Vectors, lines & planes (HL) – IB Maths exam questions with mark scheme

1CoreCalculatorHL only5 marks

Given $\mathbf a=\begin{pmatrix}2\\-1\\3\end{pmatrix}$, $\mathbf b=\begin{pmatrix}1\\4\\-2\end{pmatrix}$ and $\mathbf c=\begin{pmatrix}k\\2\\1\end{pmatrix}$:

(a) Find the angle between $\mathbf a$ and $\mathbf b$. [3]

(b) Given that $\mathbf a$ is perpendicular to $\mathbf c$, find $k$. [2]

$\mathbf a\cdot\mathbf b=2-4-6=-8$M1
$\cos\theta=\dfrac{-8}{\sqrt{14}\sqrt{21}}$M1
$\theta=118^\circ$ (obtuse; accept $117.8^\circ$ or $2.06$ radians)A1
$\mathbf a\cdot\mathbf c=0$: $2k-2+3=0$M1
$k=-\dfrac12$A1
Answer: (a) $118^\circ$ ($117.8^\circ$, or $2.06$ rad) (b) $k=-\tfrac12$
Examiner tip: A negative scalar product means an obtuse angle, so check your final angle agrees with its sign.

2StandardCalculatorHL only7 marks

Two lines have equations $L_1:\ \mathbf r=\begin{pmatrix}1\\2\\0\end{pmatrix}+s\begin{pmatrix}1\\1\\2\end{pmatrix}$ and $L_2:\ \mathbf r=\begin{pmatrix}0\\7\\7\end{pmatrix}+t\begin{pmatrix}1\\-1\\-1\end{pmatrix}$.

(a) Show that the lines intersect and find the coordinates of the point of intersection. [5]

(b) Find the acute angle between the lines. [2]

Equates components: $1+s=t$, $2+s=7-t$, $2s=7-t$ (two used to solve)M1
Solves two equations: $s=2$, $t=3$A1
Checks the third equation: $2(2)=4$ and $7-3=4$M1
Consistent, so the lines intersect (they are not parallel)R1
Point $(3,\,4,\,4)$A1
$\cos\theta=\dfrac{|1-1-2|}{\sqrt6\sqrt3}$ using the direction vectorsM1
$\theta=61.9^\circ$ (accept $1.08$ radians)A1
Answer: (a) $(3,4,4)$ (b) $61.9^\circ$ ($1.08$ rad)
Examiner tip: Always substitute into the third equation: two equations always give a solution, but only the third tells you whether the lines meet or are skew.

3StandardCalculatorHL only6 marks

The points $A$, $B$ and $C$ have coordinates $A(1,0,2)$, $B(3,1,1)$ and $C(2,4,0)$.

(a) Find $\overrightarrow{AB}$ and $\overrightarrow{AC}$. [2]

(b) Find $\overrightarrow{AB}\times\overrightarrow{AC}$. [2]

$\overrightarrow{AB}=\begin{pmatrix}2\\1\\-1\end{pmatrix}$A1
$\overrightarrow{AC}=\begin{pmatrix}1\\4\\-2\end{pmatrix}$A1
Cross product with at least two components correctM1
$\overrightarrow{AB}\times\overrightarrow{AC}=\begin{pmatrix}2\\3\\7\end{pmatrix}$A1
Area $=\tfrac12\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|=\tfrac12\sqrt{2^2+3^2+7^2}$ using the magnitude of their cross productM1
$\tfrac12\sqrt{62}=3.94$ (accept $\tfrac{\sqrt{62}}{2}$)A1
Answer: (a) $\overrightarrow{AB}=\begin{pmatrix}2\\1\\-1\end{pmatrix}$, $\overrightarrow{AC}=\begin{pmatrix}1\\4\\-2\end{pmatrix}$ (b) $\begin{pmatrix}2\\3\\7\end{pmatrix}$ (c) $\tfrac12\sqrt{62}\approx3.94$
Examiner tip: Subtract in the order (end $-$ start) and remember the factor $\tfrac12$: the magnitude of the cross product is the area of the parallelogram.

4HardCalculatorHL only7 marks

The plane $\Pi$ passes through the points $A(1,0,0)$, $B(0,2,0)$ and $C(0,0,3)$.

(a) Find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$. [1]

(b) Find a normal vector to $\Pi$ and hence show that the Cartesian equation of $\Pi$ is $6x+3y+2z=6$. [4]

$\overrightarrow{AB}=\begin{pmatrix}-1\\2\\0\end{pmatrix}$, $\overrightarrow{AC}=\begin{pmatrix}-1\\0\\3\end{pmatrix}$A1
Cross product $\overrightarrow{AB}\times\overrightarrow{AC}$ attemptedM1
Normal $=\begin{pmatrix}6\\3\\2\end{pmatrix}$ (any multiple)A1
Uses $\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n$, i.e. writes $6x+3y+2z=d$ and substitutes a point of $\Pi$ ($A$, $B$ or $C$)M1
$d=6(1)+3(0)+2(0)=6$ (or equivalent with $B$ or $C$), so $6x+3y+2z=6$ AGA1
$d=\dfrac{|6(1)+3(1)+2(1)-6|}{\sqrt{6^2+3^2+2^2}}$ (or: line through $P$ with direction $(6,3,2)$ meets $\Pi$ at $t=-\tfrac5{49}$, distance $=|t|\,|\mathbf n|$)M1
$d=\dfrac57\ (\approx0.714)$A1
Answer: (b) $6x+3y+2z=6$ (c) $\dfrac57$
Examiner tip: Substitute a point of the plane to find the constant on the right; check by testing $B$ and $C$ as well.

5HardCalculatorHL only7 marks

The line $L$ has equation $\mathbf r=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-1\end{pmatrix}$ and the plane $\Pi$ has equation $x+2y+2z=10$.

(a) Find the coordinates of the point where $L$ meets $\Pi$. [4]

(b) Find the acute angle between $L$ and $\Pi$. [3]

Substitutes $x=1+2\lambda,\ y=\lambda,\ z=2-\lambda$ into the planeM1
$(1+2\lambda)+2\lambda+2(2-\lambda)=10$, i.e. $2\lambda+5=10$A1
$\lambda=\dfrac52$A1
Point $\left(6,\ \tfrac52,\ -\tfrac12\right)$A1
Uses $\sin\phi=\dfrac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}$ with $\mathbf d=(2,1,-1)$, $\mathbf n=(1,2,2)$M1
$\sin\phi=\dfrac{2}{3\sqrt6}$A1
$\phi=15.8^\circ$ (accept $0.276$ radians)A1
Answer: (a) $(6,\,2.5,\,-0.5)$ (b) $15.8^\circ$ ($0.276$ rad)
Examiner tip: The angle with the plane uses $\sin$ of the direction/normal formula; using $\cos$ gives the angle to the normal instead.

6StretchCalculatorHL only8 marks

Two lines have equations $L_1:\ \mathbf r=\begin{pmatrix}1\\2\\3\end{pmatrix}+s\begin{pmatrix}2\\1\\0\end{pmatrix}$ and $L_2:\ \mathbf r=\begin{pmatrix}0\\1\\-1\end{pmatrix}+t\begin{pmatrix}1\\0\\1\end{pmatrix}$.

(a) Show that $L_1$ and $L_2$ do not intersect and are not parallel. [4]

(b) Find the shortest distance between the two lines. [4]

Direction vectors $(2,1,0)$ and $(1,0,1)$ are not scalar multiples, so not parallelR1
Equates components: $1+2s=t,\ 2+s=1,\ 3=-1+t$M1
$s=-1$ from the second and $t=4$ from the thirdA1
First equation: $1+2(-1)=-1\ne4$, inconsistent, so no intersection (skew)R1
Common perpendicular direction $\mathbf n=\mathbf d_1\times\mathbf d_2=\begin{pmatrix}1\\-2\\-1\end{pmatrix}$ (any multiple)M1
$\mathbf a_2-\mathbf a_1=(-1,-1,-4)$ and scalar product with $\mathbf n$: $-1+2+4=5$M1
Divides by $|\mathbf n|=\sqrt6$M1
Shortest distance $=\dfrac5{\sqrt6}=2.04$A1
Answer: $\dfrac5{\sqrt6}\approx2.04$
Examiner tip: Two lines in 3D that fail to meet are not automatically parallel: show both facts before claiming they are skew.

7HardCalculatorHL only8 marks

The line $L$ has equation $\mathbf r=\begin{pmatrix}0\\1\\1\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\2\end{pmatrix}$ and the point $P$ has coordinates $(2,1,4)$.

(a) Find the value of $\lambda$ at the point $F$ on $L$ where $PF$ is perpendicular to $L$. [4]

(b) Hence find the coordinates of $F$ and the shortest distance from $P$ to $L$. [3]

Writes $F$ as a general point on $L$, $F=(\lambda,\ 1+2\lambda,\ 1+2\lambda)$, and attempts $\overrightarrow{PF}$M1
$\overrightarrow{PF}=(\lambda-2,\ 2\lambda,\ 2\lambda-3)$ (accept $\overrightarrow{FP}$)A1
$\overrightarrow{PF}\cdot(1,2,2)=0$: $(\lambda-2)+4\lambda+2(2\lambda-3)=0$M1
$9\lambda=8$, $\lambda=\dfrac89$A1
$F=\left(\tfrac89,\ \tfrac{25}9,\ \tfrac{25}9\right)$ (ft their $\lambda$)A1
Distance $=|\overrightarrow{PF}|=\sqrt{(-\tfrac{10}9)^2+(\tfrac{16}9)^2+(-\tfrac{11}9)^2}$ (ft their $F$; accept $\dfrac{|\mathbf d\times\overrightarrow{AP}|}{|\mathbf d|}$ with $A(0,1,1)$, $\mathbf d=(1,2,2)$)M1
Distance $=\dfrac{\sqrt{477}}9=\dfrac{\sqrt{53}}3$ (accept $2.43$)A1
Reflection $=2\overrightarrow{OF}-\overrightarrow{OP}=\left(-\tfrac{2}9,\ \tfrac{41}9,\ \tfrac{14}9\right)$ (ft their $F$; accept $(-0.222,\ 4.56,\ 1.56)$)A1
Answer: (a) $\lambda=\tfrac89$ (b) $F=(\tfrac89,\tfrac{25}9,\tfrac{25}9)$, distance $=\tfrac{\sqrt{53}}3\approx2.43$ (c) $(-\tfrac29,\tfrac{41}9,\tfrac{14}9)$
Examiner tip: Set the scalar product of $\overrightarrow{PF}$ with the direction of the line to zero; students often use the position vector by mistake.

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