IB Maths AA HL: Vectors Practice Questions with Examiner's Mark Schemes

Seven vectors questions, easy to hard, mark scheme included

Vectors is one of the most mark-heavy topics on IB Maths AA HL Paper 1, and marks get lost less often through not knowing the method than through small, specific errors I see year after year as an examiner. If you want a refresher on the dot product, cross product, and 3D lines first, start with our Vectors in IB Maths HL AA guide — 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:

$$\mathbf{a}=\begin{pmatrix}2\\-1\\3\end{pmatrix} \quad \text{and} \quad \mathbf{b}=\begin{pmatrix}4\\p\\-2\end{pmatrix}, \quad p\in\mathbb{R}$$

(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 $\hat{\mathbf{a}}=\dfrac{1}{\sqrt{14}}\begin{pmatrix}2\\-1\\3\end{pmatrix}$ [A1]

Examiner's note: students frequently lose marks in part (b) through arithmetic errors squaring negative components under the root, or by forgetting that a "unit vector" requires dividing by the magnitude — leaving just $\sqrt{14}$ as a final answer scores no accuracy mark.

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) $\overrightarrow{AB}=\begin{pmatrix}2\\-2\\-1\end{pmatrix}$, $\overrightarrow{AC}=\begin{pmatrix}1\\2\\-2\end{pmatrix}$ [M1]
$\overrightarrow{AB}\times\overrightarrow{AC}=\begin{pmatrix}6\\3\\6\end{pmatrix}$ [A1]

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

Examiner's note: the most common mistake is omitting the $\tfrac{1}{2}$ factor in part (b), which gives the area of the parallelogram spanned by the two vectors rather than the triangle. Sign errors in the $j$-component of the cross product are the second most common slip.

Question 3 (Medium — Paper 1, 6 marks)

Two lines are defined by the vector equations:

$$L_1: \mathbf{r}_1=\begin{pmatrix}2\\0\\-1\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\1\end{pmatrix} \qquad L_2: \mathbf{r}_2=\begin{pmatrix}3\\2\\0\end{pmatrix}+\mu\begin{pmatrix}1\\-1\\1\end{pmatrix}$$

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]
$2+\lambda=3+\mu \Rightarrow \lambda-\mu=1$; $\quad 2\lambda=2-\mu \Rightarrow 2\lambda+\mu=2$; $\quad -1+\lambda=\mu$

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]

Examiner's note: candidates frequently lose the reasoning mark (R1) by solving two equations for $\lambda$ and $\mu$ and immediately writing down the intersection point, without checking that those values also satisfy the third equation. Without that check you cannot distinguish intersecting lines from skew lines.

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

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

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

Mark scheme

(a) Substitute $x=2+\lambda, y=1+2\lambda$ into $\Pi$: [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 $\mathbf{d}=\begin{pmatrix}1\\2\\2\end{pmatrix}$, normal $\mathbf{n}=\begin{pmatrix}3\\-4\\0\end{pmatrix}$: [M1]
$\sin\theta=\dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}=\dfrac{|-5|}{(3)(5)}=\dfrac{1}{3}$ [M1]
$\theta=\arcsin\left(\tfrac{1}{3}\right)\approx 19.5°$ [A1]

Examiner's note: the single biggest mark-loser on line–plane angle questions is using $\cos\theta$ and stopping there — that gives the angle between the line and the plane's normal (70.5°), not the angle between the line and the plane itself (19.5°). Angle between a line and a plane always uses $\sin\theta$.

Question 5 (Hard — Paper 1, 7 marks)

Two planes are given by $\Pi_1: x-y+z=4$ and $\Pi_2: 2x+y-z=5$. Find a vector equation of $L$, the line of intersection of $\Pi_1$ and $\Pi_2$, 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]
$\mathbf{n}_1=\begin{pmatrix}1\\-1\\1\end{pmatrix}$, $\mathbf{n}_2=\begin{pmatrix}2\\1\\-1\end{pmatrix}$
$\mathbf{b}=\begin{pmatrix}0\\3\\3\end{pmatrix}$, simplify to $\begin{pmatrix}0\\1\\1\end{pmatrix}$ [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 $\mathbf{a}=\begin{pmatrix}3\\0\\1\end{pmatrix}$

$\mathbf{r}=\begin{pmatrix}3\\0\\1\end{pmatrix}+t\begin{pmatrix}0\\1\\1\end{pmatrix}$ [A1]

Examiner's note: students often compute $\mathbf{n}_1\times\mathbf{n}_2$ correctly but stall on finding a point on the line. Fixing one coordinate (e.g. $x=3$) turns the two plane equations into a simple 2-variable system. A second common error is dropping "$\mathbf{r}=$" or the parameter $t$, leaving two disconnected vectors instead of a valid line equation.

Question 6 (Hard — Paper 1, 7 marks)

A plane $\Pi$ 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 $\Pi$. [2 marks]
(b) Find the coordinates of $F$, the foot of the perpendicular from $A$ to $\Pi$. [3 marks]
(c) Hence find the coordinates of $A'$, the reflection of $A$ in $\Pi$. [2 marks]

Mark scheme

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

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

Examiner's note: in part (c), candidates often confuse the foot of the perpendicular $F$ with the reflection $A'$ and stop there. Another frequent error is adding $\overrightarrow{AF}$ to $A$ instead of $F$, which just returns $F$ again rather than stepping all the way across the plane.

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:

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

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 $=\tfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|=\tfrac{1}{2}|(\mathbf{b}-\mathbf{a})\times(\mathbf{c}-\mathbf{a})|$: [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]
$= \mathbf{b}\times\mathbf{c}+\mathbf{a}\times\mathbf{b}+\mathbf{c}\times\mathbf{a}$
$\text{Area}=\tfrac{1}{2}|\mathbf{a}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}+\mathbf{c}\times\mathbf{a}|$ (proof complete) [A1]

Examiner's note: in formal vector proofs, students lose reasoning marks by not stating why terms simplify or change sign. Dropping $\mathbf{a}\times\mathbf{a}$ without stating $\mathbf{a}\times\mathbf{a}=\mathbf{0}$, or flipping cross-product order without citing anti-commutativity ($\mathbf{u}\times\mathbf{v}=-\mathbf{v}\times\mathbf{u}$), costs R marks on Paper 1 even when the final answer is correct.

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.

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