IB Maths AA HL: Vectors Practice Questions with Full Mark Schemes

Seven vectors questions, easy to hard, mark scheme included

In IB Maths AA HL vectors, marks get lost less often through not knowing the method than through small, specific errors that come up again and again. If you want a refresher on the dot product, cross product, and 3D lines first, start with our HL AA 3D vectors notes — this post is pure practice: seven exam-style questions, graded easy to hard, each with a full IB-style mark scheme and an examiner's note on the specific mistake that costs the most marks.

Question 1 (Easy — Paper 1, 4 marks)

The vectors $\mathbf{a}$ and $\mathbf{b}$ are given by:

$$\mathbf{a}=\left(\begin{array}{c}2\\ -1\\ 3\end{array}\right)\text{and}\mathbf{b}=\left(\begin{array}{c}4\\ p\\ -2\end{array}\right),p\in \mathbb{R}$$

(a) Given that $\mathbf{a}$ and $\mathbf{b}$ are perpendicular, find the value of $p$. [2 marks]
(b) Find a unit vector in the direction of $\mathbf{a}$. [2 marks]

Mark scheme

(a) Perpendicular vectors have a scalar product of zero, $\mathbf{a}\cdot \mathbf{b}=0$: [M1]
$(2)(4)+(-1)(p)+(3)(-2)=0\Rightarrow 8-p-6=0\Rightarrow p=2$ [A1]

(b) $|\mathbf{a}|=\sqrt{2^2+(-1{)}^2+3^2}=\sqrt{14}$ [M1]
Unit vector $\hat{\mathbf{a}}={\displaystyle \frac{1}{\sqrt{14}}}\left(\begin{array}{c}2\\ -1\\ 3\end{array}\right)$ [A1]

Question 2 (Easy–Medium — Paper 1, 5 marks)

Three points in 3D space have coordinates $A(1,2,3)$, $B(3,0,2)$, and $C(2,4,1)$.

(a) Find the vector product $\overrightarrow{AB}\times \overrightarrow{AC}$. [3 marks]
(b) Hence, calculate the exact area of triangle $ABC$. [2 marks]

Mark scheme

(a) $\overrightarrow{AB}=\left(\begin{array}{c}2\\ -2\\ -1\end{array}\right)$, $\overrightarrow{AC}=\left(\begin{array}{c}1\\ 2\\ -2\end{array}\right)$ [M1]
$\overrightarrow{AB}\times \overrightarrow{AC}=\left(\begin{array}{c}6\\ 3\\ 6\end{array}\right)$ [A1]

(b) Area $=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$: [M1]
$|\overrightarrow{AB}\times \overrightarrow{AC}|=\sqrt{6^2+3^2+6^2}=\sqrt{81}=9$
Area $=\frac{1}{2}\times 9=4.5$ square units [A1]

Question 3 (Medium — Paper 1, 6 marks)

Two lines are defined by the vector equations:

$$L_1:{\mathbf{r}}_1=\left(\begin{array}{c}2\\ 0\\ -1\end{array}\right)+\lambda \left(\begin{array}{c}1\\ 2\\ 1\end{array}\right)L_2:{\mathbf{r}}_2=\left(\begin{array}{c}3\\ 2\\ 0\end{array}\right)+\mu \left(\begin{array}{c}1\\ -1\\ 1\end{array}\right)$$

Show algebraically that $L_1$ and $L_2$ intersect, and find the coordinates of their point of intersection. [6 marks]

Mark scheme

Equate parametric equations for $x,y,z$: [M1]
$2+\lambda =3+\mu \Rightarrow \lambda -\mu =1$; $2\lambda =2-\mu \Rightarrow 2\lambda +\mu =2$; $-1+\lambda =\mu$

Solving the first two equations: $\lambda =1$, $\mu =0$ [M1 A1]
Check consistency in the third equation: [R1] $-1+(1)=0=\mu$ ✓ — consistent, so the lines intersect.
Substitute back: intersection point is $(3,2,0)$ [M1 A1]

Question 4 (Medium–Hard — Paper 2, 6 marks)

A line $L$ and a plane $\mathrm{\Pi }$ are defined by $L:\mathbf{r}=\left(\begin{array}{c}2\\ 1\\ 0\end{array}\right)+\lambda \left(\begin{array}{c}1\\ 2\\ 2\end{array}\right)$, $\lambda \in \mathbb{R}$, and $\mathrm{\Pi }:3x-4y=10$.

(a) Find the coordinates of the point of intersection of $L$ and $\mathrm{\Pi }$. [3 marks]
(b) Calculate the acute angle between $L$ and $\mathrm{\Pi }$, correct to one decimal place. [3 marks]

Mark scheme

(a) Substitute $x=2+\lambda ,y=1+2\lambda$ into $\mathrm{\Pi }$: [M1]
$3(2+\lambda )-4(1+2\lambda )=10\Rightarrow 2-5\lambda =10\Rightarrow \lambda =-1.6$ [A1]
Point: $(0.4,-2.2,-3.2)$ [A1]

(b) Direction $\mathbf{d}=\left(\begin{array}{c}1\\ 2\\ 2\end{array}\right)$, normal $\mathbf{n}=\left(\begin{array}{c}3\\ -4\\ 0\end{array}\right)$: [M1]
$\sin \theta ={\displaystyle \frac{|\mathbf{d}\cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}}={\displaystyle \frac{|-5|}{(3)(5)}}={\displaystyle \frac{1}{3}}$ [M1]
$\theta =\arcsin \left(\frac{1}{3}\right)\approx 19.5°$ [A1]

Question 5 (Hard — Paper 1, 7 marks)

Two planes are given by ${\mathrm{\Pi }}_1:x-y+z=4$ and ${\mathrm{\Pi }}_2:2x+y-z=5$. Find a vector equation of $L$, the line of intersection of ${\mathrm{\Pi }}_1$ and ${\mathrm{\Pi }}_2$, in the form $\mathbf{r}=\mathbf{a}+t\mathbf{b}$, $t\in \mathbb{R}$. [7 marks]

Mark scheme

Direction $\mathbf{b}$ is perpendicular to both normals, so $\mathbf{b}={\mathbf{n}}_1\times {\mathbf{n}}_2$: [M1]
${\mathbf{n}}_1=\left(\begin{array}{c}1\\ -1\\ 1\end{array}\right)$, ${\mathbf{n}}_2=\left(\begin{array}{c}2\\ 1\\ -1\end{array}\right)$
$\mathbf{b}=\left(\begin{array}{c}0\\ 3\\ 3\end{array}\right)$, simplify to $\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)$ [M1 A1]

Fix $x=3$ to find a point: $-y+z=1$ and $y-z=-1$ (same equation) — choose $y=0\Rightarrow z=1$: [M1 A1]
Point $\mathbf{a}=\left(\begin{array}{c}3\\ 0\\ 1\end{array}\right)$

$\mathbf{r}=\left(\begin{array}{c}3\\ 0\\ 1\end{array}\right)+t\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)$ [A1]

Question 6 (Hard — Paper 1, 7 marks)

A plane $\mathrm{\Pi }$ has Cartesian equation $x+y+z=1$, and $A$ is the point $(2,3,-1)$.

(a) Find a vector equation of the line $L$ through $A$, perpendicular to $\mathrm{\Pi }$. [2 marks]
(b) Find the coordinates of $F$, the foot of the perpendicular from $A$ to $\mathrm{\Pi }$. [3 marks]
(c) Hence find the coordinates of $A^{\prime }$, the reflection of $A$ in $\mathrm{\Pi }$. [2 marks]

Mark scheme

(a) Normal to $\mathrm{\Pi }$ is $\mathbf{n}=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$: [A1]
$L:\mathbf{r}=\left(\begin{array}{c}2\\ 3\\ -1\end{array}\right)+\lambda \left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$ [A1]

(b) Substitute into $\mathrm{\Pi }$: $(2+\lambda )+(3+\lambda )+(-1+\lambda )=1\Rightarrow \lambda =-1$ [M1 A1]
$F=(1,2,-2)$ [A1]

(c) $\overrightarrow{O{A}^{\prime }}=2\overrightarrow{OF}-\overrightarrow{OA}$: [M1]
$A^{\prime }=(0,1,-3)$ [A1]

Question 7 (Very hard / proof — Paper 1, 7 marks)

Let $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ be the position vectors of the vertices $A$, $B$, $C$ of a triangle relative to an origin $O$.

(a) Show that $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ and $\overrightarrow{AC}=\mathbf{c}-\mathbf{a}$. [1 mark]
(b) Using properties of the vector product, prove that the area of triangle $ABC$ is given by:

$$\text{Area}=\frac{1}{2}|\mathbf{a}\times \mathbf{b}+\mathbf{b}\times \mathbf{c}+\mathbf{c}\times \mathbf{a}|\text{[6 marks]}$$

Mark scheme

(a) $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\mathbf{b}-\mathbf{a}$, $\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=\mathbf{c}-\mathbf{a}$ [A1]

(b) Area $=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|=\frac{1}{2}|(\mathbf{b}-\mathbf{a})\times (\mathbf{c}-\mathbf{a})|$: [M1]
Expand: $(\mathbf{b}-\mathbf{a})\times (\mathbf{c}-\mathbf{a})=\mathbf{b}\times \mathbf{c}-\mathbf{b}\times \mathbf{a}-\mathbf{a}\times \mathbf{c}+\mathbf{a}\times \mathbf{a}$ [M1 A1]
Apply $\mathbf{a}\times \mathbf{a}=\mathbf{0}$: [R1]
Apply anti-commutativity, $-\mathbf{b}\times \mathbf{a}=\mathbf{a}\times \mathbf{b}$ and $-\mathbf{a}\times \mathbf{c}=\mathbf{c}\times \mathbf{a}$: [R1]
$=\mathbf{b}\times \mathbf{c}+\mathbf{a}\times \mathbf{b}+\mathbf{c}\times \mathbf{a}$
$\text{Area}=\frac{1}{2}|\mathbf{a}\times \mathbf{b}+\mathbf{b}\times \mathbf{c}+\mathbf{c}\times \mathbf{a}|$ (proof complete) [A1]

Where to go from here

If any of these seven caught you out, that is normal — vectors rewards practising the specific traps above until checking for them becomes automatic. The full practice engine on this site has hundreds more vectors questions marked against real IB schemes, with instant feedback on exactly where marks were dropped.