Which of the following vectors is NOT parallel to pmatrix 2 \ -3 pmatrix?

Picture the vectors: sketch the vectors Recall the key idea: recall the relevant rule Restate the question in your own words: What is being asked? Picture the vectors: sketch the vectors Recall the key idea: recall the relevant rule Restate the question in your own words: What is being asked? Recall the parallel test: pmatrix x \ y pmatrix = kpmatrix 2 \ -3 pmatrix Test the multiples: pmatrix 4 \ -6 pmatrix=2×, pmatrix -2 \ 3 pmatrix=-1×, pmatrix 6 \ -9 pmatrix=3× Test the odd one out: pmatrix 4 \ -5 pmatrix: (-5)/(4) (-3)/(2) Conclude: pmatrix 4 \ -5 pmatrix is not parallel Connect to earlier topics: link to prior learning State the units or form: check units / form Check the answer is sensible: does the answer look reasonable? Connect to earlier topics: link to prior learning State the units or form: check units / form Answer: pmatrix 4 \ -5 pmatrix

Points A(1,2) and B(7,10) are given. A point P has position vector pmatrix 1 \ 2 pmatrix + tpmatrix 6 \ 8 pmatrix. Find the value of t for which P is the midpoint of AB.

Picture the vectors: sketch the vectors Recall the key idea: recall the relevant rule Restate the question in your own words: What is being asked? Picture the vectors: sketch the vectors Recall the key idea: recall the relevant rule Restate the question in your own words: What is being asked? Recognise the path of P: P = A + t AB Find the midpoint of AB: M = ( (1+7)/(2), (2+10)/(2) ) = (4, 6) Write P's coordinates: P = (1 + 6t, 2 + 8t) Set the x-coordinates equal: 1 + 6t = 4 6t = 3 Solve for t: t = 12 Check with the y-coordinate: 2 + 8(12) = 2 + 4 = 6 Connect to earlier topics: link to prior learning State the units or form: check units / form Check the answer is sensible: does the answer look reasonable? Answer: t = 12

The vector a + b, where a=pmatrix 2 \ 6 pmatrix and b=pmatrix 1 \ 1 pmatrix, is parallel to the x-axis. Which statement gives the correct value of and reasoning?

Picture the vectors: sketch the vectors Recall the key idea: recall the relevant rule Restate the question in your own words: What is being asked? Picture the vectors: sketch the vectors Recall the key idea: recall the relevant rule Restate the question in your own words: What is being asked? Form the vector: a + b = pmatrix 2 + \ 6 + pmatrix State the condition for parallel to x-axis: y-component = 0 Set the y-component to zero: 6 + = 0 Solve: = -6 Connect to earlier topics: link to prior learning State the units or form: check units / form Check the answer is sensible: does the answer look reasonable? Connect to earlier topics: link to prior learning State the units or form: check units / form Answer: = -6, because the y-component 6+ must be zero for the vector to be horizontal

Arrange p=pmatrix 2 \ 2 pmatrix, q=pmatrix 3 \ 1 pmatrix, r=pmatrix 1 \ 4 pmatrix in order of increasing magnitude.

Picture the vectors: sketch the vectors Recall the key idea: recall the relevant rule Restate the question in your own words: What is being asked? Picture the vectors: sketch the vectors Recall the key idea: recall the relevant rule Restate the question in your own words: What is being asked? Find |p|: √(2^2 + 2^2) = √(8) = 2√(2) 2.83 Find |q|: √(3^2 + 1^2) = √(10) 3.16 Find |r|: √(1^2 + 4^2) = √(17) 4.12 Compare the values: √(8) < √(10) < √(17) Write the order: p, q, r Connect to earlier topics: link to prior learning State the units or form: check units / form Check the answer is sensible: does the answer look reasonable? Connect to earlier topics: link to prior learning Answer: p, q, r

Given a = pmatrix 3 \ 4 pmatrix and b = pmatrix 1 \ 0 pmatrix, find the value of t that minimises |a + tb|.

Picture the vectors: sketch the vectors Recall the key idea: recall the relevant rule Restate the question in your own words: What is being asked? Picture the vectors: sketch the vectors Recall the key idea: recall the relevant rule Restate the question in your own words: What is being asked? Form the combined vector: a + tb = pmatrix 3 + t \ 4 pmatrix Write the squared magnitude: |a+tb|^2 = (3+t)^2 + 4^2 Expand: = t^2 + 6t + 9 + 16 = t^2 + 6t + 25 Complete the square: = (t+3)^2 + 16 Identify the minimum: minimum when (t+3)^2 = 0 Solve for t: t = -3 Connect to earlier topics: link to prior learning State the units or form: check units / form Check the answer is sensible: does the answer look reasonable? Answer: t = -3

Hard A-Level Vector basics Questions

Challenging, exam-style A-Level Vector basics questions with worked solutions. Stretch yourself on the hardest unit-vector, scalar-multiplication, magnitude, collinear problems.

unit-vectorscalar-multiplicationmagnitudecollinearparallel-vectorsreasoning

A-Level34 questionsStep-by-step solutions

Question 1

5 markschallenging

Which of the following vectors is NOT parallel to

\begin{pmatrix} 2 \\ -3 \end{pmatrix}

Show worked solution

Worked solution

Picture the vectors
$\text{sketch the vectors}$
Drawing a quick sketch of the vectors as arrows makes the arithmetic easier to picture and helps catch mistakes.
Recall the key idea
$\text{recall the relevant rule}$
Bring to mind the vector rule needed here. Linking new work to a rule you already know builds confidence.
Restate the question in your own words
$\text{What is being asked?}$
Before starting, it helps to say clearly what the question wants. This keeps the working focused on the goal.
Picture the vectors
$\text{sketch the vectors}$
Drawing a quick sketch of the vectors as arrows makes the arithmetic easier to picture and helps catch mistakes.
Recall the key idea
$\text{recall the relevant rule}$
Bring to mind the vector rule needed here. Linking new work to a rule you already know builds confidence.
Restate the question in your own words
$\text{What is being asked?}$
Before starting, it helps to say clearly what the question wants. This keeps the working focused on the goal.
Recall the parallel test
$\begin{pmatrix} x \\ y \end{pmatrix} = k\begin{pmatrix} 2 \\ -3 \end{pmatrix}$
A vector is parallel when it is a scalar multiple of $\begin{pmatrix} 2 \\ -3 \end{pmatrix}$ , i.e. the ratio $y/x$ equals $-3/2$ .
Test the multiples
$\begin{pmatrix} 4 \\ -6 \end{pmatrix}=2\times,\ \begin{pmatrix} -2 \\ 3 \end{pmatrix}=-1\times,\ \begin{pmatrix} 6 \\ -9 \end{pmatrix}=3\times$
Each of these is a clean scalar multiple, so they are all parallel.
Test the odd one out
$\begin{pmatrix} 4 \\ -5 \end{pmatrix}:\ \frac{-5}{4} \ne \frac{-3}{2}$
The ratio does not match, so this vector is not a scalar multiple.
Conclude
$\begin{pmatrix} 4 \\ -5 \end{pmatrix}\ \text{is not parallel}$
This is the vector that fails the parallel test.
Connect to earlier topics
$\text{link to prior learning}$
Notice how this uses Pythagoras and coordinate geometry from earlier work. Spotting these links makes vectors feel familiar.
State the units or form
$\text{check units / form}$
Make sure the final answer is written in the correct form (vector, length, angle) with any units. Presentation earns marks.
Check the answer is sensible
$\text{does the answer look reasonable?}$
Finally, sanity-check the result against the diagram or rough estimates. A quick check catches slips before they cost marks.
Connect to earlier topics
$\text{link to prior learning}$
Notice how this uses Pythagoras and coordinate geometry from earlier work. Spotting these links makes vectors feel familiar.
State the units or form
$\text{check units / form}$
Make sure the final answer is written in the correct form (vector, length, angle) with any units. Presentation earns marks.

Answer

\begin{pmatrix} 4 \\ -5 \end{pmatrix}

Question 2

6 markschallenging

Points

A(1,2)

and

B(7,10)

are given. A point

P

has position vector

\begin{pmatrix} 1 \\ 2 \end{pmatrix} + t\begin{pmatrix} 6 \\ 8 \end{pmatrix}

. Find the value of

t

for which

P

is the midpoint of

AB

Show worked solution

Worked solution

Picture the vectors
$\text{sketch the vectors}$
Drawing a quick sketch of the vectors as arrows makes the arithmetic easier to picture and helps catch mistakes.
Recall the key idea
$\text{recall the relevant rule}$
Bring to mind the vector rule needed here. Linking new work to a rule you already know builds confidence.
Restate the question in your own words
$\text{What is being asked?}$
Before starting, it helps to say clearly what the question wants. This keeps the working focused on the goal.
Picture the vectors
$\text{sketch the vectors}$
Drawing a quick sketch of the vectors as arrows makes the arithmetic easier to picture and helps catch mistakes.
Recall the key idea
$\text{recall the relevant rule}$
Bring to mind the vector rule needed here. Linking new work to a rule you already know builds confidence.
Restate the question in your own words
$\text{What is being asked?}$
Before starting, it helps to say clearly what the question wants. This keeps the working focused on the goal.
Recognise the path of P
$P = A + t\,\overrightarrow{AB}$
Since $\begin{pmatrix} 6 \\ 8 \end{pmatrix} = \overrightarrow{AB}$ , the point $P$ moves along the line from $A$ to $B$ as $t$ varies.
Find the midpoint of AB
$M = \left( \frac{1+7}{2},\ \frac{2+10}{2} \right) = (4,\ 6)$
Average the coordinates of $A$ and $B$ to find the midpoint.
Write P's coordinates
$P = (1 + 6t,\ 2 + 8t)$
Expand the position vector of $P$ into coordinate form.
Set the x-coordinates equal
$1 + 6t = 4 \ \Rightarrow\ 6t = 3$
For $P$ to be the midpoint, its coordinates must match $M$ .
Solve for t
$t = \tfrac{1}{2}$
Divide to find $t$ .
Check with the y-coordinate
$2 + 8(\tfrac12) = 2 + 4 = 6 \ \checkmark$
The $y$ -coordinate also gives $6$ , confirming $t=\tfrac12$ is correct. This makes sense: the midpoint is halfway along.
Connect to earlier topics
$\text{link to prior learning}$
Notice how this uses Pythagoras and coordinate geometry from earlier work. Spotting these links makes vectors feel familiar.
State the units or form
$\text{check units / form}$
Make sure the final answer is written in the correct form (vector, length, angle) with any units. Presentation earns marks.
Check the answer is sensible
$\text{does the answer look reasonable?}$
Finally, sanity-check the result against the diagram or rough estimates. A quick check catches slips before they cost marks.

Answer

t = \tfrac{1}{2}

Question 3

5 markschallenging

The vector

\mathbf{a} + \lambda\mathbf{b}

, where

\mathbf{a}=\begin{pmatrix} 2 \\ 6 \end{pmatrix}

and

\mathbf{b}=\begin{pmatrix} 1 \\ 1 \end{pmatrix}

, is parallel to the

x

-axis. Which statement gives the correct value of

\lambda

and reasoning?

Show worked solution

Worked solution

Picture the vectors
$\text{sketch the vectors}$
Drawing a quick sketch of the vectors as arrows makes the arithmetic easier to picture and helps catch mistakes.
Recall the key idea
$\text{recall the relevant rule}$
Bring to mind the vector rule needed here. Linking new work to a rule you already know builds confidence.
Restate the question in your own words
$\text{What is being asked?}$
Before starting, it helps to say clearly what the question wants. This keeps the working focused on the goal.
Picture the vectors
$\text{sketch the vectors}$
Drawing a quick sketch of the vectors as arrows makes the arithmetic easier to picture and helps catch mistakes.
Recall the key idea
$\text{recall the relevant rule}$
Bring to mind the vector rule needed here. Linking new work to a rule you already know builds confidence.
Restate the question in your own words
$\text{What is being asked?}$
Before starting, it helps to say clearly what the question wants. This keeps the working focused on the goal.
Form the vector
$\mathbf{a} + \lambda\mathbf{b} = \begin{pmatrix} 2 + \lambda \\ 6 + \lambda \end{pmatrix}$
Add $\lambda$ lots of $\mathbf{b}$ to $\mathbf{a}$ , leaving $\lambda$ unknown.
State the condition for parallel to x-axis
$\text{y-component} = 0$
A vector parallel to the $x$ -axis has no vertical part, so its $y$ -component must be zero.
Set the y-component to zero
$6 + \lambda = 0$
Apply the condition to the vertical component.
Solve
$\lambda = -6$
This makes the vector $\begin{pmatrix} -4 \\ 0 \end{pmatrix}$ , which is horizontal.
Connect to earlier topics
$\text{link to prior learning}$
Notice how this uses Pythagoras and coordinate geometry from earlier work. Spotting these links makes vectors feel familiar.
State the units or form
$\text{check units / form}$
Make sure the final answer is written in the correct form (vector, length, angle) with any units. Presentation earns marks.
Check the answer is sensible
$\text{does the answer look reasonable?}$
Finally, sanity-check the result against the diagram or rough estimates. A quick check catches slips before they cost marks.
Connect to earlier topics
$\text{link to prior learning}$
Notice how this uses Pythagoras and coordinate geometry from earlier work. Spotting these links makes vectors feel familiar.
State the units or form
$\text{check units / form}$
Make sure the final answer is written in the correct form (vector, length, angle) with any units. Presentation earns marks.

Answer

\lambda = -6

, because the

y

-component

6+\lambda

must be zero for the vector to be horizontal

Question 4

5 markschallenging

Arrange

\mathbf{p}=\begin{pmatrix} 2 \\ 2 \end{pmatrix}

\mathbf{q}=\begin{pmatrix} 3 \\ 1 \end{pmatrix}

\mathbf{r}=\begin{pmatrix} 1 \\ 4 \end{pmatrix}

in order of increasing magnitude.

Show worked solution

Worked solution

Picture the vectors
$\text{sketch the vectors}$
Drawing a quick sketch of the vectors as arrows makes the arithmetic easier to picture and helps catch mistakes.
Recall the key idea
$\text{recall the relevant rule}$
Bring to mind the vector rule needed here. Linking new work to a rule you already know builds confidence.
Restate the question in your own words
$\text{What is being asked?}$
Before starting, it helps to say clearly what the question wants. This keeps the working focused on the goal.
Picture the vectors
$\text{sketch the vectors}$
Drawing a quick sketch of the vectors as arrows makes the arithmetic easier to picture and helps catch mistakes.
Recall the key idea
$\text{recall the relevant rule}$
Bring to mind the vector rule needed here. Linking new work to a rule you already know builds confidence.
Restate the question in your own words
$\text{What is being asked?}$
Before starting, it helps to say clearly what the question wants. This keeps the working focused on the goal.
Find |p|
$\sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \approx 2.83$
Work out each magnitude, simplifying surds and using decimals to compare.
Find |q|
$\sqrt{3^2 + 1^2} = \sqrt{10} \approx 3.16$
Continue with the second vector.
Find |r|
$\sqrt{1^2 + 4^2} = \sqrt{17} \approx 4.12$
And the third vector.
Compare the values
$\sqrt{8} < \sqrt{10} < \sqrt{17}$
Since $8 < 10 < 17$ , the square roots are in the same order.
Write the order
$\mathbf{p},\ \mathbf{q},\ \mathbf{r}$
Replace each magnitude with its vector to give the increasing order.
Connect to earlier topics
$\text{link to prior learning}$
Notice how this uses Pythagoras and coordinate geometry from earlier work. Spotting these links makes vectors feel familiar.
State the units or form
$\text{check units / form}$
Make sure the final answer is written in the correct form (vector, length, angle) with any units. Presentation earns marks.
Check the answer is sensible
$\text{does the answer look reasonable?}$
Finally, sanity-check the result against the diagram or rough estimates. A quick check catches slips before they cost marks.
Connect to earlier topics
$\text{link to prior learning}$
Notice how this uses Pythagoras and coordinate geometry from earlier work. Spotting these links makes vectors feel familiar.

Answer

\mathbf{p},\ \mathbf{q},\ \mathbf{r}

Question 5

6 markschallenging

Given

\mathbf{a} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}

and

\mathbf{b} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}

, find the value of

t

that minimises

|\mathbf{a} + t\mathbf{b}|

Show worked solution

Worked solution

Picture the vectors
$\text{sketch the vectors}$
Drawing a quick sketch of the vectors as arrows makes the arithmetic easier to picture and helps catch mistakes.
Recall the key idea
$\text{recall the relevant rule}$
Bring to mind the vector rule needed here. Linking new work to a rule you already know builds confidence.
Restate the question in your own words
$\text{What is being asked?}$
Before starting, it helps to say clearly what the question wants. This keeps the working focused on the goal.
Picture the vectors
$\text{sketch the vectors}$
Drawing a quick sketch of the vectors as arrows makes the arithmetic easier to picture and helps catch mistakes.
Recall the key idea
$\text{recall the relevant rule}$
Bring to mind the vector rule needed here. Linking new work to a rule you already know builds confidence.
Restate the question in your own words
$\text{What is being asked?}$
Before starting, it helps to say clearly what the question wants. This keeps the working focused on the goal.
Form the combined vector
$\mathbf{a} + t\mathbf{b} = \begin{pmatrix} 3 + t \\ 4 \end{pmatrix}$
Add $t$ lots of $\mathbf{b}$ to $\mathbf{a}$ . Only the $x$ -component changes because $\mathbf{b}$ is horizontal.
Write the squared magnitude
$|\mathbf{a}+t\mathbf{b}|^2 = (3+t)^2 + 4^2$
Minimising the length is the same as minimising its square, which avoids the square root.
Expand
$= t^2 + 6t + 9 + 16 = t^2 + 6t + 25$
Multiply out and collect the constant terms.
Complete the square
$= (t+3)^2 + 16$
Rewriting in completed-square form shows the smallest possible value.
Identify the minimum
$\text{minimum when } (t+3)^2 = 0$
A square is smallest (zero) when its bracket is zero.
Solve for t
$t = -3$
Setting $t+3=0$ gives the minimising value. (The minimum length is then $\sqrt{16}=4$ .)
Connect to earlier topics
$\text{link to prior learning}$
Notice how this uses Pythagoras and coordinate geometry from earlier work. Spotting these links makes vectors feel familiar.
State the units or form
$\text{check units / form}$
Make sure the final answer is written in the correct form (vector, length, angle) with any units. Presentation earns marks.
Check the answer is sensible
$\text{does the answer look reasonable?}$
Finally, sanity-check the result against the diagram or rough estimates. A quick check catches slips before they cost marks.

Answer

t = -3

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