-2\end{pmatrix}.

Question 18Challenge3 marks

a=\begin{pmatrix}4\\-1\end{pmatrix} and b=\begin{pmatrix}-2\\3\end{pmatrix}. A learner says 2a-3b=\begin{pmatrix}2\\-11\end{pmatrix}.

(a)

Which component of the claimed result is wrong?

1 mark

(b)

Give the correct result as (horizontal, vertical).

2 marks

Write your answer as (x, y)

Hint

Calculate both components independently and compare them with the claim.

Worked solution

a=\begin{pmatrix}4\\-1\end{pmatrix} and b=\begin{pmatrix}-2\\3\end{pmatrix}. A learner says 2a-3b=\begin{pmatrix}2\\-11\end{pmatrix}.

Part (a)

Which component of the claimed result is wrong?

  1. The top component only
  2. The bottom component only
  3. Both components
  1. The top component should be 2(4)-3(-2)=8+6=14, not 2.
  2. The bottom component is 2(-1)-3(3)=-2-9=-11, as claimed.

Part (b)

Give the correct result as (horizontal, vertical).

  1. The corrected vector is \begin{pmatrix}14\\-11\end{pmatrix}.
  2. Enter (14,-11).

Question 19Challenge3 marks

Write \overrightarrow{AE} in terms of a and b. Simplify your answer.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Follow the arrows from A all the way to E.

Worked solution

Write \overrightarrow{AE} in terms of a and b. Simplify your answer.

  1. Add the four directed steps: a+b-2a+3b.
  2. Combine matching terms: \overrightarrow{AE}=-a+4b.

Question 20Challenge3 marks

ABCD is a parallelogram. O is the midpoint of AC. \overrightarrow{AB}=\begin{pmatrix}6\\2\end{pmatrix} and \overrightarrow{AD}=\begin{pmatrix}-2\\4\end{pmatrix}. Find \overrightarrow{AO}. Enter the two components as (\text{horizontal},\text{vertical}).

Write your answer as (x, y)

Hint

First find the whole diagonal vector using a route along two sides.

Worked solution

ABCD is a parallelogram. O is the midpoint of AC. \overrightarrow{AB}=\begin{pmatrix}6\\2\end{pmatrix} and \overrightarrow{AD}=\begin{pmatrix}-2\\4\end{pmatrix}. Find \overrightarrow{AO}. Enter the two components as (\text{horizontal},\text{vertical}).

  1. The diagonal vector is \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}.
  2. Opposite sides of a parallelogram give \overrightarrow{BC}=\overrightarrow{AD}, so \overrightarrow{AC}=\begin{pmatrix}4\\6\end{pmatrix}.
  3. O is the midpoint, so \overrightarrow{AO}=\dfrac12\overrightarrow{AC}=\begin{pmatrix}2\\3\end{pmatrix}.

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