Vector equations of lines (intersecting, parallel and skew lines): practice worksheet

1. Question 1warm-up · no calculator

   The points $A$ and $B$ have coordinates $(2,\,-1,\,4)$ and $(5,\,1,\,0)$ respectively. The line $L$ passes through $A$ and $B$.

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

   Markscheme

   $\overrightarrow{AB} = \begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}$ (or any scalar multiple)	M1
   $\mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}$	A1

   Award A0 if "$\mathbf{r} =$" is not seen. Accept the position vector of $B$ as the point and any non-zero scalar multiple of the direction vector.

   Answer: $\mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}$

   (b)Write down the equation of $L$ in Cartesian form.[1]

   Markscheme

   $\dfrac{x-2}{3} = \dfrac{y+1}{2} = \dfrac{z-4}{-4}$

   A1

   Accept $\dfrac{4-z}{4}$ for the third fraction. Follow through from their direction vector in part (a).

   Answer: $\dfrac{x-2}{3} = \dfrac{y+1}{2} = \dfrac{z-4}{-4}$

2. Question 2warm-up · no calculator

   The lines $L_1$ and $L_2$ have equations $$L_1:\ \mathbf{r} = \begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix},\ \lambda \in \mathbb{R} \qquad L_2:\ \mathbf{r} = \begin{pmatrix} 0 \\ 1 \\ 6 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix},\ \mu \in \mathbb{R}.$$

   (a)Show that the point $P(3,\,7,\,2)$ lies on $L_1$.[2]

   Markscheme

   equating one component, e.g. $1 + \lambda = 3 \Rightarrow \lambda = 2$	M1
   verifying the other two components: $3 + 2(2) = 7$ and $-2 + 2(2) = 2$, so $P$ lies on $L_1$	A1
   	AG

   All three components must be checked for the A1.

   Answer: $\lambda = 2$ gives $(1+2,\ 3+4,\ -2+4) = (3,\,7,\,2)$, so $P$ lies on $L_1$.

   (b)Find the cosine of the acute angle between $L_1$ and $L_2$.[2]

   Markscheme

   $\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} = 2 - 2 + 4 = 4$ and $|\mathbf{b}_1| = |\mathbf{b}_2| = 3$	M1
   $\cos\theta = \dfrac{4}{3 \times 3} = \dfrac{4}{9}$	A1

   Answer: $\cos\theta = \dfrac{4}{9}$

3. Question 3exam-style · no calculator

   The lines $L_1$ and $L_2$ have equations $$L_1:\ \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix},\ \lambda \in \mathbb{R} \qquad L_2:\ \mathbf{r} = \begin{pmatrix} 1 \\ -1 \\ 8 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix},\ \mu \in \mathbb{R}.$$

   (a)Show that $L_1$ and $L_2$ are not parallel.[1]

   Markscheme

   direction vectors are not scalar multiples of each other (e.g. $k = \tfrac{1}{2}$ from the $x$-components but $k = -1$ from the $y$-components), so not parallel	R1
   	AG

   A statement that the vectors are not parallel without justification scores R0.

   Answer: $\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$ is not a scalar multiple of $\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ (the ratios of the first two components are $\tfrac{1}{2}$ and $-1$), so the lines are not parallel.

   (b)Find the coordinates of the point of intersection of $L_1$ and $L_2$.[4]

   Markscheme

   equating components: $1 + \lambda = 1 + 2\mu$, $\ 2 - \lambda = -1 + \mu$, $\ 3 + 2\lambda = 8 - \mu$	M1
   solving two equations: $\mu = 1$, $\lambda = 2$	A1
   checking the third equation: $3 + 4 = 7 = 8 - 1$, so the lines intersect	A1
   point of intersection $(3,\,0,\,7)$	A1

   Award M0 if the same parameter is used for both lines. Accept the answer as a position vector.

   Answer: $(3,\,0,\,7)$

   (c)Find the acute angle between $L_1$ and $L_2$, giving your answer in the form $\arccos k$ where $k \in \mathbb{Q}$.[2]

   Markscheme

   $\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} = -1$, $\ |\mathbf{b}_1| = |\mathbf{b}_2| = \sqrt{6}$	M1
   $\cos\theta = \dfrac{|-1|}{6} \Rightarrow \theta = \arccos\left(\dfrac{1}{6}\right)$	A1

   Award A0 for $\arccos\left(-\tfrac{1}{6}\right)$ as the final answer since the acute angle is required.

   Answer: $\theta = \arccos\left(\dfrac{1}{6}\right)$

4. Question 4exam-style · no calculator

   The lines $L_1$ and $L_2$ have equations $$L_1:\ \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + s \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix},\ s \in \mathbb{R} \qquad L_2:\ \mathbf{r} = \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix},\ t \in \mathbb{R}.$$

   (a)Show that $L_1$ and $L_2$ are not parallel.[1]

   Markscheme

   direction vectors are not scalar multiples (e.g. $k = 2$ from the $x$-components, $k = -1$ from the $y$-components), hence not parallel	R1
   	AG

   Answer: $\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} \neq k\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$ for any $k$ (e.g. $k = 2$ from $x$ but $k = -1$ from $y$), so the lines are not parallel.

   (b)Show that $L_1$ and $L_2$ do not intersect, and hence deduce that $L_1$ and $L_2$ are skew.[4]

   Markscheme

   equating components: $1 + 2s = 3 + t$, $\ s = 4 - t$, $\ 2 + s = -1 + 2t$	M1
   solving the first two equations: $t = 2$, $s = 2$	A1
   substituting into the third equation: $2 + 2 = 4$ but $-1 + 4 = 3$, $4 \neq 3$, so there is no common point and the lines do not intersect	R1
   the lines are not parallel (from part (a)) and do not intersect, so they are skew	R1
   	AG

   Award the final R1 only if both conditions (not parallel and not intersecting) are explicitly stated. Any pair of equations may be solved first, with the remaining equation used for the contradiction.

   Answer: From the $x$- and $y$-components $s = t = 2$, but then the $z$-components give $4 \neq 3$; the lines do not intersect and, being non-parallel, are skew.

   (c)The line $L_3$ has equation $\mathbf{r} = \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ k \end{pmatrix}$, $t \in \mathbb{R}$, where $k \in \mathbb{R}$. Find the value of $k$ for which $L_1$ and $L_3$ intersect.[3]

   Markscheme

   the $x$- and $y$-components are unchanged, so $s = t = 2$ (or re-solving $1 + 2s = 3 + t$, $s = 4 - t$)	M1
   substituting into the $z$-components: $2 + 2 = -1 + 2k$	M1
   $k = \dfrac{5}{2}$	A1

   Note that $\begin{pmatrix} 1 \\ -1 \\ k \end{pmatrix}$ is never parallel to $\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$, so this value of $k$ gives a unique point of intersection.

   Answer: $k = \dfrac{5}{2}$

5. Question 5exam-style · calculator

   Two drones, $A$ and $B$, fly in straight lines with constant velocities. At time $t$ seconds after launch, their position vectors, in metres relative to a fixed origin, are given by $$\mathbf{r}_A = \begin{pmatrix} 0 \\ 5 \\ 10 \end{pmatrix} + t \begin{pmatrix} 4 \\ 2 \\ 1 \end{pmatrix} \qquad \mathbf{r}_B = \begin{pmatrix} 28 \\ -9 \\ 7 \end{pmatrix} + t \begin{pmatrix} -1 \\ 3 \\ 1 \end{pmatrix}, \qquad t \geq 0.$$

   (a)Write down the speed of drone $A$.[1]

   Markscheme

   $\sqrt{4^2 + 2^2 + 1^2} = \sqrt{21}$ ($= 4.58$ m s$^{-1}$)

   A1

   Units are not required for the mark.

   Answer: $\sqrt{21} \approx 4.58$ m s$^{-1}$

   (b)Find the acute angle between the flight paths of the two drones.[3]

   Markscheme

   $\begin{pmatrix} 4 \\ 2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 3 \\ 1 \end{pmatrix} = -4 + 6 + 1 = 3$	M1
   $\cos\theta = \dfrac{3}{\sqrt{21}\sqrt{11}}$ $\left(= \dfrac{3}{\sqrt{231}}\right)$	A1
   $\theta = 78.6^\circ$ (accept $1.37$ radians)	A1

   Accept $78.62^\circ$. Award A1A0 for $101^\circ$ unless the acute angle is subsequently stated.

   Answer: $\theta = 78.6^\circ$ (exact: $\arccos\dfrac{3}{\sqrt{231}}$)

   (c)Show that the flight paths of the two drones intersect, and find the coordinates of the point of intersection.[4]

   Markscheme

   using different parameters for the two paths: $4\lambda = 28 - \mu$, $\ 5 + 2\lambda = -9 + 3\mu$, $\ 10 + \lambda = 7 + \mu$	M1
   solving two of the equations (e.g. $\mu = \lambda + 3$ then $5\lambda = 25$): $\lambda = 5$, $\mu = 8$	A1
   checking the remaining equation: $5 + 10 = 15$ and $-9 + 24 = 15$, consistent, so the paths intersect	R1
   point of intersection $(20,\,15,\,15)$	A1

   Award M0 if the same parameter $t$ is used for both drones, as this only tests for a collision rather than an intersection of paths.

   Answer: $(20,\,15,\,15)$

   (d)Explain whether or not the two drones collide.[1]

   Markscheme

   $A$ is at the point of intersection when $t = 5$ and $B$ when $t = 8$; since the times are different the drones do not collide

   R1

   Follow through from their parameter values in part (c).

   Answer: They do not collide: $A$ reaches the intersection point at $t = 5$ but $B$ reaches it at $t = 8$.

6. Question 6challenge · no calculator

   The lines $L_1$ and $L_2$ have equations $$L_1:\ \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix},\ \lambda \in \mathbb{R} \qquad L_2:\ \mathbf{r} = \begin{pmatrix} 3 \\ 0 \\ 3 \end{pmatrix} + \mu \begin{pmatrix} a \\ 2 \\ 2 \end{pmatrix},\ \mu \in \mathbb{R},$$ where $a \in \mathbb{R}$.

   (a)Show that $L_1$ and $L_2$ are not parallel for any value of $a$.[2]

   Markscheme

   supposing $\begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} = k\begin{pmatrix} a \\ 2 \\ 2 \end{pmatrix}$ for some $k$	M1
   $y$-components give $k = -1$, $z$-components give $k = \tfrac{1}{2}$, contradiction independent of $a$, so not parallel for any $a$	R1
   	AG

   The argument must not depend on the $x$-component, since that is the only component involving $a$.

   Answer: If $\begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} = k\begin{pmatrix} a \\ 2 \\ 2 \end{pmatrix}$ then $k = -1$ from the $y$-components and $k = \tfrac{1}{2}$ from the $z$-components, a contradiction, so the lines are never parallel.

   (b)Find the value of $a$ for which $L_1$ and $L_2$ intersect, and find the coordinates of the point of intersection.[5]

   Markscheme

   equating components: $1 + 2\lambda = 3 + a\mu$, $\ 2 - 2\lambda = 2\mu$, $\ -1 + \lambda = 3 + 2\mu$	M1
   from the $y$-components $\lambda = 1 - \mu$	A1
   substituting into the $z$-components: $-\mu = 3 + 2\mu \Rightarrow \mu = -1$, $\lambda = 2$	A1
   substituting into the $x$-components: $5 = 3 - a \Rightarrow a = -2$	A1
   point of intersection $(5,\,-2,\,1)$	A1

   Since the lines are never parallel (part (a)), $a = -2$ gives a unique point of intersection. Accept the point as a position vector.

   Answer: $a = -2$; point of intersection $(5,\,-2,\,1)$

   (c)Find the value of $a$ for which $L_1$ and $L_2$ are perpendicular.[2]

   Markscheme

   $\begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} a \\ 2 \\ 2 \end{pmatrix} = 0 \Rightarrow 2a - 4 + 2 = 0$	M1
   $a = 1$	A1

   Answer: $a = 1$

   (d)In the case $a = -2$, find the Cartesian equation of the plane $\Pi$ that contains both $L_1$ and $L_2$. Give your answer in the form $px + qy + rz = d$ where $p, q, r, d \in \mathbb{Z}$.[4]

   Markscheme

   normal vector $\mathbf{n} = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} \times \begin{pmatrix} -2 \\ 2 \\ 2 \end{pmatrix}$	M1
   $\mathbf{n} = \begin{pmatrix} -6 \\ -6 \\ 0 \end{pmatrix}$ (or $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$)	A1
   substituting a point on the plane, e.g. $(1, 2, -1)$ or $(3, 0, 3)$ or $(5, -2, 1)$: $x + y = 1 + 2$	M1
   $x + y = 3$ (accept $6x + 6y = 18$)	A1

   Accept any correct integer scalar multiple. Award the first M1 for any valid attempt at the cross product, including determinant notation.

   Answer: $x + y = 3$

Original questions and notes written by Pietro Meloni, PhD, from the IB syllabus. Not endorsed by or affiliated with the International Baccalaureate Organization. Found a mistake? Please tell me privately at pietro@pietromeloni.com. More questions on this topic: Geometry and trigonometry practice.