Vectors, scalar product and vector equations of lines – 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.

3StretchCalculatorHL 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.

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4HardCalculatorHL 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.

5CoreNon-calculatorHL only7 marks

The points $A$ and $B$ have coordinates $A(3,\,-2,\,5)$ and $B(9,\,1,\,-1)$.

(a) Find $\overrightarrow{AB}$. [1]

(b) Find a unit vector in the direction of $\overrightarrow{AB}$. [2]

The vector $\mathbf c=\begin{pmatrix}m+2\\ m\\ -4\end{pmatrix}$ is parallel to $\overrightarrow{AB}$.

(d) Find the value of $m$. [3]

$\overrightarrow{AB}=\begin{pmatrix}6\\3\\-6\end{pmatrix}$ (accept $6\mathbf i+3\mathbf j-6\mathbf k$)A1
Attempt to find magnitude: $\sqrt{6^2+3^2+(-6)^2}=9$ (follow through their $\overrightarrow{AB}$)M1
$\dfrac19\begin{pmatrix}6\\3\\-6\end{pmatrix}=\begin{pmatrix}2/3\\1/3\\-2/3\end{pmatrix}$ (accept unsimplified $\tfrac{1}{9}(6,3,-6)$)A1
$\begin{pmatrix}-10\\-5\\10\end{pmatrix}$ (ft $-15\times$ their unit vector; A0 if direction not reversed)A1
Sets $\mathbf c=k\overrightarrow{AB}$ (or uses equal ratios of components), e.g. $-4=-6k$M1
$k=\tfrac23$ (or a correct ratio equation such as $\dfrac{m}{3}=\dfrac{-4}{-6}$)A1
$m=2$ (check: $m+2=4=\tfrac23\times6$ consistent) — A0 for a value that fits only one componentA1
Answer: (a) $(6,3,-6)$ (b) $\left(\tfrac23,\tfrac13,-\tfrac23\right)$ (c) $(-10,-5,10)$ (d) $m=2$
Examiner tip: In (c) students often scale up correctly but forget to reverse the direction; 'opposite direction' needs the minus sign.

6StandardCalculatorHL only9 marks

The line $L$ has Cartesian equation $$\frac{x-4}{2}=\frac{y+1}{-3}=\frac{z-2}{6}.$$

(a) Write down a vector equation of $L$. [2]

(b) Show that the point $P(10,\,-10,\,20)$ lies on $L$. [2]

(d) Find the acute angle between $L$ and the $z$-axis. [2]

Position vector of a point on $L$, e.g. $\begin{pmatrix}4\\-1\\2\end{pmatrix}$ (sign of $y$-component must be $-1$)A1
Correct complete equation $\mathbf r=\begin{pmatrix}4\\-1\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\-3\\6\end{pmatrix}$ (must include $\mathbf r=$ and a parameter; accept any correct equivalent)A1
Substitutes $P$ into the equation, e.g. $\dfrac{10-4}{2}=\dfrac{-10+1}{-3}=\dfrac{20-2}{6}$ or solves $4+2\lambda=10$M1
All three give $3$ (or $\lambda=3$ checked in all three components), so $P$ lies on $L$A1
Sets $z$-component to zero: $2+6\lambda=0$M1
$\lambda=-\tfrac13$A1
$\left(\tfrac{10}{3},\,0,\,0\right)$ (accept $(3.33,0,0)$)A1
Uses scalar product of $\begin{pmatrix}2\\-3\\6\end{pmatrix}$ with $\begin{pmatrix}0\\0\\1\end{pmatrix}$: $\cos\theta=\dfrac{6}{7\times1}$M1
$\theta=31.0^\circ$ (accept $0.541$ rad)A1
Answer: (a) $\mathbf r=(4,-1,2)+\lambda(2,-3,6)$ (b) $\lambda=3$ in every component (c) $\left(\tfrac{10}{3},0,0\right)$ (d) $31.0^\circ$
Examiner tip: The commonest slip in (a) is reading the point as $(4,1,2)$: $y+1$ means the $y$-coordinate is $-1$.

7StandardCalculatorHL only9 marks

In this question, distances are in kilometres and time $t$ is in hours after 12:00. The position vector of ship $A$ relative to a harbour at the origin is $$\mathbf r_A=\begin{pmatrix}2\\-3\end{pmatrix}+t\begin{pmatrix}4\\3\end{pmatrix}.$$

(a) Find the speed of ship $A$. [2]

A lighthouse is at the point $H(14,\,1)$.

(b) Find the time at which ship $A$ is closest to $H$, and the distance between them at that time. [4]

Ship $B$ has position vector $\mathbf r_B=\begin{pmatrix}20\\-6\end{pmatrix}+t\begin{pmatrix}-3\\6\end{pmatrix}$.

Attempt to find magnitude of the velocity vector: $\sqrt{4^2+3^2}$M1
$5\ \text{km h}^{-1}$A1
$\overrightarrow{HA}=\begin{pmatrix}4t-12\\3t-4\end{pmatrix}$ and sets $\overrightarrow{HA}\cdot\begin{pmatrix}4\\3\end{pmatrix}=0$ (or minimises $|\overrightarrow{HA}|^2=25t^2-120t+160$, including by GDC)M1
$t=2.4$, i.e. at 14:24 (accept $t=2.4$ hours)A1
Substitutes their $t$ to find $\overrightarrow{HA}=\begin{pmatrix}-2.4\\3.2\end{pmatrix}$ and its magnitudeM1
Distance $=4$ kmA1
Equates the paths using different parameters: $2+4t=20-3s$ and $-3+3t=-6+6s$M1
$t=3,\ s=2$, giving crossing point $(14,\,6)$A1
Ship $A$ reaches $(14,6)$ at $t=3$ but ship $B$ is there at $t=2$; different times so they do not collide (R1 needs the two times compared)R1
Answer: (a) $5$ km/h (b) $t=2.4$ (14:24), $4$ km (c) paths meet at $(14,6)$ at $t=3$ for $A$ and $t=2$ for $B$
Examiner tip: In (c), using the same letter $t$ for both ships forces the times to be equal and finds no crossing point; use a separate parameter to test whether the paths cross.

8HardNon-calculatorHL only9 marks

The lines $L_1$ and $L_2$ have equations $$L_1:\ \mathbf r=\begin{pmatrix}1\\-2\\3\end{pmatrix}+s\begin{pmatrix}2\\1\\-1\end{pmatrix},\qquad L_2:\ \mathbf r=\begin{pmatrix}-1\\-3\\4\end{pmatrix}+t\begin{pmatrix}-4\\-2\\2\end{pmatrix}.$$

(a) Show that $L_1$ and $L_2$ are coincident. [3]

The line $L_3$ has equation $\mathbf r=\begin{pmatrix}-1\\6\\a\end{pmatrix}+u\begin{pmatrix}1\\-1\\2\end{pmatrix}$, where $a\in\mathbb R$. The lines $L_1$ and $L_3$ intersect.

(b) Find the value of $a$ and the coordinates of the point of intersection. [4]

$\begin{pmatrix}-4\\-2\\2\end{pmatrix}=-2\begin{pmatrix}2\\1\\-1\end{pmatrix}$, so the lines are parallelR1
Attempt to show a point of $L_2$ lies on $L_1$, e.g. $1+2s=-1,\ -2+s=-3,\ 3-s=4$M1
$s=-1$ satisfies all three equations, so $(-1,-3,4)$ is on both lines; parallel with a common point $\Rightarrow$ coincidentA1
Equates $L_1$ and $L_3$ components with different parameters: $1+2s=-1+u,\ -2+s=6-u,\ 3-s=a+2u$M1
Solves the first two equations: $s=2,\ u=6$A1
$a=-11$A1
Point of intersection $(5,\,0,\,1)$A1
Direction vectors $\begin{pmatrix}2\\1\\-1\end{pmatrix}$ and $\begin{pmatrix}1\\-1\\2\end{pmatrix}$ are not scalar multiples, so the lines are not parallel (and they only meet when $a=-11$)R1
Lines are skew (A1 dependent on R1)A1
Answer: (a) parallel directions and common point $(-1,-3,4)$ (b) $a=-11$, $(5,0,1)$ (c) skew
Examiner tip: Parallel direction vectors only prove parallel: to earn the final mark in (a) you must also show a shared point; 'not intersecting' alone is not enough for 'skew' in (c).

9HardCalculatorHL only11 marks

The diagram shows a parallelogram $OACB$ with $\overrightarrow{OA}=\mathbf a$ and $\overrightarrow{OB}=\mathbf b$, where $|\mathbf a|=3$, $|\mathbf b|=4$ and $A\hat{O}B=60^\circ$. The diagonals $[OC]$ and $[AB]$ are shown dashed.

(a) Find $\mathbf a\cdot\mathbf b$. [2]

(b) Find the exact length of the diagonal $[OC]$. [3]

(d) Show that if instead $|\mathbf a|=|\mathbf b|$, the diagonals are perpendicular for any angle $A\hat{O}B$. [2]

$\mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos60^\circ=3\times4\times\tfrac12$M1
$\mathbf a\cdot\mathbf b=6$A1
$\overrightarrow{OC}=\mathbf a+\mathbf b$ and $|\mathbf a+\mathbf b|^2=\mathbf a\cdot\mathbf a+2\,\mathbf a\cdot\mathbf b+\mathbf b\cdot\mathbf b$ (or cosine rule in triangle $OAC$ with angle $120^\circ$)M1
$=9+12+16=37$A1
$OC=\sqrt{37}$ (A0 for $6.08$ only, as exact answer required)A1
$\overrightarrow{AB}=\mathbf b-\mathbf a$ and $|\mathbf b-\mathbf a|^2=9-12+16$ giving $|\overrightarrow{AB}|=\sqrt{13}$M1
$(\mathbf a+\mathbf b)\cdot(\mathbf b-\mathbf a)=|\mathbf b|^2-|\mathbf a|^2=7$A1
Uses $\cos\theta=\dfrac{7}{\sqrt{37}\sqrt{13}}$ (accept $-7$ in the numerator leading to the obtuse angle $108.6^\circ$)M1
$\theta=71.4^\circ$ (accept $1.25$ rad; A0 for $108.6^\circ$ as final answer)A1
$(\mathbf a+\mathbf b)\cdot(\mathbf b-\mathbf a)=\mathbf b\cdot\mathbf b-\mathbf a\cdot\mathbf a$ (cross terms cancel, using $\mathbf a\cdot\mathbf b=\mathbf b\cdot\mathbf a$)M1
$=|\mathbf b|^2-|\mathbf a|^2=0$ when $|\mathbf a|=|\mathbf b|$, so the diagonals are perpendicular (independent of the angle)R1
Answer: (a) $6$ (b) $\sqrt{37}$ (c) $71.4^\circ$ (d) $(\mathbf a+\mathbf b)\cdot(\mathbf b-\mathbf a)=|\mathbf b|^2-|\mathbf a|^2=0$
Examiner tip: In (b) and (c) the $2\,\mathbf a\cdot\mathbf b$ term is often dropped, treating $|\mathbf a+\mathbf b|$ as $|\mathbf a|+|\mathbf b|=7$.

10StretchCalculatorHL only11 marks

The line $L$ has equation $\mathbf r=\begin{pmatrix}1\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\2\end{pmatrix}$ and the point $A$ has coordinates $(11,\,9,\,-3)$.

(a) Find the coordinates of the two points $P_1$ and $P_2$ on $L$ for which $AP=15$. [6]

(b) Use a scalar product to find the angle $P_1\hat{A}P_2$. [3]

Expresses $\overrightarrow{AP}$ in terms of $\lambda$: $\overrightarrow{AP}=\begin{pmatrix}2\lambda-10\\-\lambda-7\\2\lambda+2\end{pmatrix}$ (allow sign reversal)M1
Sets $|\overrightarrow{AP}|^2=225$: $(2\lambda-10)^2+(\lambda+7)^2+(2\lambda+2)^2=225$M1
$9\lambda^2-18\lambda+153=225$ or $\lambda^2-2\lambda-8=0$A1
$\lambda=4$ or $\lambda=-2$A1
One correct point, e.g. $(9,\,-2,\,7)$A1
Both points: $(9,\,-2,\,7)$ and $(-3,\,4,\,-5)$A1
$\overrightarrow{AP_1}=\begin{pmatrix}-2\\-11\\10\end{pmatrix}$, $\overrightarrow{AP_2}=\begin{pmatrix}-14\\-5\\-2\end{pmatrix}$ and attempt at their scalar productM1
$\cos P_1\hat{A}P_2=\dfrac{63}{15\times15}=0.28$A1
$P_1\hat{A}P_2=73.7^\circ$ (accept $1.29$ rad)A1
Uses $\tfrac12\times15\times15\times\sin(\text{their angle})$ (or $\tfrac12\times P_1P_2\times$ perpendicular height, with $P_1P_2=18$, height $12$)M1
Area $=108$ (accept $108.0$; units$^2$ not required)A1
Answer: (a) $(9,-2,7)$ and $(-3,4,-5)$ (b) $73.7^\circ$ (c) $108$
Examiner tip: Working with $|\overrightarrow{AP}|=15$ without squaring leads to messy algebra; square first and the quadratic factorises.

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