Quick Recall
Quick check — pick the correct option in each question before moving on.
In $\vec{r} = \vec{a} + \lambda \vec{d}$, the vector multiplying the parameter $\lambda$ is the direction vector, so here it is $\begin{pmatrix}2\\-1\\4\end{pmatrix}$.
Parallel lines have direction vectors that are scalar multiples of each other, e.g. $\vec{d}_1 = k\vec{d}_2$ for some constant $k$.
In 2D, any two non-parallel lines must meet at exactly one point. In 3D this rule breaks — non-parallel lines can miss each other entirely, and those lines are called skew.
The Wall
Consider these two lines in 3D:
$$\vec{r}_1 = \begin{pmatrix}1\\0\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\0\end{pmatrix}, \qquad \vec{r}_2 = \begin{pmatrix}0\\0\\1\end{pmatrix} + \mu\begin{pmatrix}1\\0\\1\end{pmatrix}$$
Their direction vectors $\begin{pmatrix}1\\1\\0\end{pmatrix}$ and $\begin{pmatrix}1\\0\\1\end{pmatrix}$ are not parallel. In 2D you would conclude: “they must intersect”. But do they? Try equating the $x$ and $y$ components, then check the $z$ component. The 2D rule has broken — in 3D, non-parallel lines can miss each other entirely.
The Need
We need a systematic test to classify any pair of 3D lines as intersecting, parallel, or skew, and a way to find the intersection point and the angle between two lines.
💡 The Bridge (click to reveal)
Set the two line equations equal: $\vec{a}_1 + \lambda\vec{d}_1 = \vec{a}_2 + \mu\vec{d}_2$. This gives three component equations (for $x$, $y$, $z$) in two unknowns ($\lambda$, $\mu$).
- Solve two equations for $\lambda$ and $\mu$. If the third equation is consistent, the lines intersect.
- If the third equation fails, the lines are skew.
- If $\vec{d}_1$ and $\vec{d}_2$ are parallel (one is a scalar multiple of the other), the lines are parallel (or coincident if they share a point).
The angle between two lines uses only their direction vectors: $\cos\theta = \dfrac{|\vec{d}_1 \cdot \vec{d}_2|}{|\vec{d}_1|\,|\vec{d}_2|}$.
Given two lines $\vec{r}_1 = \vec{a}_1 + \lambda\vec{d}_1$ and $\vec{r}_2 = \vec{a}_2 + \mu\vec{d}_2$, we want to decide whether they intersect, are parallel, or are skew; find the intersection point if it exists; and find the angle between them.
Two lines intersect when there exist values of $\lambda$ and $\mu$ giving the same point. Equate the equations and solve three component equations:
$$\boxed{\;\vec{a}_1 + \lambda\vec{d}_1 = \vec{a}_2 + \mu\vec{d}_2 \quad\Longleftrightarrow\quad \begin{cases} a_{1x} + \lambda d_{1x} = a_{2x} + \mu d_{2x} \\ a_{1y} + \lambda d_{1y} = a_{2y} + \mu d_{2y} \\ a_{1z} + \lambda d_{1z} = a_{2z} + \mu d_{2z} \end{cases}\;}$$
Solve any two equations for $\lambda$ and $\mu$, then check the third. If it holds, the lines intersect at $\vec{a}_1 + \lambda\vec{d}_1$.
First compare the direction vectors:
- If $\vec{d}_1$ and $\vec{d}_2$ are parallel (i.e. $\vec{d}_1 = k\vec{d}_2$ for some scalar $k$), the lines are parallel. If they also share a point, they are coincident (the same line).
- If $\vec{d}_1$ and $\vec{d}_2$ are not parallel, solve the system from Key relation 1. Consistent $\Rightarrow$ intersect; inconsistent $\Rightarrow$ skew.
$$\boxed{\;\text{Skew lines: non-parallel lines in 3D that do NOT intersect.}\;}$$
Skew lines exist only in 3D. In 2D, any two non-parallel lines must meet.
The angle $\theta$ between two lines depends only on their direction vectors (not on where the lines sit). Using the scalar product:
$$\boxed{\;\cos\theta = \dfrac{|\vec{d}_1 \cdot \vec{d}_2|}{|\vec{d}_1|\,|\vec{d}_2|}, \qquad 0^{\circ} \le \theta \le 90^{\circ}\;}$$
We take the absolute value of the dot product because the angle between two lines is the acute (or right) angle. This formula works whether the lines intersect or are skew — we only need their directions.
Four rounds, alternating teams. Teams must decide whether lines are parallel, intersecting or skew before the 30-second timer runs out. A correct classification scores 10 points.
Five classification and angle questions, 90 seconds, one attempt each. Immediate feedback and automatic progression. 10 points per correct answer.
Decide whether each of these pairs of lines intersect, are parallel or are skew. If the lines intersect, find the coordinates of the point of intersection.
(i) $\vec{r} = \begin{pmatrix}1 \\ -6 \\ -1\end{pmatrix} + \lambda\begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix}$ and $\vec{r} = \begin{pmatrix}9 \\ 7 \\ 2\end{pmatrix} + \mu\begin{pmatrix}2 \\ 3 \\ -1\end{pmatrix}$
(ii) $\vec{r} = \begin{pmatrix}1 \\ -6 \\ 0\end{pmatrix} + \lambda\begin{pmatrix}6 \\ -9 \\ -3\end{pmatrix}$ and $\vec{r} = \begin{pmatrix}-5 \\ 3 \\ 0\end{pmatrix} + \mu\begin{pmatrix}2 \\ -3 \\ -1\end{pmatrix}$
(iii) $\vec{r} = \begin{pmatrix}6 \\ -4 \\ 2\end{pmatrix} + \lambda\begin{pmatrix}1 \\ -2 \\ 5\end{pmatrix}$ and $\vec{r} = \begin{pmatrix}1 \\ 4 \\ -17\end{pmatrix} + \mu\begin{pmatrix}1 \\ -1 \\ 2\end{pmatrix}$
(iv) $\vec{r} = \begin{pmatrix}-1 \\ 2 \\ 4\end{pmatrix} + \lambda\begin{pmatrix}2 \\ 0 \\ 3\end{pmatrix}$ and $\vec{r} = \begin{pmatrix}-4 \\ 4 \\ 6\end{pmatrix} + \mu\begin{pmatrix}5 \\ -2 \\ 1\end{pmatrix}$
(v) $\vec{r} = \begin{pmatrix}0 \\ -1 \\ 4\end{pmatrix} + \lambda\begin{pmatrix}5 \\ 3 \\ -3\end{pmatrix}$ and $\vec{r} = \begin{pmatrix}2 \\ 5 \\ -1\end{pmatrix} + \mu\begin{pmatrix}4 \\ -3 \\ 2\end{pmatrix}$
(vi) $\vec{r} = \begin{pmatrix}9 \\ 3 \\ -4\end{pmatrix} + \lambda\begin{pmatrix}1 \\ 2 \\ -3\end{pmatrix}$ and $\vec{r} = \begin{pmatrix}1 \\ -4 \\ 5\end{pmatrix} + \mu\begin{pmatrix}1 \\ -1 \\ 2\end{pmatrix}$
(vii) $\vec{r} = \begin{pmatrix}2 \\ 3 \\ 1\end{pmatrix} + \lambda\begin{pmatrix}1 \\ 1 \\ -2\end{pmatrix}$ and $\vec{r} = \begin{pmatrix}-1 \\ -3 \\ -1\end{pmatrix} + \mu\begin{pmatrix}1 \\ 3 \\ 2\end{pmatrix}$
👁 Reveal Solution
Directions $\begin{pmatrix}1\\2\\3\end{pmatrix}$, $\begin{pmatrix}2\\3\\-1\end{pmatrix}$ not parallel. Equate: $1+\lambda=9+2\mu$, $-6+2\lambda=7+3\mu$, $-1+3\lambda=2-\mu$.
From eq3: $3\lambda+\mu=3 \Rightarrow \mu=3-3\lambda$. Sub eq1: $1+\lambda=9+2(3-3\lambda)=15-6\lambda \Rightarrow 7\lambda=14 \Rightarrow \lambda=2$, $\mu=-3$.
Check eq2: $-6+2(2)=-2$; $7+3(-3)=-2$ ✓. Point $= (1+2,\,-6+4,\,-1+6) = \boxed{(3,-2,5)}$.
$\vec{d}_1=\begin{pmatrix}6\\-9\\-3\end{pmatrix}=3\begin{pmatrix}2\\-3\\-1\end{pmatrix}=3\vec{d}_2$, so parallel. Test point $(1,-6,0)$ of line 1 on line 2: $-5+2\mu=1\Rightarrow\mu=3$; $3-3(3)=-6$ ✓; $0-(3)=-3\neq0$ ✗. Not on line 2, so $\boxed{\text{parallel (distinct)}}$.
Not parallel. Equate: $6+\lambda=1+\mu$, $-4-2\lambda=4-\mu$, $2+5\lambda=-17+2\mu$.
From eq1: $\mu=5+\lambda$. Sub eq2: $-4-2\lambda=4-(5+\lambda)=-1-\lambda \Rightarrow -\lambda=3 \Rightarrow \lambda=-3$, $\mu=2$.
Check eq3: $2+5(-3)=-13$; $-17+2(2)=-13$ ✓. Point $=(6-3,\,-4+6,\,2-15)=\boxed{(3,2,-13)}$.
Not parallel. Equate: $-1+2\lambda=-4+5\mu$, $2=4-2\mu$, $4+3\lambda=6+\mu$.
From eq2: $2\mu=2 \Rightarrow \mu=1$. Sub eq1: $-1+2\lambda=-4+5=1 \Rightarrow \lambda=1$.
Check eq3: $4+3(1)=7$; $6+1=7$ ✓. Point $=(-1+2,\,2,\,4+3)=\boxed{(1,2,7)}$.
Not parallel. Equate: $5\lambda=2+4\mu$, $-1+3\lambda=5-3\mu$, $4-3\lambda=-1+2\mu$.
From eq3: $3\lambda+2\mu=5 \Rightarrow \mu=\dfrac{5-3\lambda}{2}$. Sub eq1: $5\lambda=2+4\cdot\dfrac{5-3\lambda}{2}=2+2(5-3\lambda)=12-6\lambda \Rightarrow 11\lambda=12 \Rightarrow \lambda=\dfrac{12}{11}$, $\mu=\dfrac{19}{22}$.
Check eq2: LHS $=-1+3\cdot\dfrac{12}{11}=\dfrac{25}{11}=\dfrac{50}{22}$; RHS $=5-3\cdot\dfrac{19}{22}=\dfrac{53}{22}$. $\dfrac{50}{22}\neq\dfrac{53}{22}$ — inconsistent. $\boxed{\text{Skew}}$.
Not parallel. Equate: $9+\lambda=1+\mu$, $3+2\lambda=-4-\mu$, $-4-3\lambda=5+2\mu$.
From eq1: $\mu=8+\lambda$. Sub eq2: $3+2\lambda=-4-(8+\lambda)=-12-\lambda \Rightarrow 3\lambda=-15 \Rightarrow \lambda=-5$, $\mu=3$.
Check eq3: $-4-3(-5)=11$; $5+2(3)=11$ ✓. Point $=(9-5,\,3-10,\,-4+15)=\boxed{(4,-7,11)}$.
Not parallel. Equate: $2+\lambda=-1+\mu$, $3+\lambda=-3+3\mu$, $1-2\lambda=-1+2\mu$.
From eq1: $\mu=3+\lambda$. Sub eq2: $3+\lambda=-3+3(3+\lambda)=6+3\lambda \Rightarrow -2\lambda=3 \Rightarrow \lambda=-\dfrac{3}{2}$, $\mu=\dfrac{3}{2}$.
Check eq3: LHS $=1-2\left(-\dfrac{3}{2}\right)=4$; RHS $=-1+2\left(\dfrac{3}{2}\right)=2$. $4\neq2$ — inconsistent. $\boxed{\text{Skew}}$.
(i) intersect $(3,-2,5)$ · (ii) parallel distinct · (iii) intersect $(3,2,-13)$ · (iv) intersect $(1,2,7)$ · (v) skew · (vi) intersect $(4,-7,11)$ · (vii) skew.
Find the angle between the pair of lines: $\vec{r} = \begin{pmatrix}2 \\ 1 \\ 3\end{pmatrix} + s\begin{pmatrix}1 \\ 4 \\ 0\end{pmatrix}$ and $\vec{r} = \begin{pmatrix}6 \\ 10 \\ 4\end{pmatrix} + t\begin{pmatrix}2 \\ 1 \\ 1\end{pmatrix}$
👁 Reveal Solution
$\vec{d}_1 = \begin{pmatrix}1\\4\\0\end{pmatrix}$, $\vec{d}_2 = \begin{pmatrix}2\\1\\1\end{pmatrix}$.
$\vec{d}_1\cdot\vec{d}_2 = (1)(2)+(4)(1)+(0)(1) = 2+4+0 = 6$.
$|\vec{d}_1| = \sqrt{1^2+4^2+0^2} = \sqrt{17}$, $|\vec{d}_2| = \sqrt{2^2+1^2+1^2} = \sqrt{6}$.
$\cos\theta = \dfrac{|\vec{d}_1\cdot\vec{d}_2|}{|\vec{d}_1|\,|\vec{d}_2|} = \dfrac{6}{\sqrt{17}\,\sqrt{6}} = \dfrac{6}{\sqrt{102}} \approx 0.5941$.
$$\boxed{\theta = \arccos\!\left(\dfrac{6}{\sqrt{102}}\right) \approx 53.6^{\circ} \text{ (to 1 d.p.)}}$$
Test your understanding of all four lesson objectives. Each question targets one objective — try each before moving on!
Two lines have directions $\vec{d}_1 = \begin{pmatrix}1\\2\\3\end{pmatrix}$ and $\vec{d}_2 = \begin{pmatrix}2\\4\\6\end{pmatrix}$. What can you conclude?
Since $\vec{d}_2 = 2\vec{d}_1$, the direction vectors are parallel. The lines are either parallel (distinct) or coincident. To determine which, check if a point on one line lies on the other.
Line 1: $\vec{r} = \begin{pmatrix}1\\0\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\1\end{pmatrix}$, Line 2: $\vec{r} = \begin{pmatrix}0\\1\\0\end{pmatrix} + \mu\begin{pmatrix}1\\0\\1\end{pmatrix}$. Do they intersect?
Equating components: $1+\lambda = \mu$, $\lambda = 1$, $\lambda = \mu$. From the second equation $\lambda = 1$, then $\mu = 2$ from the third. But the first gives $\mu = 2$ while $\lambda = 1$, which is consistent. Wait: $1+\lambda = \mu \Rightarrow 1+1=2=\mu$, and $\lambda = \mu \Rightarrow 1=2$, which is a contradiction. The system is inconsistent, so the lines are skew.
Two lines have direction vectors $\vec{d}_1 = \begin{pmatrix}1\\2\\2\end{pmatrix}$ and $\vec{d}_2 = \begin{pmatrix}2\\1\\0\end{pmatrix}$. What is the acute angle between them?
$\vec{d}_1 \cdot \vec{d}_2 = 1(2) + 2(1) + 2(0) = 4$. $|\vec{d}_1| = 3$, $|\vec{d}_2| = \sqrt{5}$. $\cos\theta = \dfrac{|4|}{3\sqrt{5}} = \dfrac{4}{3\sqrt{5}} = \dfrac{4\sqrt{5}}{15}$. Hmm, let me recalculate: $\dfrac{4}{3\sqrt{5}} = \dfrac{4}{3 \times 2.236} = \dfrac{4}{6.708} = \dfrac{4\sqrt{5}}{15}$. Actually $\dfrac{4}{3\sqrt{5}} \neq \dfrac{4}{15}$. The correct answer is $\cos^{-1}\!\left(\dfrac{4}{3\sqrt{5}}\right) = \cos^{-1}\!\left(\dfrac{4\sqrt{5}}{15}\right)$.
Two boats move along straight paths. Boat A: $\vec{r}_A = \begin{pmatrix}0\\0\\0\end{pmatrix} + t\begin{pmatrix}3\\4\\0\end{pmatrix}$, Boat B: $\vec{r}_B = \begin{pmatrix}5\\0\\0\end{pmatrix} + s\begin{pmatrix}0\\5\\0\end{pmatrix}$. At what time do they collide (if at all)?
Equate: $3t = 5$ gives $t = 5/3$. Then $4t = 5s$ gives $s = 4/3$. But $0 = 0$ (z-component). Check: position of A at $t=5/3$: $(5, 20/3, 0)$. Position of B at $s=4/3$: $(5, 20/3, 0)$. Wait, they do collide! Actually: $3t = 5 \Rightarrow t = 5/3$, $4t = 5s \Rightarrow s = 4t/5 = 4/3$. Position A: $(5, 20/3, 0)$, Position B: $(5, 20/3, 0)$. They collide at $t = 5/3$. So the correct answer is "At $t = 5/3$, they collide".
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.
Find the position vector of the point of intersection of each of these pairs of lines.
(i) $\vec{r} = \begin{pmatrix}2 \\ 1\end{pmatrix} + \lambda\begin{pmatrix}1 \\ 0\end{pmatrix}$ and $\vec{r} = \begin{pmatrix}3 \\ 0\end{pmatrix} + \mu\begin{pmatrix}1 \\ 1\end{pmatrix}$
(ii) $\vec{r} = \begin{pmatrix}2 \\ -1\end{pmatrix} + \lambda\begin{pmatrix}1 \\ 2\end{pmatrix}$ and $\vec{r} = \mu\begin{pmatrix}1 \\ 1\end{pmatrix}$
💡 Hint
Equate the $x$ and $y$ components to get two equations in $\lambda$ and $\mu$. Solve, then substitute back into either line equation for the position vector.
👁 Reveal Solution
$2+\lambda = 3+\mu$ and $1 = \mu$.
So $\mu = 1$ and $\lambda = 3+\mu-2 = 3+1-2 = 2$.
Intersection: $\vec{r} = \begin{pmatrix}2\\1\end{pmatrix} + 2\begin{pmatrix}1\\0\end{pmatrix} = \boxed{\begin{pmatrix}4\\1\end{pmatrix}}$. (Check on line 2: $\begin{pmatrix}3\\0\end{pmatrix}+1\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}4\\1\end{pmatrix}$ ✓.)
$2+\lambda = \mu$ and $-1+2\lambda = \mu$.
Set equal: $2+\lambda = -1+2\lambda \Rightarrow \lambda = 3$, so $\mu = 2+3 = 5$.
Intersection: $\vec{r} = \begin{pmatrix}2\\-1\end{pmatrix} + 3\begin{pmatrix}1\\2\end{pmatrix} = \boxed{\begin{pmatrix}5\\5\end{pmatrix}}$. (Check on line 2: $5\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}5\\5\end{pmatrix}$ ✓.)
In this question the origin is taken to be at a harbour and the unit vectors $\hat{i}$ and $\hat{j}$ to have lengths of 1 km in the directions E and N. A cargo vessel leaves the harbour and its position vector $t$ hours later is given by $\vec{r}_1 = 12t\hat{i} + 16t\hat{j}$. A fishing boat is trawling nearby and its position at time $t$ is given by $\vec{r}_2 = (10 - 3t)\hat{i} + (8 + 4t)\hat{j}$.
(i) How far apart are the two boats when the cargo vessel leaves harbour?
(ii) How fast is each boat travelling?
(iii) What happens to the boats?
💡 Hint 1
(i) At $t=0$ the cargo vessel is at the harbour. Substitute $t=0$ into both position vectors and find the distance between the two points.
(ii) Speed is the magnitude of the velocity vector $\dfrac{\text{d}\vec{r}}{\text{d}t}$.
💡 Hint 2
(iii) Set $\vec{r}_1 = \vec{r}_2$ and solve for $t$. If a valid $t \ge 0$ exists, the boats meet at that time. Find the position too.
👁 Reveal Solution
At $t=0$: cargo vessel at $\vec{r}_1 = \begin{pmatrix}0\\0\end{pmatrix}$ (the harbour); fishing boat at $\vec{r}_2 = \begin{pmatrix}10\\8\end{pmatrix}$.
$$\text{Distance} = \sqrt{(10-0)^2 + (8-0)^2} = \sqrt{100+64} = \sqrt{164} = \boxed{2\sqrt{41} \approx 12.8 \text{ km}}.$$
Cargo velocity: $\dfrac{\text{d}\vec{r}_1}{\text{d}t} = 12\hat{i} + 16\hat{j}$, so speed $= \sqrt{12^2+16^2} = \sqrt{144+256} = \sqrt{400} = \boxed{20 \text{ km/h}}$.
Fishing velocity: $\dfrac{\text{d}\vec{r}_2}{\text{d}t} = -3\hat{i} + 4\hat{j}$, so speed $= \sqrt{(-3)^2+4^2} = \sqrt{9+16} = \sqrt{25} = \boxed{5 \text{ km/h}}$.
Set $\vec{r}_1 = \vec{r}_2$: $12t = 10-3t \Rightarrow 15t = 10 \Rightarrow t = \dfrac{2}{3}$.
Check $y$: $16t = 8+4t \Rightarrow 12t = 8 \Rightarrow t = \dfrac{2}{3}$ ✓ (consistent).
So at $t = \dfrac{2}{3}$ h $= 40$ min, both boats are at $\vec{r} = 12\cdot\dfrac{2}{3}\hat{i} + 16\cdot\dfrac{2}{3}\hat{j} = \boxed{8\hat{i} + \dfrac{32}{3}\hat{j}}$. $\boxed{\text{The two boats meet (collide) at that time and position.}}$
$\vec{r} = \begin{pmatrix}4 \\ 2 \\ -1\end{pmatrix} + s\begin{pmatrix}3 \\ 7 \\ -4\end{pmatrix}$ and $\vec{r} = \begin{pmatrix}5 \\ 1 \\ 0\end{pmatrix} + t\begin{pmatrix}2 \\ 8 \\ -5\end{pmatrix}$
Find the angle between the pair of lines.
💡 Hint
Use $\cos\theta = \dfrac{|\vec{d}_1\cdot\vec{d}_2|}{|\vec{d}_1|\,|\vec{d}_2|}$ with the two direction vectors. The points on the lines are irrelevant for the angle.
👁 Reveal Solution
$\vec{d}_1 = \begin{pmatrix}3\\7\\-4\end{pmatrix}$, $\vec{d}_2 = \begin{pmatrix}2\\8\\-5\end{pmatrix}$.
$\vec{d}_1\cdot\vec{d}_2 = (3)(2)+(7)(8)+(-4)(-5) = 6+56+20 = 82$.
$|\vec{d}_1| = \sqrt{9+49+16} = \sqrt{74}$, $|\vec{d}_2| = \sqrt{4+64+25} = \sqrt{93}$.
$\cos\theta = \dfrac{|82|}{\sqrt{74}\,\sqrt{93}} = \dfrac{82}{\sqrt{6882}} \approx \dfrac{82}{82.96} \approx 0.9885$.
$$\boxed{\theta = \arccos\!\left(\dfrac{82}{\sqrt{6882}}\right) \approx 8.7^{\circ} \text{ (to 1 d.p.)}}$$
The line $\vec{r} = \begin{pmatrix}-4 \\ 4 \\ -12\end{pmatrix} + q\begin{pmatrix}2 \\ -10 \\ 11\end{pmatrix}$ meets $\vec{r} = \begin{pmatrix}4 \\ -15 \\ -16\end{pmatrix} + s\begin{pmatrix}2 \\ -3 \\ -5\end{pmatrix}$ at $A$ and meets $\vec{r} = \begin{pmatrix}-1 \\ -29 \\ -3\end{pmatrix} + t\begin{pmatrix}1 \\ 1 \\ 8\end{pmatrix}$ at $B$. Find the coordinates of $A$ and the length of $AB$.
💡 Hint 1
Let the first line be $L_1$ (parameter $q$). To find $A$, equate $L_1$ with the second line and solve for $q$ and $s$. To find $B$, equate $L_1$ with the third line and solve for $q$ and $t$.
💡 Hint 2
Once you have the parameter values $q_A$ and $q_B$, substitute back into $L_1$ to get the coordinates of $A$ and $B$. Then $|AB| = \sqrt{(x_B-x_A)^2 + (y_B-y_A)^2 + (z_B-z_A)^2}$.
Shortcut: since $A$ and $B$ both lie on $L_1$ with direction $\vec{d}_1$, $|AB| = |q_B - q_A|\,|\vec{d}_1|$.
👁 Reveal Solution
Equate $L_1=\begin{pmatrix}-4+2q\\4-10q\\-12+11q\end{pmatrix}$ with $L_2=\begin{pmatrix}4+2s\\-15-3s\\-16-5s\end{pmatrix}$:
$-4+2q = 4+2s \Rightarrow q-s=4$ (eq1)
$4-10q = -15-3s \Rightarrow 10q-3s=19$ (eq2)
$-12+11q = -16-5s \Rightarrow 11q+5s=-4$ (eq3)
From eq1: $s=q-4$. Sub eq2: $10q-3(q-4)=19 \Rightarrow 7q+12=19 \Rightarrow q=1$, so $s=-3$.
Check eq3: $11(1)+5(-3)=11-15=-4$ ✓.
$A = \begin{pmatrix}-4+2(1)\\4-10(1)\\-12+11(1)\end{pmatrix} = \boxed{(-2,\,-6,\,-1)}$.
Equate $L_1$ with $L_3=\begin{pmatrix}-1+t\\-29+t\\-3+8t\end{pmatrix}$:
$-4+2q = -1+t \Rightarrow 2q-t=3$ (eq1)
$4-10q = -29+t \Rightarrow 10q+t=33$ (eq2)
$-12+11q = -3+8t \Rightarrow 11q-8t=9$ (eq3)
From eq1: $t=2q-3$. Sub eq2: $10q+2q-3=33 \Rightarrow 12q=36 \Rightarrow q=3$, so $t=3$.
Check eq3: $11(3)-8(3)=33-24=9$ ✓.
$B = \begin{pmatrix}-4+2(3)\\4-10(3)\\-12+11(3)\end{pmatrix} = \boxed{(2,\,-26,\,21)}$.
$|AB| = \sqrt{(2-(-2))^2 + (-26-(-6))^2 + (21-(-1))^2} = \sqrt{4^2+(-20)^2+22^2}$.
$= \sqrt{16+400+484} = \sqrt{900} = \boxed{30 \text{ units}}$.
Shortcut check: $q_B - q_A = 3-1 = 2$, and $|\vec{d}_1| = \sqrt{2^2+(-10)^2+11^2} = \sqrt{4+100+121} = \sqrt{225} = 15$, so $|AB| = 2 \times 15 = 30$ ✓.
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.