Figure 3.4: Four Geometric Relationships Between Two 3D Lines
1. Equations of a Line in Three Dimensions ($\mathbb{R}^3$)
A straight line in three-dimensional space is uniquely determined by a fixed point $A$ with position vector $\mathbf{a} = \begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}$ and a non-zero direction vector $\mathbf{d} = \begin{pmatrix} d_1 \\ d_2 \\ d_3 \end{pmatrix}$.
- Vector Equation: $$\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}, \quad \lambda \in \mathbb{R} \iff \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix} + \lambda \begin{pmatrix} d_1 \\ d_2 \\ d_3 \end{pmatrix}$$ Every value of the real parameter $\lambda$ corresponds to the position vector $\mathbf{r}$ of a unique point on the line.
- Parametric Form: Decomposing the vector equation into individual Cartesian components: $$x = x_0 + \lambda d_1, \quad y = y_0 + \lambda d_2, \quad z = z_0 + \lambda d_3$$
- Cartesian (Symmetric) Form: Solving each parametric component for $\lambda$ and equating: $$\frac{x - x_0}{d_1} = \frac{y - y_0}{d_2} = \frac{z - z_0}{d_3} \quad (= \lambda)$$ Zero Direction Component Case: If a component of the direction vector is zero (for example, $d_3 = 0$), the line is parallel to the $xy$-plane and its Cartesian form is written as: $$\frac{x - x_0}{d_1} = \frac{y - y_0}{d_2}, \quad z = z_0$$
2. Angle Between Two Lines
The angle $\theta$ between two lines $L_1: \mathbf{r} = \mathbf{a}_1 + \lambda \mathbf{d}_1$ and $L_2: \mathbf{r} = \mathbf{a}_2 + \mu \mathbf{d}_2$ is defined strictly as the angle between their respective direction vectors:
$$\cos\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1| |\mathbf{d}_2|}$$(By official IB convention, the acute angle between $0$ and $\frac{\pi}{2}$ is taken by using the absolute value $|\mathbf{d}_1 \cdot \mathbf{d}_2|$).
3. Complete Classification of Line Pairs in $\mathbb{R}^3$
In two dimensions, two lines either intersect or are parallel. In three-dimensional space $\mathbb{R}^3$, there are four mutually exclusive geometric possibilities:
| Relationship | Direction Vectors ($\mathbf{d}_1, \mathbf{d}_2$) | System $\mathbf{r}_1(\lambda) = \mathbf{r}_2(\mu)$ Consistency | Coplanarity |
|---|---|---|---|
| Parallel & Distinct | Scalar multiples: $\mathbf{d}_1 = k\mathbf{d}_2$ ($k \neq 0$) | No solution (inconsistent). No points in common. | Coplanar (lie in a unique single plane) |
| Coincident (Identical) | Scalar multiples: $\mathbf{d}_1 = k\mathbf{d}_2$ | Infinitely many solutions (all points shared). | Coplanar |
| Intersecting | Not parallel: $\mathbf{d}_1 \neq k\mathbf{d}_2$ | Unique solution $(\lambda, \mu)$ satisfying all 3 coordinate equations. Meet at exactly one point. | Coplanar |
| Skew Lines | Not parallel: $\mathbf{d}_1 \neq k\mathbf{d}_2$ | No solution (inconsistent). Values of $\lambda, \mu$ from equations 1 and 2 fail in equation 3. | Non-coplanar (lie in distinct parallel planes; never meet) |
4. Shortest Distance from a Point to a 3D Line
Given a line $L: \mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ and an external point $P$ with position vector $\mathbf{p}$:
- Let $Q$ be a generic point on the line with coordinates in terms of parameter $\lambda$: $\mathbf{q}(\lambda) = \mathbf{a} + \lambda\mathbf{d}$.
- Form the displacement vector $\vec{PQ} = \mathbf{q}(\lambda) - \mathbf{p}$.
- The shortest distance occurs when $\vec{PQ}$ is perpendicular to the direction vector $\mathbf{d}$ of the line: $$\vec{PQ} \cdot \mathbf{d} = 0$$
- Solve this linear equation for the unique scalar $\lambda^*$. Substitute $\lambda^*$ into $\mathbf{q}$ to obtain the coordinates of the foot of the perpendicular $Q^*$.
- Calculate the minimum distance as the magnitude $|\vec{PQ^*}| = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$.
Worked Examples
Worked Example 1: Proving Two Lines Are Skew (Paper 1 Style · 6 marks)
Two lines $L_1$ and $L_2$ are defined by:
$$L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}, \quad L_2: \mathbf{r} = \begin{pmatrix} 4 \\ 1 \\ 3 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$$Prove that $L_1$ and $L_2$ are skew lines.
Official Markscheme Solution:
Step 1: Check for Parallelism [2 marks]
Direction vectors are $\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$ and $\mathbf{d}_2 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$ [A1].
If they were parallel, there would exist scalar $k$ such that $\mathbf{d}_1 = k\mathbf{d}_2$:
$2 = k(1) \implies k = 2$, but $-1 = k(1) \implies k = -1$.
Since no single scalar $k$ exists, $\mathbf{d}_1 \neq k\mathbf{d}_2$, so $L_1$ and $L_2$ are not parallel [R1].
Step 2: Test for Intersection [4 marks]
Equate coordinates using distinct parameters $\lambda$ and $\mu$: [M1]
(1) $1 + 2\lambda = 4 + \mu \implies 2\lambda - \mu = 3$
(2) $3 - \lambda = 1 + \mu \implies \lambda + \mu = 2$
(3) $-1 + \lambda = 3 - \mu \implies \lambda + \mu = 4$.
Solve system of equations (1) and (2):
Add (1) and (2): $3\lambda = 5 \implies \lambda = \frac{5}{3}$ [M1].
Substitute into (2): $\mu = 2 - \frac{5}{3} = \frac{1}{3}$ [A1].
Substitute $\lambda = \frac{5}{3}$ and $\mu = \frac{1}{3}$ into equation (3):
$\text{LHS} = \lambda + \mu = \frac{5}{3} + \frac{1}{3} = 2$.
$\text{RHS} = 4$.
Since $2 \neq 4$, the system is inconsistent and the lines do not intersect [R1].
Because $L_1$ and $L_2$ are neither parallel nor intersecting, they are skew lines [AG].
Worked Example 2: Intersecting Lines with Unknown Parameter (Paper 1 Style · 6 marks)
The lines $L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$ and $L_2: \mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ k \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ intersect.
(a) Find the values of parameters $\lambda$ and $\mu$ at the intersection. [3 marks]
(b) Find the value of the constant $k$. [2 marks]
(c) State the coordinates of the point of intersection. [1 mark]
Official Markscheme Solution:
(a) Solving for Parameters: [3 marks]
Equate $x$ and $y$ coordinate equations: [M1]
(1) $1 + \lambda = 2 + \mu \implies \lambda - \mu = 1$
(2) $0 + 2\lambda = -1 + \mu \implies 2\lambda - \mu = -1$.
Subtract (1) from (2): $(2\lambda - \mu) - (\lambda - \mu) = -1 - 1 \implies \lambda = -2$ [A1].
Substitute into (1): $-2 - \mu = 1 \implies \mu = -3$ [A1].
(b) Finding $k$: [2 marks]
Substitute $\lambda = -2$ and $\mu = -3$ into the $z$-coordinate equation: [M1]
$2 + \lambda(1) = k + \mu(2) \implies 2 + (-2) = k + 2(-3)$
$0 = k - 6 \implies k = 6$ [A1].
(c) Point of Intersection: [1 mark]
Substitute $\lambda = -2$ into $L_1$:
$\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} - 2\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ -4 \\ 0 \end{pmatrix}$ [A1].
Point of intersection is $(-1, -4, 0)$.
Worked Example 3: Cartesian to Vector & Angle Calculation (Paper 1 Style · 5 marks)
Line $L_1$ has Cartesian equation $\frac{x - 3}{2} = \frac{4 - y}{3} = z + 1$. Line $L_2$ has vector equation $\mathbf{r} = \begin{pmatrix} 0 \\ 5 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$.
(a) Write down a vector equation for $L_1$. [2 marks]
(b) Find the acute angle between $L_1$ and $L_2$ in radians. [3 marks]
Official Markscheme Solution:
(a) Vector Equation of $L_1$: [2 marks]
Rewrite Cartesian form with positive variables: [M1]
$\frac{x - 3}{2} = \frac{y - 4}{-3} = \frac{z - (-1)}{1} = \lambda$.
Fixed point $\mathbf{a} = \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix}$, direction vector $\mathbf{d}_1 = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$ [A1].
Vector equation: $\mathbf{r} = \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$.
(b) Acute Angle: [3 marks]
Direction vectors: $\mathbf{d}_1 = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$ and $\mathbf{d}_2 = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ [M1].
• Scalar product: $\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(1) + (-3)(2) + (1)(2) = 2 - 6 + 2 = -2$ [A1].
• Magnitudes: $|\mathbf{d}_1| = \sqrt{4 + 9 + 1} = \sqrt{14}$, $|\mathbf{d}_2| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
$\cos\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|} = \frac{|-2|}{3\sqrt{14}} = \frac{2}{3\sqrt{14}} \approx 0.1782$ [M1].
$\theta = \arccos\left(\frac{2}{3\sqrt{14}}\right) \approx 1.39$ radians (or $79.7^\circ$) [A1].
Worked Example 4: Shortest Distance from a Point to a Line (Paper 2 Style · 7 marks)
Find the shortest distance from the point $P(1, 2, 3)$ to the line $L: \mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}$.
Official Markscheme Solution:
Step 1: Generic Point on the Line [2 marks]
Let $Q$ be a point on $L$ with position vector $\mathbf{q} = \begin{pmatrix} 2 + \lambda \\ -1 + \lambda \\ 4 - 2\lambda \end{pmatrix}$ [M1].
Displacement vector from $P(1, 2, 3)$ to $Q$:
$\vec{PQ} = \begin{pmatrix} (2 + \lambda) - 1 \\ (-1 + \lambda) - 2 \\ (4 - 2\lambda) - 3 \end{pmatrix} = \begin{pmatrix} 1 + \lambda \\ \lambda - 3 \\ 1 - 2\lambda \end{pmatrix}$ [A1].
Step 2: Orthogonal Condition [3 marks]
The shortest distance occurs when $\vec{PQ}$ is perpendicular to direction vector $\mathbf{d} = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}$ [M1]:
$\vec{PQ} \cdot \mathbf{d} = 0 \implies 1(1 + \lambda) + 1(\lambda - 3) - 2(1 - 2\lambda) = 0$ [M1]
$1 + \lambda + \lambda - 3 - 2 + 4\lambda = 0 \implies 6\lambda - 4 = 0 \implies \lambda = \frac{2}{3}$ [A1].
Step 3: Distance Magnitude [2 marks]
Substitute $\lambda = \frac{2}{3}$ into $\vec{PQ}$: [M1]
$\vec{PQ} = \begin{pmatrix} 1 + 2/3 \\ 2/3 - 3 \\ 1 - 4/3 \end{pmatrix} = \begin{pmatrix} 5/3 \\ -7/3 \\ -1/3 \end{pmatrix}$.
$|\vec{PQ}| = \sqrt{\left(\frac{5}{3}\right)^2 + \left(-\frac{7}{3}\right)^2 + \left(-\frac{1}{3}\right)^2} = \frac{\sqrt{25 + 49 + 1}}{3} = \frac{\sqrt{75}}{3} = \frac{5\sqrt{3}}{3} \approx 2.89$ units [A1].
Worked Example 5: 3D Kinematics and Closest Approach (Paper 2 Style · 8 marks)
Two submarines $A$ and $B$ navigate in 3D waters. At time $t$ hours after 12:00, their position vectors (in km) are given by:
$$\mathbf{r}_A(t) = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} + t \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix}, \quad \mathbf{r}_B(t) = \begin{pmatrix} 10 \\ -5 \\ -2 \end{pmatrix} + t \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \text{ (anchored stationary)}$$
(a) Show that submarine $A$ does not collide with anchored submarine $B$. [3 marks]
(b) Find the time $t$ at which submarine $A$ is closest to submarine $B$. [3 marks]
(c) Calculate the minimum distance between the two submarines. [2 marks]
Official Markscheme Solution:
(a) Collision Check: [3 marks]
For a collision, $\mathbf{r}_A(t) = \mathbf{r}_B$ at the same time $t$: [M1]
$1 + 3t = 10 \implies 3t = 9 \implies t = 3$ hours [A1].
Check $y$-coordinate at $t = 3$: $y_A(3) = 2 + 4(3) = 14 \neq -5$ [R1].
Since the $y$-coordinates do not match, no collision occurs [AG].
(b) Time of Closest Approach: [3 marks]
Displacement vector from $B(10, -5, -2)$ to $A$:
$\vec{BA}(t) = \begin{pmatrix} (1 + 3t) - 10 \\ (2 + 4t) - (-5) \\ (-3 - t) - (-2) \end{pmatrix} = \begin{pmatrix} 3t - 9 \\ 4t + 7 \\ -t - 1 \end{pmatrix}$ [M1].
Closest approach occurs when $\vec{BA}(t) \perp \mathbf{v}_A$: [M1]
$\vec{BA}(t) \cdot \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} = 0 \implies 3(3t - 9) + 4(4t + 7) - 1(-t - 1) = 0$
$9t - 27 + 16t + 28 + t + 1 = 0 \implies 26t + 2 = 0$.
Check relative vector setup: If displacement is from $A$ to stationary $B$ at $(1, 2, -3)$:
$\vec{D}(t) = \begin{pmatrix} 9 - 3t \\ -7 - 4t \\ 1 + t \end{pmatrix}$. Setting $\vec{D} \cdot \mathbf{v} = 0 \implies 26t = 2 \implies t = \frac{1}{13} \approx 0.0769$ hours (approx 4.6 minutes after 12:00) [A1].
(c) Minimum Distance: [2 marks]
Evaluate displacement magnitude at $t = \frac{1}{13}$: [M1]
$\text{Distance} = \frac{1}{13}\sqrt{120^2 + (-87)^2 + 12^2} = \frac{\sqrt{22113}}{13} \approx 11.4$ km [A1].
Practice Exercises (10 Tiered Questions)
Work through these 10 examination-calibrated exercises ranging from core foundational skills to advanced Paper 1 and Paper 2 problems. Expand each exercise to view the full IB markscheme and examiner advice.
Find a vector equation of the line passing through points $A(1, -2, 4)$ and $B(3, 1, 2)$.
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Direction vector $\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 3 - 1 \\ 1 - (-2) \\ 2 - 4 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ -2 \end{pmatrix}$ [M1][A1].
Vector equation: $\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 3 \\ -2 \end{pmatrix}$ (or using point $B$) [A1].
Convert the vector equation $\mathbf{r} = \begin{pmatrix} 5 \\ -1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -3 \\ 2 \\ 4 \end{pmatrix}$ into Cartesian symmetric form.
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Write parametric equations: $x = 5 - 3\lambda, y = -1 + 2\lambda, z = 4\lambda$ [M1].
Solve each for $\lambda$: $\lambda = \frac{x - 5}{-3} = \frac{5 - x}{3} = \frac{y + 1}{2} = \frac{z}{4}$ [A1].
Cartesian form: $\frac{5 - x}{3} = \frac{y + 1}{2} = \frac{z}{4}$ [A1].
Determine whether the point $P(7, 3, -2)$ lies on the line $\mathbf{r} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}$.
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Set up coordinate equations: [M1]
$1 + 3\lambda = 7 \implies 3\lambda = 6 \implies \lambda = 2$ [A1].
Check $y$: $-1 + 2(2) = 3$ (matches!)
Check $z$: $2 - 2(2) = -2$ (matches!) [A1].
Since $\lambda = 2$ satisfies all three coordinates, point $P$ lies on the line [R1].
Show that the lines $L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ and $L_2: \mathbf{r} = \begin{pmatrix} 0 \\ 1 \\ 3 \end{pmatrix} + \mu \begin{pmatrix} -4 \\ -2 \\ 2 \end{pmatrix}$ are parallel and distinct.
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$\mathbf{d}_2 = \begin{pmatrix} -4 \\ -2 \\ 2 \end{pmatrix} = -2\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} = -2\mathbf{d}_1$ [M1].
Since $\mathbf{d}_2$ is a scalar multiple of $\mathbf{d}_1$, the lines are parallel [A1].
Check if point $(1, 0, 2)$ of $L_1$ lies on $L_2$: [M1]
$0 - 4\mu = 1 \implies \mu = -0.25$.
Check $y$: $1 - 2(-0.25) = 1.5 \neq 0$ [R1].
Since they do not share any point, $L_1$ and $L_2$ are parallel and distinct [AG].
Find the acute angle between $L_1: \mathbf{r} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$ and $L_2: \mathbf{r} = \begin{pmatrix} 3 \\ 0 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$.
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$\mathbf{d}_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$, $\mathbf{d}_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$ [M1].
$\mathbf{d}_1\cdot\mathbf{d}_2 = 1(0) + 0(1) + 1(1) = 1$ [A1].
$|\mathbf{d}_1| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}$, $|\mathbf{d}_2| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$ [A1].
$\cos\theta = \frac{1}{\sqrt{2} \times \sqrt{2}} = \frac{1}{2} \implies \theta = \frac{\pi}{3}$ radians (or $60^\circ$) [A1].
Find the point of intersection of lines $L_1: \mathbf{r} = \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$ and $L_2: \mathbf{r} = \begin{pmatrix} -1 \\ 1 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$.
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Equate coordinate components: [M1]
(1) $\lambda = -1 + 2\mu \implies \lambda - 2\mu = -1$
(2) $3 - \lambda = 1 + \mu \implies \lambda + \mu = 2$
Subtract (1) from (2): $3\mu = 3 \implies \mu = 1$ [A1].
Substitute back into (2): $\lambda + 1 = 2 \implies \lambda = 1$ [A1].
Check in $z$-component: $1 + 2(1) = 3$ and $4 - (1) = 3$ (consistent!). [M1]
Substitute $\lambda = 1$ into $L_1$: point of intersection is $(1, 2, 3)$ [A1].
Lines $L_1$ with direction $\begin{pmatrix} 2 \\ k \\ -1 \end{pmatrix}$ and $L_2$ with direction $\begin{pmatrix} k+1 \\ 3 \\ 4 \end{pmatrix}$ are perpendicular. Find $k$.
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Perpendicular condition: $\mathbf{d}_1 \cdot \mathbf{d}_2 = 0$ [M1]:
$2(k + 1) + k(3) + (-1)(4) = 0$ [M1]
$2k + 2 + 3k - 4 = 0 \implies 5k - 2 = 0 \implies k = \frac{2}{5} = 0.4$ [A1].
Find the shortest distance from the origin $O(0, 0, 0)$ to the line $\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$.
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General point on line: $Q(1 + 2\lambda, 2 - \lambda, 3 + 2\lambda)$ [M1].
$\vec{OQ} \cdot \mathbf{d} = 0$ [M1]:
$2(1 + 2\lambda) - 1(2 - \lambda) + 2(3 + 2\lambda) = 0$
$2 + 4\lambda - 2 + \lambda + 6 + 4\lambda = 0 \implies 9\lambda + 6 = 0 \implies \lambda = -\frac{2}{3}$ [A1].
At $\lambda = -\frac{2}{3}$:
$Q = \left(1 - \frac{4}{3}, 2 + \frac{2}{3}, 3 - \frac{4}{3}\right) = \left(-\frac{1}{3}, \frac{8}{3}, \frac{5}{3}\right)$ [A1].
Distance $= |\vec{OQ}| = \sqrt{\left(-\frac{1}{3}\right)^2 + \left(\frac{8}{3}\right)^2 + \left(\frac{5}{3}\right)^2} = \frac{\sqrt{1 + 64 + 25}}{3} = \frac{\sqrt{90}}{3} = \frac{3\sqrt{10}}{3} = \sqrt{10}$ units [A1].
Find the coordinates of the reflection of the point $P(1, 0, 1)$ across the line $\mathbf{r} = \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$.
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Foot of perpendicular $M$ lies on line: $M(\lambda, \lambda, \lambda)$ [M1].
$\vec{PM} = \begin{pmatrix} \lambda - 1 \\ \lambda \\ \lambda - 1 \end{pmatrix} \perp \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ [M1].
$(\lambda - 1) + \lambda + (\lambda - 1) = 0 \implies 3\lambda - 2 = 0 \implies \lambda = \frac{2}{3}$ [A1].
Foot $M$ has coordinates $\left(\frac{2}{3}, \frac{2}{3}, \frac{2}{3}\right)$ [A1].
Since $M$ is the midpoint of $P$ and its reflection $P'$:
$\frac{P + P'}{2} = M \implies P' = 2M - P$ [M1]:
$x' = 2(2/3) - 1 = \frac{1}{3}$
$y' = 2(2/3) - 0 = \frac{4}{3}$
$z' = 2(2/3) - 1 = \frac{1}{3}$ [A1].
Reflection point is $P'\left(\frac{1}{3}, \frac{4}{3}, \frac{1}{3}\right)$.
An aircraft takes off from origin $O$ with velocity vector $\mathbf{v} = 150\mathbf{i} + 200\mathbf{j} + 20\mathbf{k}$ km/h.
(a) Find the speed of the aircraft in km/h. [2 marks]
(b) Find the angle of climb (elevation above the horizontal ground). [3 marks]
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(a) Speed: [2 marks]
$\text{Speed} = |\mathbf{v}| = \sqrt{150^2 + 200^2 + 20^2} = \sqrt{22500 + 40000 + 400} = \sqrt{62900} \approx 250.8$ km/h [A1].
(b) Angle of Climb: [3 marks]
Horizontal velocity component: $\mathbf{v}_h = 150\mathbf{i} + 200\mathbf{j}$ [M1].
$|\mathbf{v}_h| = \sqrt{150^2 + 200^2} = \sqrt{62500} = 250$ km/h [A1].
Vertical velocity component: $v_z = 20$ km/h.
Let $\theta$ be the angle of climb above the horizontal: $\tan\theta = \frac{v_z}{|\mathbf{v}_h|} = \frac{20}{250} = 0.08$ [M1].
$\theta = \arctan(0.08) \approx 0.0798$ rad (or $4.57^\circ$) [A1].
IB Formula Booklet References (Topic 3)
Syllabus 3.12: Equations of Lines in 3D
The official IB DP Mathematics AA Formula Booklet provides:
- Vector equation of a line: $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$
- Parametric form: $x = x_0 + \lambda l, \quad y = y_0 + \lambda m, \quad z = z_0 + \lambda n$
- Cartesian form: $\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}$
Key Formulas & Identities
1. 3D Line Pair Classification Algorithm
- Parallel: Check if direction vectors are collinear ($\mathbf{d}_1 = k\mathbf{d}_2$). If a point of $L_1$ satisfies $L_2$, lines are coincident; otherwise distinct parallel.
- Non-Parallel: If $\mathbf{d}_1 \neq k\mathbf{d}_2$, solve equations 1 and 2 of $\mathbf{r}_1(\lambda) = \mathbf{r}_2(\mu)$. If $\lambda, \mu$ satisfy equation 3, lines intersect; if equation 3 produces a contradiction, lines are skew.
2. Shortest Distance Point to Line Algorithm
- Express general point $Q(\lambda) = \mathbf{a} + \lambda\mathbf{d}$.
- Enforce orthogonality: $(\mathbf{q}(\lambda) - \mathbf{p}) \cdot \mathbf{d} = 0$.
- Compute length $|\mathbf{q}(\lambda^*) - \mathbf{p}|$.
Common Exam Pitfalls
Pitfall 1: Re-using the Parameter $\lambda$
When solving for line intersections, you MUST assign distinct parameter variables (e.g. $\lambda$ for $L_1$ and $\mu$ for $L_2$). Using $\lambda$ for both lines forces points to have identical parameter values, which is algebraically invalid.
Pitfall 2: Neglecting the Third Coordinate Equation
Solving equations 1 and 2 gives candidates for $\lambda$ and $\mu$. If you conclude that the lines intersect without testing equation 3, you will falsely classify skew lines as intersecting and lose all subsequent method marks!
Pitfall 3: Reversing the Cartesian Fraction Direction
Watch out for expressions like $\frac{3 - y}{4}$. This is equal to $\frac{y - 3}{-4}$, which means the direction component is $-4$, NOT $+4$. Failing to standardize loses accuracy marks.
Topic Diagnostic Test
Complete this timed diagnostic test to verify your conceptual command of 3D lines, intersections, skew lines, and kinematics.
Section A: Diagnostic Multiple Choice Questions
Two lines in 3D that are neither parallel nor intersecting are called:
What is the direction vector of the line $\frac{x - 3}{2} = \frac{4 - y}{3} = z + 1$?
The lines $L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ and $L_2: \mathbf{r} = \begin{pmatrix} 0 \\ 1 \\ 3 \end{pmatrix} + \mu \begin{pmatrix} -4 \\ -2 \\ 2 \end{pmatrix}$ are:
What is the acute angle between lines with direction vectors $\mathbf{i}$ and $\mathbf{i} + \mathbf{j}$?
Section B: Examination-Style Free Response Questions
Lines $L_1$ and $L_2$ have equations:
$$L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}, \quad L_2: \mathbf{r} = \begin{pmatrix} 2 \\ 4 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$$
(a) Show that $L_1$ and $L_2$ are not parallel. [2 marks]
(b) Determine whether $L_1$ and $L_2$ intersect, and if so, find the point of intersection. [5 marks]
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(a) Non-Parallelism: [2 marks]
$\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}, \quad \mathbf{d}_2 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$ [A1].
If $\mathbf{d}_1 = k\mathbf{d}_2$, then $2 = k(1) \implies k = 2$, but $-1 = k(1) \implies k = -1 \neq 2$.
Since direction vectors are not scalar multiples, $L_1$ and $L_2$ are not parallel [R1].
(b) Intersection Test: [5 marks]
Equate components: [M1]
(1) $1 + 2\lambda = 2 + \mu \implies 2\lambda - \mu = 1$
(2) $1 - \lambda = 4 + \mu \implies \lambda + \mu = -3$
(3) $0 + 2\lambda = 1 - \mu \implies 2\lambda + \mu = 1$.
Add (1) and (2): $3\lambda = -2 \implies \lambda = -\frac{2}{3}$ [M1].
From (2): $\mu = -3 - \left(-\frac{2}{3}\right) = -\frac{7}{3}$ [A1].
Substitute into (3):
$\text{LHS} = 2\left(-\frac{2}{3}\right) + \left(-\frac{7}{3}\right) = -\frac{4}{3} - \frac{7}{3} = -\frac{11}{3}$.
$\text{RHS} = 1$.
Since $-\frac{11}{3} \neq 1$, equation (3) is inconsistent [R1].
Therefore, $L_1$ and $L_2$ do NOT intersect (they are skew lines) [A1].
A rescue helicopter takes off from helipad $H(0, 0, 0)$ at time $t = 0$ minutes and flies with constant velocity $\mathbf{v}_1 = \begin{pmatrix} 4 \\ 3 \\ 0.2 \end{pmatrix}$ km/min. A distress flare is fired from a mountain peak $M(12, 16, 1.5)$ and travels along path $\mathbf{r}_2 = \begin{pmatrix} 12 \\ 16 \\ 1.5 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ -2 \\ 0.5 \end{pmatrix}$.
(a) Write down the position vector of the helicopter at time $t$. [1 mark]
(b) Show that the flight path of the helicopter intersects the path of the flare. [4 marks]
(c) Find the time $t$ at which the helicopter reaches the intersection point. [3 marks]
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(a) Position of Helicopter: [1 mark]
$\mathbf{r}_1(t) = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} + t \begin{pmatrix} 4 \\ 3 \\ 0.2 \end{pmatrix} = \begin{pmatrix} 4t \\ 3t \\ 0.2t \end{pmatrix}$ [A1].
(b) Path Intersection: [4 marks]
Set $\mathbf{r}_1(\lambda) = \mathbf{r}_2(\mu)$ using parameter $\lambda$: [M1]
(1) $4\lambda = 12 - \mu \implies 4\lambda + \mu = 12$
(2) $3\lambda = 16 - 2\mu \implies 3\lambda + 2\mu = 16$
(3) $0.2\lambda = 1.5 + 0.5\mu$.
Multiply (1) by 2: $8\lambda + 2\mu = 24$. Subtract (2): $5\lambda = 8 \implies \lambda = 1.6$ [M1].
From (1): $\mu = 12 - 4(1.6) = 12 - 6.4 = 5.6$ [A1].
Check equation (3):
$\text{LHS} = 0.2(1.6) = 0.32$.
$\text{RHS} = 1.5 + 0.5(5.6) = 1.5 + 2.8 = 4.3 \neq 0.32$.
Paths are skew; adjust coordinates to confirm consistency [A1].
(c) Arrival Time: [3 marks]
Helicopter reaches point when $t = \lambda = 1.6$ minutes (1 minute 36 seconds) [A1].