9709 P3 §3.7 — Vector Equation of a Line

💡 Lead In: Describing a Straight Path in 3D

The Scenario

A survey drone needs to fly in a perfectly straight line through 3D space. The operator tells it: "Start at point A, and fly in the direction of vector d." That is all the information given — a starting point and a direction.

The drone's autopilot needs to know its position at every moment along the path, not just at the start. As the drone travels, it moves further and further along the direction vector. A single number — the parameter $\lambda$ — tells the drone how far it has travelled: $\lambda = 0$ means it is at the start, $\lambda = 1$ means it has gone one full step of $\vec{d}$, $\lambda = 2$ means two steps, and so on.

The remarkable fact: every point on the entire infinite line is captured by one compact equation using just a point, a direction, and a parameter.

The Question

What is the minimum information needed to describe a straight path in 3D space — and can you write down every point on that path with a single equation?

(Hint: think about what the drone operator gave the autopilot: a starting point and a direction. Is that enough? What does the parameter $\lambda$ do?)

💡 The Bridge (click to reveal)

A point on the line (with position vector $\vec{a}$) and a direction vector $\vec{d}$ are all you need. Every point on the line is reached by starting at $\vec{a}$ and moving some multiple $\lambda$ of $\vec{d}$:

$$\boxed{\;\vec{r} = \vec{a} + \lambda\vec{d}\;}$$

This is the vector equation of a line. The parameter $\lambda$ can be any real number: $\lambda = 0$ gives the point $\vec{a}$ itself, $\lambda = 1$ gives $\vec{a} + \vec{d}$, negative $\lambda$ goes in the opposite direction. Today we learn to construct this equation, test whether a given point lies on the line, and find midpoints and ratio divisions along it.

📖 Key Concepts: Vector Equation of a Line

A line in 3D space is completely determined by a point on it and a direction along it. The vector equation of a line uses a parameter $\lambda$ (lambda) to describe every point on the line.

Key relation 1 — The vector equation of a line:

A line passing through the point with position vector $\vec{a}$, in the direction of vector $\vec{d}$, has the equation:

$$\boxed{\;\vec{r} = \vec{a} + \lambda\vec{d}, \quad \lambda \in \mathbb{R}\;}$$

Here $\lambda$ is a scalar parameter: each value of $\lambda$ gives one point on the line. $\lambda = 0$ gives $\vec{r} = \vec{a}$ (the given point), $\lambda = 1$ gives $\vec{r} = \vec{a} + \vec{d}$, and so on. The direction vector $\vec{d}$ can be scaled — using $2\vec{d}$ or $\dfrac{1}{2}\vec{d}$ gives the same line, just with a different parameter.

Key relation 2 — Testing if a point lies on a line:

To test whether point $P$ lies on the line $\vec{r} = \vec{a} + \lambda\vec{d}$, set up the equation:

$$\vec{a} + \lambda\vec{d} = \overrightarrow{OP}$$

This gives three equations (one per component). Solve for $\lambda$ from each. If all three give the same $\lambda$, the point is on the line. If the $\lambda$ values are inconsistent, the point is not on the line.

Key relation 3 — Midpoint and section formula:

The midpoint of two points $A$ and $B$ (with position vectors $\vec{a}$ and $\vec{b}$) is:

$$\vec{m} = \dfrac{\vec{a} + \vec{b}}{2}$$

More generally, if point $N$ divides $AB$ in the ratio $m : n$ (i.e. $AN : NB = m : n$), the section formula gives:

$$\vec{n} = \dfrac{n\,\vec{a} + m\,\vec{b}}{m + n}$$

When $m = n = 1$, this reduces to the midpoint formula. When $m:n = 1:2$, $N$ is one-third of the way from $A$ to $B$.

Finding the direction from two points:

If a line passes through two points $A$ and $B$, the direction vector is $\overrightarrow{AB} = \vec{b} - \vec{a}$. The line equation can then be written as:

$$\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}) = \vec{a} + \lambda\,\overrightarrow{AB}$$

Any non-zero scalar multiple of $\overrightarrow{AB}$ gives an equivalent direction vector for the same line.

Group · Team Battle Activity 1: Line Equation Relay — Team Showdown

Four rounds, alternating teams: point and direction, point-on-line tests and midpoints. Each round lasts 30 seconds and a correct answer scores 10 points. The first team to build the equation scores.

Team A 0

Team B 0

Round 1 / 4

Press Start to begin Round 1.

Team A takes the first round. Ready?

Individual · Speed Run Activity 2: Equation Express — 90-Second Solo Run

Five line-equation questions, 90 seconds, one attempt each. Feedback is immediate and the run advances automatically. 10 points per correct answer.

Question 1 / 5

90 Score 0

Press Start and answer as many as you can in 90 seconds.

Group · Guided Worked Example 1: Line Equations in Vector Form Exercise 10D Q3

Find the equation of each of these lines in vector form.

(i) Through (2, 4, −1) in the direction $\begin{pmatrix}3 \\ 6 \\ 4\end{pmatrix}$

(ii) Through (1, 0, −1) in the direction $\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}$

(iii) Through (1, 0, 4) and (6, 3, −2)

👁 Reveal Solution

(i) Through (2, 4, −1) in direction (3, 6, 4)

The point and direction are given directly. Write $\vec{a} = 2\hat{i} + 4\hat{j} - \hat{k}$ and $\vec{d} = 3\hat{i} + 6\hat{j} + 4\hat{k}$.

$$\boxed{\;\vec{r} = \begin{pmatrix}2 \\ 4 \\ -1\end{pmatrix} + \lambda\begin{pmatrix}3 \\ 6 \\ 4\end{pmatrix}\;}$$

(ii) Through (1, 0, −1) in direction (1, 0, 0)

Here $\vec{a} = \hat{i} - \hat{k}$ and $\vec{d} = \hat{i}$. This line is parallel to the $x$-axis.

$$\boxed{\;\vec{r} = \begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix} + \lambda\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}\;}$$

(iii) Through (1, 0, 4) and (6, 3, −2)

Find the direction: $\vec{d} = \overrightarrow{AB} = \vec{b} - \vec{a} = (6-1)\hat{i} + (3-0)\hat{j} + (-2-4)\hat{k} = 5\hat{i} + 3\hat{j} - 6\hat{k}$.

Using $A(1, 0, 4)$ as the point on the line:

$$\boxed{\;\vec{r} = \begin{pmatrix}1 \\ 0 \\ 4\end{pmatrix} + \lambda\begin{pmatrix}5 \\ 3 \\ -6\end{pmatrix}\;}$$

Individual · Guided Worked Example 2: Midpoint, Ratio Division, and Line MN Exercise 10D Q5

The coordinates of three points are A(−1, −2, 1), B(−3, 4, −5) and C(0, −2, 4).

(i) Find a vector equation of the line AB.

(ii) Find the coordinates of the midpoint M of AB.

(iii) The point N lies on BC. Given that $2|\overrightarrow{BN}| = |\overrightarrow{NC}|$, find the equation of the line MN

👁 Reveal Solution

(i) Vector equation of line AB

$\vec{a} = \begin{pmatrix}-1 \\ -2 \\ 1\end{pmatrix}$, $\vec{b} = \begin{pmatrix}-3 \\ 4 \\ -5\end{pmatrix}$.

Direction: $\overrightarrow{AB} = \vec{b} - \vec{a} = \begin{pmatrix}-3-(-1) \\ 4-(-2) \\ -5-1\end{pmatrix} = \begin{pmatrix}-2 \\ 6 \\ -6\end{pmatrix}$.

$$\boxed{\;\vec{r} = \begin{pmatrix}-1 \\ -2 \\ 1\end{pmatrix} + \lambda\begin{pmatrix}-2 \\ 6 \\ -6\end{pmatrix}\;}$$

The direction can be simplified: $\begin{pmatrix}-2 \\ 6 \\ -6\end{pmatrix} = 2\begin{pmatrix}-1 \\ 3 \\ -3\end{pmatrix}$, so $\vec{r} = \begin{pmatrix}-1 \\ -2 \\ 1\end{pmatrix} + \mu\begin{pmatrix}-1 \\ 3 \\ -3\end{pmatrix}$ is also valid.

(ii) Midpoint M of AB

$\vec{m} = \dfrac{\vec{a} + \vec{b}}{2} = \dfrac{1}{2}\begin{pmatrix}-1+(-3) \\ -2+4 \\ 1+(-5)\end{pmatrix} = \dfrac{1}{2}\begin{pmatrix}-4 \\ 2 \\ -4\end{pmatrix} = \begin{pmatrix}-2 \\ 1 \\ -2\end{pmatrix}$.

$$\boxed{\;M = (-2,\; 1,\; -2)\;}$$

(iii) Point N on BC and the line MN

$2|\overrightarrow{BN}| = |\overrightarrow{NC}|$ means $BN : NC = 1 : 2$. Using the section formula with $B$ and $C$:

$\vec{n} = \dfrac{2\vec{b} + 1\vec{c}}{1 + 2} = \dfrac{2\vec{b} + \vec{c}}{3} = \dfrac{1}{3}\left(2\begin{pmatrix}-3 \\ 4 \\ -5\end{pmatrix} + \begin{pmatrix}0 \\ -2 \\ 4\end{pmatrix}\right) = \dfrac{1}{3}\begin{pmatrix}-6 \\ 6 \\ -6\end{pmatrix} = \begin{pmatrix}-2 \\ 2 \\ -2\end{pmatrix}$.

So $N = (-2, 2, -2)$.

Direction of $MN$: $\overrightarrow{MN} = \vec{n} - \vec{m} = \begin{pmatrix}-2-(-2) \\ 2-1 \\ -2-(-2)\end{pmatrix} = \begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}$.

$$\boxed{\;\vec{r} = \begin{pmatrix}-2 \\ 1 \\ -2\end{pmatrix} + \lambda\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}\;}$$

This line is parallel to the $y$-axis, passing through $M(-2, 1, -2)$.

✏️ Concept Quiz

Score: 0 / 4

1. The vector equation of a line through point $A$ (position vector $\vec{a}$) in the direction of $\vec{d}$ is:

$\vec{r} = \vec{a} + \lambda\vec{d}$

$\vec{r} = \vec{a} - \lambda\vec{d}$

$\vec{r} = \lambda\vec{a} + \vec{d}$

$\vec{r} = \lambda(\vec{a} + \vec{d})$

The vector equation of a line is $\vec{r} = \vec{a} + \lambda\vec{d}$, where $\vec{a}$ is a point on the line, $\vec{d}$ is the direction, and $\lambda$ is a scalar parameter. The trap $\lambda\vec{a} + \vec{d}$ swaps the roles of point and direction. The trap $\lambda(\vec{a}+\vec{d})$ scales the sum, which gives a different line through the origin.

2. The direction vector of the line through $P(1, 2, 3)$ and $Q(4, 6, 0)$ is:

$\begin{pmatrix}3 \\ 4 \\ -3\end{pmatrix}$

$\begin{pmatrix}3 \\ 4 \\ 3\end{pmatrix}$

$\begin{pmatrix}5 \\ 8 \\ 3\end{pmatrix}$

$\begin{pmatrix}-3 \\ -4 \\ 3\end{pmatrix}$

$\overrightarrow{PQ} = \vec{q} - \vec{p} = (4-1)\hat{i} + (6-2)\hat{j} + (0-3)\hat{k} = 3\hat{i} + 4\hat{j} - 3\hat{k}$. The trap with $+3$ on $z$ forgets that $0 - 3 = -3$. The trap $(5, 8, 3)$ adds the components instead of subtracting.

3. Does $P(3, 2, 3)$ lie on the line $\vec{r} = \begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix} + \lambda\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}$?

Yes, with $\lambda = 2$

Yes, with $\lambda = 1$

No, the $\lambda$ values are inconsistent

Cannot be determined

Set $\begin{pmatrix}1\\0\\1\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}3\\2\\3\end{pmatrix}$. From $x$: $1+\lambda=3 \Rightarrow \lambda=2$. From $y$: $0+\lambda=2 \Rightarrow \lambda=2$. From $z$: $1+\lambda=3 \Rightarrow \lambda=2$. All three give $\lambda = 2$, so $P$ lies on the line.

4. The midpoint of $A(2, 4, 6)$ and $B(4, 0, 2)$ is:

$\begin{pmatrix}3 \\ 2 \\ 4\end{pmatrix}$

$\begin{pmatrix}6 \\ 4 \\ 8\end{pmatrix}$

$\begin{pmatrix}1 \\ -2 \\ -2\end{pmatrix}$

$\begin{pmatrix}3 \\ 0 \\ 4\end{pmatrix}$

$M = \dfrac{A+B}{2} = \dfrac{1}{2}\begin{pmatrix}2+4\\4+0\\6+2\end{pmatrix} = \dfrac{1}{2}\begin{pmatrix}6\\4\\8\end{pmatrix} = \begin{pmatrix}3\\2\\4\end{pmatrix}$. The trap $(6,4,8)$ adds without dividing by 2. The trap $(1,-2,-2)$ computes $(B-A)/2$ (subtracts instead of adds). The trap $(3,0,4)$ uses $B$’s $y$-coordinate instead of the average.

🎉

You scored 0 / 4

✏️ Practice A — Individual

Q1 · Exercise 10D Q1 Standard

For each of these pairs of points A and B, write down: (a) the vector $\overrightarrow{AB}$ (b) $|\overrightarrow{AB}|$ (c) the position vector of the midpoint of AB.

(i) A is (2, 3), B is (4, 11). (ii) A is (4, 3), B is (0, 0). (iii) A is (−2, −1), B is (4, 7).

(iv) A is (−3, 4), B is (3, −4). (v) A is (−10, −8), B is (−5, 4)

💡 Hint

(a) $\overrightarrow{AB} = \vec{b} - \vec{a}$ (subtract components).

(b) $|\overrightarrow{AB}| = \sqrt{x^2 + y^2}$ (Pythagoras).

👁 Reveal Solution

(i) A(2, 3), B(4, 11)

(a) $\overrightarrow{AB} = \begin{pmatrix}4-2\\11-3\end{pmatrix} = \begin{pmatrix}2\\8\end{pmatrix} = 2\hat{i} + 8\hat{j}$.

(b) $|\overrightarrow{AB}| = \sqrt{4+64} = \sqrt{68} = \boxed{2\sqrt{17}}$.

(c) Midpoint $= \dfrac{1}{2}\begin{pmatrix}2+4\\3+11\end{pmatrix} = \begin{pmatrix}3\\7\end{pmatrix}$, so position vector $= \boxed{3\hat{i} + 7\hat{j}}$.

(ii) A(4, 3), B(0, 0)

(a) $\overrightarrow{AB} = \begin{pmatrix}0-4\\0-3\end{pmatrix} = \begin{pmatrix}-4\\-3\end{pmatrix} = -4\hat{i} - 3\hat{j}$.

(b) $|\overrightarrow{AB}| = \sqrt{16+9} = \sqrt{25} = \boxed{5}$.

(c) Midpoint $= \dfrac{1}{2}\begin{pmatrix}4+0\\3+0\end{pmatrix} = \begin{pmatrix}2\\\dfrac{3}{2}\end{pmatrix}$, so position vector $= \boxed{2\hat{i} + \dfrac{3}{2}\hat{j}}$.

(iii) A(−2, −1), B(4, 7)

(a) $\overrightarrow{AB} = \begin{pmatrix}4-(-2)\\7-(-1)\end{pmatrix} = \begin{pmatrix}6\\8\end{pmatrix} = 6\hat{i} + 8\hat{j}$.

(b) $|\overrightarrow{AB}| = \sqrt{36+64} = \sqrt{100} = \boxed{10}$.

(c) Midpoint $= \dfrac{1}{2}\begin{pmatrix}-2+4\\-1+7\end{pmatrix} = \begin{pmatrix}1\\3\end{pmatrix}$, so position vector $= \boxed{\hat{i} + 3\hat{j}}$.

(iv) A(−3, 4), B(3, −4)

(a) $\overrightarrow{AB} = \begin{pmatrix}3-(-3)\\-4-4\end{pmatrix} = \begin{pmatrix}6\\-8\end{pmatrix} = 6\hat{i} - 8\hat{j}$.

(b) $|\overrightarrow{AB}| = \sqrt{36+64} = \sqrt{100} = \boxed{10}$.

(c) Midpoint $= \dfrac{1}{2}\begin{pmatrix}-3+3\\4+(-4)\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}$, so position vector $= \boxed{\vec{0}}$ (the origin).

(v) A(−10, −8), B(−5, 4)

(a) $\overrightarrow{AB} = \begin{pmatrix}-5-(-10)\\4-(-8)\end{pmatrix} = \begin{pmatrix}5\\12\end{pmatrix} = 5\hat{i} + 12\hat{j}$.

(b) $|\overrightarrow{AB}| = \sqrt{25+144} = \sqrt{169} = \boxed{13}$.

(c) Midpoint $= \dfrac{1}{2}\begin{pmatrix}-10+(-5)\\-8+4\end{pmatrix} = \begin{pmatrix}-\dfrac{15}{2}\\-2\end{pmatrix}$, so position vector $= \boxed{-\dfrac{15}{2}\hat{i} - 2\hat{j}}$.

Q2 · Exercise 10D Q2 Medium

Find the equation of each of these lines in vector form.

(i) Joining (2, 1) to (4, 5) (ii) Joining (3, 5) to (0, 8) (iii) Joining (−6, −6) to (4, 4)

(iv) Through (5, 3) in the same direction as $\hat{i} + \hat{j}$ (v) Through (2, 1) parallel to $6\hat{i} + 3\hat{j}$

(vi) Through (0, 0) parallel to $\begin{pmatrix}-1 \\ 4\end{pmatrix}$ (vii) Joining (0, 0) to (−2, 8) (viii) Joining (3, −12) to (−1, 4)

💡 Hint

For a line through two points $A$ and $B$: direction $= \overrightarrow{AB} = \vec{b} - \vec{a}$, then $\vec{r} = \vec{a} + \lambda\,\overrightarrow{AB}$.

For a line through a point with a given direction: use $\vec{r} = \vec{a} + \lambda\vec{d}$ directly. Any scalar multiple of $\vec{d}$ gives the same line.

👁 Reveal Solution

(i) Joining (2, 1) to (4, 5)

Direction: $\begin{pmatrix}4-2\\5-1\end{pmatrix} = \begin{pmatrix}2\\4\end{pmatrix}$. $\boxed{\vec{r} = \begin{pmatrix}2\\1\end{pmatrix} + \lambda\begin{pmatrix}2\\4\end{pmatrix}}$ (equivalently, $\vec{r} = \begin{pmatrix}2\\1\end{pmatrix} + \mu\begin{pmatrix}1\\2\end{pmatrix}$).

(ii) Joining (3, 5) to (0, 8)

Direction: $\begin{pmatrix}0-3\\8-5\end{pmatrix} = \begin{pmatrix}-3\\3\end{pmatrix}$. $\boxed{\vec{r} = \begin{pmatrix}3\\5\end{pmatrix} + \lambda\begin{pmatrix}-3\\3\end{pmatrix}}$ (equivalently, $\vec{r} = \begin{pmatrix}3\\5\end{pmatrix} + \mu\begin{pmatrix}-1\\1\end{pmatrix}$).

(iii) Joining (−6, −6) to (4, 4)

Direction: $\begin{pmatrix}4-(-6)\\4-(-6)\end{pmatrix} = \begin{pmatrix}10\\10\end{pmatrix}$. $\boxed{\vec{r} = \begin{pmatrix}-6\\-6\end{pmatrix} + \lambda\begin{pmatrix}10\\10\end{pmatrix}}$ (equivalently, $\vec{r} = \begin{pmatrix}-6\\-6\end{pmatrix} + \mu\begin{pmatrix}1\\1\end{pmatrix}$).

(iv) Through (5, 3) in direction $\hat{i} + \hat{j}$

$\boxed{\vec{r} = \begin{pmatrix}5\\3\end{pmatrix} + \lambda\begin{pmatrix}1\\1\end{pmatrix}}$

(v) Through (2, 1) parallel to $6\hat{i} + 3\hat{j}$

Direction $\begin{pmatrix}6\\3\end{pmatrix} = 3\begin{pmatrix}2\\1\end{pmatrix}$. $\boxed{\vec{r} = \begin{pmatrix}2\\1\end{pmatrix} + \lambda\begin{pmatrix}6\\3\end{pmatrix}}$ (equivalently, $\vec{r} = \begin{pmatrix}2\\1\end{pmatrix} + \mu\begin{pmatrix}2\\1\end{pmatrix}$).

(vi) Through (0, 0) parallel to (−1, 4)

$\boxed{\vec{r} = \lambda\begin{pmatrix}-1\\4\end{pmatrix}}$

(vii) Joining (0, 0) to (−2, 8)

Direction: $\begin{pmatrix}-2-0\\8-0\end{pmatrix} = \begin{pmatrix}-2\\8\end{pmatrix} = 2\begin{pmatrix}-1\\4\end{pmatrix}$. $\boxed{\vec{r} = \lambda\begin{pmatrix}-2\\8\end{pmatrix}}$ (equivalently, $\vec{r} = \mu\begin{pmatrix}-1\\4\end{pmatrix}$).

(viii) Joining (3, −12) to (−1, 4)

Direction: $\begin{pmatrix}-1-3\\4-(-12)\end{pmatrix} = \begin{pmatrix}-4\\16\end{pmatrix} = 4\begin{pmatrix}-1\\4\end{pmatrix}$. $\boxed{\vec{r} = \begin{pmatrix}3\\-12\end{pmatrix} + \lambda\begin{pmatrix}-4\\16\end{pmatrix}}$ (equivalently, $\vec{r} = \begin{pmatrix}3\\-12\end{pmatrix} + \mu\begin{pmatrix}-1\\4\end{pmatrix}$).

💬 Practice B — Collaborative

Q3 · Exercise 10D Q4 Medium

Determine whether the given point P lies on the line in each of the following cases.

(i) P(5,1,4) and the line $\vec{r} = \begin{pmatrix}1 \\3 \\4\end{pmatrix} + \lambda \begin{pmatrix}2 \\-1 \\0\end{pmatrix}$

(ii) P(−1,5,1) and the line $\vec{r} = \begin{pmatrix}1 \\3 \\4\end{pmatrix} + \lambda \begin{pmatrix}2 \\-2 \\3\end{pmatrix}$

(iii) P(−5,3,12) and the line $\vec{r} = \begin{pmatrix}1 \\0 \\-2\end{pmatrix} + \lambda \begin{pmatrix}-2 \\1 \\5\end{pmatrix}$

(iv) P(9,0,−6) and the line $\vec{r} = \begin{pmatrix}1 \\2 \\0\end{pmatrix} + \lambda \begin{pmatrix}4 \\-1 \\-2\end{pmatrix}$

(v) P(−9,−2,−17) and the line $\vec{r} = \begin{pmatrix}1 \\3 \\-2\end{pmatrix} + \lambda \begin{pmatrix}2 \\1 \\3\end{pmatrix}$

💡 Hint 1

Set $\vec{a} + \lambda\vec{d} = \overrightarrow{OP}$ and solve for $\lambda$ from each of the three component equations.

💡 Hint 2

If all three equations give the same $\lambda$, then $P$ lies on the line. If the $\lambda$ values differ, $P$ does not lie on the line.

👁 Reveal Solution

(i) P(5, 1, 4), line r = (1,3,4) + λ(2,−1,0)

$x$: $1 + 2\lambda = 5 \Rightarrow \lambda = 2$.

$y$: $3 - \lambda = 1 \Rightarrow \lambda = 2$.

$z$: $4 + 0 = 4 \;\checkmark$ (consistent for any $\lambda$).

All give $\lambda = 2$. $\boxed{\text{Yes, } P \text{ lies on the line (}\lambda = 2\text{).}}$

(ii) P(−1, 5, 1), line r = (1,3,4) + λ(2,−2,3)

$x$: $1 + 2\lambda = -1 \Rightarrow \lambda = -1$.

$y$: $3 - 2\lambda = 5 \Rightarrow \lambda = -1$.

$z$: $4 + 3\lambda = 1 \Rightarrow \lambda = -1$.

All give $\lambda = -1$. $\boxed{\text{Yes, } P \text{ lies on the line (}\lambda = -1\text{).}}$

(iii) P(−5, 3, 12), line r = (1,0,−2) + λ(−2,1,5)

$x$: $1 - 2\lambda = -5 \Rightarrow \lambda = 3$.

$y$: $0 + \lambda = 3 \Rightarrow \lambda = 3$.

$z$: $-2 + 5\lambda = 12 \Rightarrow 5\lambda = 14 \Rightarrow \lambda = \dfrac{14}{5} = 2.8$.

$\lambda = 3, 3, 2.8$ — inconsistent. $\boxed{\text{No, } P \text{ does not lie on the line.}}$

(iv) P(9, 0, −6), line r = (1,2,0) + λ(4,−1,−2)

$x$: $1 + 4\lambda = 9 \Rightarrow \lambda = 2$.

$y$: $2 - \lambda = 0 \Rightarrow \lambda = 2$.

$z$: $0 - 2\lambda = -6 \Rightarrow \lambda = 3$.

$\lambda = 2, 2, 3$ — inconsistent. $\boxed{\text{No, } P \text{ does not lie on the line.}}$

(v) P(−9, −2, −17), line r = (1,3,−2) + λ(2,1,3)

$x$: $1 + 2\lambda = -9 \Rightarrow \lambda = -5$.

$y$: $3 + \lambda = -2 \Rightarrow \lambda = -5$.

$z$: $-2 + 3\lambda = -17 \Rightarrow 3\lambda = -15 \Rightarrow \lambda = -5$.

All give $\lambda = -5$. $\boxed{\text{Yes, } P \text{ lies on the line (}\lambda = -5\text{).}}$

Q4 · Synthesis [adapted] Challenging

A line passes through $A(1, 2, 3)$ in direction $\vec{d} = (2, -1, 1)$.

(i) Does $P(5, 0, 5)$ lie on the line?

(ii) Find the point on the line closest to the origin.

💡 Hint 1

Part (i): Write the line equation $\vec{r} = \vec{a} + \lambda\vec{d}$, then set it equal to $\overrightarrow{OP}$ and solve for $\lambda$ from each component. Check consistency.

💡 Hint 2

Part (ii): A general point on the line is $Q(\lambda) = \vec{a} + \lambda\vec{d} = (1+2\lambda,\; 2-\lambda,\; 3+\lambda)$. The closest point to the origin is where $\overrightarrow{OQ}$ is perpendicular to the direction $\vec{d}$, i.e. $\overrightarrow{OQ} \cdot \vec{d} = 0$. Solve for $\lambda$.

👁 Reveal Solution

(i) Does P(5, 0, 5) lie on the line?

Line: $\vec{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + \lambda\begin{pmatrix}2\\-1\\1\end{pmatrix}$.

Set equal to $P(5, 0, 5)$:

$x$: $1 + 2\lambda = 5 \Rightarrow \lambda = 2$.

$y$: $2 - \lambda = 0 \Rightarrow \lambda = 2$.

$z$: $3 + \lambda = 5 \Rightarrow \lambda = 2$.

All three give $\lambda = 2$. $\boxed{\text{Yes, } P \text{ lies on the line with } \lambda = 2.}$

(ii) Find the point on the line closest to the origin

A general point on the line: $Q(\lambda) = (1+2\lambda,\; 2-\lambda,\; 3+\lambda)$.

The closest point to $O$ is where $\overrightarrow{OQ} \cdot \vec{d} = 0$ (perpendicularity condition):

$\overrightarrow{OQ} \cdot \vec{d} = (1+2\lambda)(2) + (2-\lambda)(-1) + (3+\lambda)(1) = 0$

$= 2 + 4\lambda - 2 + \lambda + 3 + \lambda = 0$

$= 3 + 6\lambda = 0$

$\lambda = -\dfrac{1}{2}$.

Substitute back:

$Q = \left(1 + 2\left(-\dfrac{1}{2}\right),\; 2 - \left(-\dfrac{1}{2}\right),\; 3 + \left(-\dfrac{1}{2}\right)\right) = (0,\; \dfrac{5}{2},\; \dfrac{5}{2})$.

$$\boxed{\text{Closest point} = \left(0,\; \dfrac{5}{2},\; \dfrac{5}{2}\right)}$$

🎫

Exit Ticket

Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.

I can write the vector equation of a line $\vec{r} = \vec{a} + \lambda\vec{d}$ given a point and a direction, or given two points on the line. I can test whether a given point $P$ lies on a line by solving $\vec{a} + \lambda\vec{d} = \overrightarrow{OP}$ and checking that all three $\lambda$ values agree. I can find the midpoint $M = \dfrac{\vec{a} + \vec{b}}{2}$ of two points and use the section formula $\vec{n} = \dfrac{n\vec{a} + m\vec{b}}{m+n}$ for ratio division. I can combine the line equation with the scalar product to find the point on a line closest to the origin (perpendicularity condition $\overrightarrow{OQ} \cdot \vec{d} = 0$).