3d Vectors: Equation of line

Understand and visualise the equation of a 3d vector line.

📚 Further Maths 🎯 Difficulty: ⭐⭐⭐ ⏱️ Reading time: 15 minutes 📋 Edexcel

📚 Introduction

(PRIOR KNOWLEDGE OF VECTORS IS HIGHLY RECCOMENDED) In this lesson, we will explore the equation of a line in 3D space, represented by a vector equation. We will learn how to derive the vector equation of a line from its parametric equations (functions of variables) and understand the geometric interpretation of the line in three-dimensional space.

🎯 Learning Objectives

Understand the concept of a line in 3D space
Learn how to derive the vector equation of a line from its parametric equations
Visualize the geometric interpretation of a line in three-dimensional space

You will have met vectors in normal A-Level Maths, but in Further Maths you will go deeper into the topic. This lesson is part of the Vectors CP1 module, which is a core part of the Further Maths curriculum.

I found 3D vectors to be quite challenging, hence the 4 star difficulty rating. This was mainly due to the fact that textbooks aren't the greatest for visualisations, so watching videos alongside this lesson is highly recommended. I strongly reccomend ExamSolutions on YouTube.

🔑 Key Concepts

Equation of a line in 3D

The equation of a line in 3D can be represented using a vector equation. The general form of the vector equation of a line is given by:

\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}

Where:
$\mathbf{r} \quad$ is the position vector of any point on the line
$\mathbf{a} \quad$ is the position vector of a specific point on the line
$\mathbf{b} \quad$ is the direction vector of the line
$\lambda \quad$ is a scalar parameter/quantity

A position vector ($\mathbf{r}$) is just a vector that starts at the origin, $(0,0,0)$ and goes anywhere else.

A direction vector ($\mathbf{b}$) is just the vector that shows direction.

A scalar ($\lambda$) is just a number that acts like a multiplier, increasing or decreasing the line length.

(Above) is the way everyone else teaches it, but i think this is much more intuitive:

\mathbf{r} = \overrightarrow{OA} + \lambda \overrightarrow{AB}

Where:
$A$ is a point on the line, so $\overrightarrow{OA}$ is the position vector from origin to point A
$B$ is another point on the line, so $\overrightarrow{AB}$ is from point A to point B, and is the direction vector <
$O$ is the origin, $(0,0,0)$

This means that you start at point A, and then move in the direction of vector AB, scaled by $\lambda$.
By changing the value of $\lambda$, you can move along the line in both directions.

I created a great visualisation:

In the above, the purple line is the result of $\overrightarrow{OA} + \lambda \overrightarrow{AB}$, which is $\mathbf{r}$.

🔑 More Concepts

3D must-knows

Coordinates in a 3D space is represented just like it is in 2D, but with an extra z axis in in the mix, to account for the 3rd dimension.
For example: 2D coordinates: (x, y)
3D coordinates: $(x, y, z) \quad \rightarrow \quad$ Example: $(3, -2, 5)$

Furthermore
There are multiple forms of vectors, ijk form:

In $ijk$ form, a point in 3D space is represented as: $$\mathbf{r} = xi + yj + zk$$

Column form: $$\begin{bmatrix} x \\ y \\ z \end{bmatrix}$$
For example:

\text{The position vector from origin to point } (3, -2, 5) \text{ is:}$$ $$\text{Column form: }\begin{bmatrix} 3 \\ -2 \\ 5 \end{bmatrix} \quad | \quad ijk \text{ form: } 3i - 2j + 5k

📝 Worked Examples

📋 Example 1: Simple example:

We have two points: A and B, which are $(4, 5, -7)$ and $(1, 2, 3)$ respectively. Find the equation of a line passing through these points.

Solution:

Step 1:

Recall the formula:

r = \overrightarrow{OA} + \lambda \overrightarrow{AB} \quad \text{or} \quad r = \mathbf{a} + \lambda \mathbf{b}$$ $$\text{Where: } \overrightarrow{OA} = a, \overrightarrow{AB} = b$$ $$\overrightarrow{OA} \text{ is simply the position vector from origin to point A}$$ $$\overrightarrow{OA} = \begin{bmatrix} 4 \\ 5 \\ -7 \end{bmatrix} \quad \text{or,} \quad = 4i + 5j - 7k

Step 2:

Lets now find AB.

{\small \text{To find } \overrightarrow{AB}, \text{think of it as going from point A to the origin, to point B.}}$$ $${\small \text{OA is origin to A. we want AB. so we start from A,}}$$ $${\small \text{ and go back to origin then to B.}}$$ $${\small \text{From A to origin is just negative OA: }} -\overrightarrow{OA} = -\begin{bmatrix} 4 \\ 5 \\ -7 \end{bmatrix} = \begin{bmatrix} -4 \\ -5 \\ 7 \end{bmatrix}$$ $${\small \text{Then from origin to B is just OB: }} \overrightarrow{OB} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$$ $${\small \text{So: }} \overrightarrow{AB} = -\overrightarrow{OA} + \overrightarrow{OB} \rightarrow \begin{bmatrix} -4 \\ -5 \\ 7 \end{bmatrix} + \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = \begin{bmatrix} -3 \\ -3 \\ 10 \end{bmatrix}$$ $${\small \text{(just add/subtract the x y and z components separately.)}}

Step 3:

Now we can write the full equation of the line:

r = \overrightarrow{OA} + \lambda \overrightarrow{AB}$$ $${\small \text{We found: }} \overrightarrow{OA} = \begin{bmatrix} 4 \\ 5 \\ -7 \end{bmatrix} {\small \text{ and }} \overrightarrow{AB} = \begin{bmatrix} -3 \\ -3 \\ 10 \end{bmatrix}$$ $${\small \text{So: }} r = \begin{bmatrix} 4 \\ 5 \\ -7 \end{bmatrix} + \lambda \begin{bmatrix} -3 \\ -3 \\ 10 \end{bmatrix}$$ $${\small \text{This is the equation of the line passing through points A and B.}}$$ $${\small \text{ We can also write it as: }}$$ $$\boxed{\mathbf{r} = \begin{bmatrix} 4 \\ 5 \\ -7 \end{bmatrix} + \lambda \begin{bmatrix} -3 \\ -3 \\ 10 \end{bmatrix}}$$ $$\text{Or, equivalently:}$$ $$\boxed{\mathbf{r} = \begin{bmatrix} 4 - 3\lambda \\ 5 - 3\lambda \\ -7 + 10\lambda \end{bmatrix}}

For this line to intersect a point, the point must satisfy the equation of the line. The way we can check this is by finding lambda and seeing if it is the same for all: $x$, $y$ and $z$.

This means that the point must be a multiple of the direction vector, $\overrightarrow{AB}$.

For example, the point $C \rightarrow \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$, and a direction vector is $\lambda \begin{bmatrix} -1 \\ -2 \\ -3 \end{bmatrix}$.

We can solve for $\lambda$ in each, $x$, $y$ and $z$. if $\lambda$ is the same for all three, then the point lies on the line.

In this case, $\lambda = -1$, and is consistent for each $x$,$y$ and $z$ meaning C lies on the line.

📋 Example 2: Line intersecting a point

The straight line $r = (6i + 3j + k) + \lambda (-3i - 3j + 10k)$ intersects point $C = (a, b, 4)$. Find the values of $a$ and $b$.

Solution:

Step 1:

Understand that $C$ must be a multiple of the direction vector $\lambda (-3i - 3j + 10k)$.

\text{since we have: } C = (a, b, 4)\text{, we have the z value.}$$ $$\text{set the z component of the line equal to 4 and solve.}$$ $$\lambda(10k) = 4, \rightarrow \lambda = \underline{\frac{4}{10} = \frac{2}{5}}

Step 2:

Now we have $\lambda$ we can find $a$ and $b$:

\lambda = \frac{2}{5}$$ $$\text{So plug this into the direction vector of the line:}$$ $$\lambda \begin{bmatrix} -3 \\ -3 \\ 10 \end{bmatrix} \rightarrow \begin{bmatrix} -3 \lambda \\ -3 \lambda \\ 10 \lambda \end{bmatrix} = \begin{bmatrix} -3 \times \frac{2}{5} \\ -3 \times \frac{2}{5} \\ 10 \times \frac{2}{5} \end{bmatrix} = \begin{bmatrix} -\frac{6}{5} \\ -\frac{6}{5} \\ 4 \end{bmatrix}$$ $$\text{Ensure you compute x, y and z separately.}

Step 3:

Final Answer:

\boxed{C = \begin{bmatrix} -\frac{6}{5} \\ -\frac{6}{5} \\ 4 \end{bmatrix}}

Cartesian form of Straight line

The straight line equation that we have just been working with - $ r = \mathbf{a} + \lambda \mathbf{b} $ - has Cartesian form.

The cartesian form is:

\underbrace{r = (a_1,a_2,a_3) + \lambda (b_1,b_2,b_3)}_{\text{parametric form}} \quad \rightarrow \quad \underbrace{\frac{x - a_1}{b_1} = \frac{y - a_2}{b_2} = \frac{z - a_3}{b_3}}_{\text{cartesian form}}

Each of the fractions in the cartesian form are equal to $\lambda$

Proof:

\begin{bmatrix} x\\y\\z \end{bmatrix} = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} + \lambda \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$$ $$ x = a_1 + \lambda b_1 \rightarrow \text{Rearranging for } \lambda \rightarrow \lambda = \boxed{\frac{x - a_1}{b_1}}$$ $$ y = a_2 + \lambda b_2 \rightarrow \text{Rearranging for } \lambda \rightarrow \lambda = \boxed{\frac{y - a_2}{b_2}}$$ $$ z = a_3 + \lambda b_3 \rightarrow \text{Rearranging for } \lambda \rightarrow \lambda = \boxed{\frac{z - a_3}{b_3}}$$ $$ \underline{\text{As you can see, they're all equal to } \lambda}

💡 Tip:

Most of the time, its really good practice to compute all the differrent coordinate axis's seperately. In vectors, it gets hard to track things over time so this makes it easier.

Also, you must know:

Collinear: if points are collinear, that means that a line can be made which intersects all points.
Magnitude: its basically how far the line is and is measured as: $$\sqrt{(b_1^2 + b_2^2 + b_3^2)}$$
Parallel: if two lines are parallel, their direction vectors are scalar multiples of each other.
Perpendicular: if two lines are perpendicular, the dot product of their direction vectors is zero. We will learn about this later.
Skew: if two lines are skew, they are not parallel and do not intersect.
a vector perpendicular to a plane or another vector is called the 'normal'.

⚠️ Common Mistakes

🚫 Mistake 1: Forgetting cartesian form and other forms

What students often do wrong: Forgetting the carteesian form is easy but make sure you rememeber. The direction vector is at the denominator.

Also, if the numerator is $x + a_1$, then that means $a_1$ is negative because $x + a_1 \quad = \quad x - (-a_1)$

You can write each component in $ a + \lambda b$ form to make it easier like this:
$$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} + \lambda \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} = \underline{\begin{bmatrix} a_1 + \lambda b_1 \\ a_2 + \lambda b_2 \\ a_3 + \lambda b_3 \end{bmatrix}}$$

📋 Example 3: Cartesian form problem with collinear points

Points A, B, and C have position vectors (with respect to the origin O) given by:
$$\vec{a} = \begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix}, \quad \vec{b} = \begin{bmatrix} 4 \\ 9 \\ 8 \end{bmatrix},\quad \vec{c} = \begin{bmatrix} 7 \\ 15 \\ 14 \end{bmatrix}$$
(a) Show that points A, B, and C are collinear by demonstrating that the direction vectors between them are parallel.
(b) Hence, write the Cartesian equation of the line passing through points A and B, expressing your answer in the form:
$$\frac{x - x_1}{d_1} = \frac{y - y_1}{d_2} = \frac{z - z_1}{d_3}$$

Solution:

Step 1:

Find the direction vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$:

\overrightarrow{AB}=\vec b-\vec a =\begin{bmatrix}4-1\\[4pt]9-3\\[4pt]8-2\end{bmatrix} =\begin{bmatrix}3\\[4pt]6\\[4pt]6\end{bmatrix},$$ $$\overrightarrow{AC}=\vec c-\vec a =\begin{bmatrix}7-1\\[4pt]15-3\\[4pt]14-2\end{bmatrix} =\begin{bmatrix}6\\[4pt]12\\[4pt]12\end{bmatrix}.$$ $$\text{Notice that}$$ $$\overrightarrow{AC}=2\overrightarrow{AB},$$ $$\text{So the direction vectors are parallel and points A, B, C are collinear.}

Step 2:

We found from the last question that:

\overrightarrow{AB}=\begin{bmatrix}3\\6\\6\end{bmatrix}.

Remember the form i gave you: $r = \overrightarrow{OA} + \lambda\overrightarrow{AB}$

\text{Remember that } \overrightarrow{OA} \text{ is just the position vector of point A: } \begin{bmatrix}1\\3\\2\end{bmatrix}.$$ $$\text{We have: } \overrightarrow{OA} = \begin{bmatrix}1\\3\\2\end{bmatrix} \text{ and } \overrightarrow{AB} = \begin{bmatrix}3\\6\\6\end{bmatrix}.$$ $$\text{So the vector equation of the line is: } \quad \overrightarrow{AB} $$ $$\mathbf{r} = \begin{bmatrix}1\\3\\2\end{bmatrix} + \lambda \begin{bmatrix}3\\6\\6\end{bmatrix}.

Step 3:

Lastly, lets write it in Cartesian form:

\text{Remember, from Parametric form to Cartesian form:}$$ $$\underbrace{r = (a_1,a_2,a_3) + \lambda (b_1,b_2,b_3)}_{\text{parametric form}} \rightarrow \underbrace{\frac{x - a_1}{b_1} = \frac{y - a_2}{b_2} = \frac{z - a_3}{b_3}}_{\text{cartesian form}}$$ $$\text{So we have: } a_1 = 1, a_2 = 3, a_3 = 2$$ $$\text{and } b_1 = 3, b_2 = 6, b_3 = 6$$ $$\text{So the Cartesian form is: }$$ $$\frac{x - 1}{3} = \frac{y - 3}{6} = \frac{z - 2}{6}.$$ $$\text{Equivalently, using the simplified direction vector } \langle1,2,2\rangle,$$ $$\boxed{\frac{x - 1}{1} = \frac{y - 3}{2} = \frac{z - 2}{2}.}

💡 Tip:

Know that you can simplify the direction vector. This is because it shows direction only, and lambda corrects any faulty magnitude, as it acts as a multiplier.

💪 Practice Problems

Try these problems to test your understanding:

🎯 Practice Questions A

(a) Find a vector equation of the line that passes through the following pairs of points:

i) (3, -2, 6) and (5, 0, 7)
ii) (-4, 1, 2) and (6, -3, 5)
iii) (0, 4, -5) and (3, 1, 1)
iv) (1, -6, 3) and (8, 2, -1)

(b) For each line above, write down the Cartesian equation in the form:
$$\frac{x - x_1}{d_1} = \frac{y - y_1}{d_2} = \frac{z - z_1}{d_3}$$

🔍 Show Solution

Question 1 Solution: $$i) \text{ Points: } (3, -2, 6) \text{ and } (5, 0, 7)$$ $$\text{Direction vector: } \vec{d} = \begin{bmatrix} 5 - 3 \\ 0 - (-2) \\ 7 - 6 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix}$$ $$\text{Vector equation: } \vec{r} = \begin{bmatrix} 3 \\ -2 \\ 6 \end{bmatrix} + \lambda \begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix}$$ $$ii) \text{ Points: } (-4, 1, 2) \text{ and } (6, -3, 5)$$ $$\text{Direction vector: } \vec{d} = \begin{bmatrix} 6 - (-4) \\ -3 - 1 \\ 5 - 2 \end{bmatrix} = \begin{bmatrix} 10 \\ -4 \\ 3 \end{bmatrix}$$ $$\text{Vector equation: } \vec{r} = \begin{bmatrix} -4 \\ 1 \\ 2 \end{bmatrix} + \lambda \begin{bmatrix} 10 \\ -4 \\ 3 \end{bmatrix}$$ $$iii) \text{ Points: } (0, 4, -5) \text{ and } (3, 1, 1)$$ $$\text{Direction vector: } \vec{d} = \begin{bmatrix} 3 - 0 \\ 1 - 4 \\ 1 - (-5) \end{bmatrix} = \begin{bmatrix} 3 \\ -3 \\ 6 \end{bmatrix}$$ $$\text{Vector equation: } \vec{r} = \begin{bmatrix} 0 \\ 4 \\ -5 \end{bmatrix} + \lambda \begin{bmatrix} 3 \\ -3 \\ 6 \end{bmatrix}$$ $$iv) \text{ Points: } (1, -6, 3) \text{ and } (8, 2, -1)$$ $$\text{Direction vector: } \vec{d} = \begin{bmatrix} 8 - 1 \\ 2 - (-6) \\ -1 - 3 \end{bmatrix} = \begin{bmatrix} 7 \\ 8 \\ -4 \end{bmatrix}$$ $$\text{Vector equation: } \vec{r} = \begin{bmatrix} 1 \\ -6 \\ 3 \end{bmatrix} + \lambda \begin{bmatrix} 7 \\ 8 \\ -4 \end{bmatrix}$$ $$(b) \text{ Cartesian equations of the lines}$$ $$i) \quad \frac{x - 3}{2} = \frac{y + 2}{2} = \frac{z - 6}{1}$$ $$ii) \quad \frac{x - (-4)}{10} = \frac{y - 1}{-4} = \frac{z - 2}{3}$$ $$iii) \quad \frac{x - 0}{3} = \frac{y - 4}{-3} = \frac{z - (-5)}{6}$$ $$iv) \quad \frac{x - 1}{7} = \frac{y + 6}{8} = \frac{z - 3}{-4}$$

🎯 Practice Questions B

(a) The points $P(2, 0, 4)$, $Q(a, 5, 1)$, and $R(3, 10, b)$, where $a$ and $b$ are constants, are collinear.
Find the values of $a$ and $b$.

(b) The line $l_1$ has the equation:
$$ \vec{r} = (8\mathbf{i} - 5\mathbf{j} + 4\mathbf{k}) + \lambda(3\mathbf{i} + \mathbf{j} - 6\mathbf{k}) $$ Point $A$ lies on line $l_1$ such that $\lambda = -2$.
The line $l_2$ passes through point $A$ and is parallel to the line:
$$ \vec{r} = (10\mathbf{i} + 3\mathbf{j} - 9\mathbf{k}) + \mu(2\mathbf{i} - 4\mathbf{j} + \mathbf{k}) $$ Find a vector equation for line $l_2$.

(c) The point $A$ with coordinates $(4, a, 0)$ lies on the line $L$ with vector equation:
$$ \vec{r} = (10\mathbf{i} + 8\mathbf{j} - 12\mathbf{k}) + \lambda(\mathbf{i} - \mathbf{j} + b\mathbf{k}) $$ where $a$ and $b$ are constants.
i) Find the values of $a$ and $b$.
The point $X$ lies on line $L$ where $\lambda = -1$.
ii) Find the coordinates of point $X$.

🔍 Show Solution

(a) Solution:
$$\text{Let } P = (2, 0, 4), Q = (a, 5, 1), R = (3, 10, b).$$ $$\text{For collinearity, } \overrightarrow{PQ} \text{ and } \overrightarrow{PR} \text{ must be parallel.}$$ $$\overrightarrow{PQ} = (a - 2, 5 - 0, 1 - 4) = (a - 2, 5, -3)$$ $$\overrightarrow{PR} = (3 - 2, 10 - 0, b - 4) = (1, 10, b - 4)$$ $$\text{So, } \overrightarrow{PQ} = k \cdot \overrightarrow{PR} \text{ for some scalar } k.$$ $$\text{Set up equations:}$$ $$a - 2 = k \cdot 1$$ $$5 = k \cdot 10$$ $$-3 = k \cdot (b - 4)$$ $$\text{From } 5 = 10k, k = 0.5$$ $$a - 2 = 0.5 \implies a = 2.5$$ $$-3 = 0.5(b - 4) \implies -6 = b - 4 \implies b = -2$$ $$\text{Final answers: } \boxed{a = 2.5, \quad b = -2}$$

(b) Solution:
$$\text{First, find point } A \text{ on } l_1 \text{ for } \lambda = -2:$$ $$A = (8, -5, 4) + (-2)(3, 1, -6) = (8 - 6, -5 - 2, 4 + 12) = (2, -7, 16)$$ $$\text{Direction vector for } l_2 \text{ is the same as the direction vector of the other line: } (2, -4, 1)$$ $$\text{So, equation for } l_2 \text{ is:}$$ $$\vec{r} = \begin{bmatrix} 2 \\ -7 \\ 16 \end{bmatrix} + \lambda \begin{bmatrix} 2 \\ -4 \\ 1 \end{bmatrix}$$ $$\text{Vector equation for } l_2: \boxed{\vec{r} = (2\mathbf{i} - 7\mathbf{j} + 16\mathbf{k}) + \lambda(2\mathbf{i} - 4\mathbf{j} + \mathbf{k})}$$

(c) Solution:
$$\text{i) } A = (4, a, 0) \text{ lies on } L:$$ $$\text{Let } \vec{r} = (10, 8, -12) + \lambda(1, -1, b)$$ $$\text{Set } x = 4, y = a, z = 0$$ $$4 = 10 + \lambda(1) \implies \lambda = -6$$ $$a = 8 + \lambda(-1) = 8 - (-6) = 14$$ $$0 = -12 + \lambda b \implies -12 + (-6)b = 0 \implies -6b = 12 \implies b = -2$$ $$\text{Final answers: } \boxed{a = 14, \quad b = -2}$$

$$\text{ii) For } \lambda = -1:$$ $$\vec{r} = (10, 8, -12) + (-1)(1, -1, -2) = (10 - 1, 8 + 1, -12 + 2) = (9, 9, -10)$$ $$\text{Coordinates of } X: \boxed{(9, 9, -10)}$$

📚 Summary

In this lesson, we explored the concept of vector equations of lines in 3D space. We learned how to find the vector equation of a line given two points, as well as how to determine if points are collinear by checking if the direction vectors are parallel.

We also covered the Cartesian form of a line and how to convert between parametric and Cartesian forms. Key takeaways include understanding direction vectors, position vectors, and the significance of collinearity in 3D geometry.

🔗 Further resources

For more mastery:

🎮 Interactive Practice

💡 Pro Tip: Visualize with Desmos 3D

I highly suggest you have a play around in Desmos 3D with both Cartesian and parametric forms of vectors. It's an amazing way to visualize how lines behave in 3D space and really solidify your understanding of the concepts we've covered.

Try plotting some of the examples from this lesson and experiment with different direction vectors and position vectors to see how they affect the line's orientation and position.

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