H2 Mathematics (9758) - JC1-2025 Chapter 1(a) Vectors – Vector Algebra & Scalar Product

Basic Properties of Vectors in Two- and Three Dimensions

• Vector Operations:

  • Addition of vectors involves combining their corresponding components. For example, if $\vec{a} = \begin{pmatrix} a<em>1 \ a</em>2 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} b<em>1 \ b</em>2 \end{pmatrix}$, then $\vec{a} + \vec{b} = \begin{pmatrix} a<em>1 + b</em>1 \ a<em>2 + b</em>2 \end{pmatrix}$.

  • Subtraction of vectors is similar, but you subtract the components: $\vec{a} - \vec{b} = \begin{pmatrix} a<em>1 - b</em>1 \ a<em>2 - b</em>2 \end{pmatrix}$.

  • Multiplication of a vector by a scalar changes the magnitude of the vector. If $\vec{a} = \begin{pmatrix} a<em>1 \ a</em>2 \end{pmatrix}$ and $c$ is a scalar, then $c\vec{a} = \begin{pmatrix} ca<em>1 \ ca</em>2 \end{pmatrix}$.

  • Geometrical interpretations:

  • Vector addition can be visualized using the parallelogram law or the triangle law. The resultant vector represents the diagonal of the parallelogram formed by the two vectors.

  • Scalar multiplication changes the length of the vector and may reverse its direction if the scalar is negative.

• Types of Vectors:

  • Position vectors: These specify the location of a point in space relative to the origin. For example, the position vector of point P is given by $\vec{OP}$.

  • Displacement vectors: These represent the change in position of an object. If an object moves from point A to point B, the displacement vector is $\vec{AB} = \vec{OB} - \vec{OA}$.

  • Direction vectors: These indicate the direction of a line or a line segment. They are often used to define the orientation of lines and planes in space.

• Vector Properties:

  • Magnitude of a vector: The length of the vector, calculated using the Pythagorean theorem. For a vector $\vec{v} = \begin{pmatrix} x \ y \ z \end{pmatrix}$, the magnitude is $||\vec{v}|| = \sqrt{x^2 + y^2 + z^2}$.

  • Unit vectors: A vector with a magnitude of 1. Any vector can be normalized to a unit vector by dividing it by its magnitude: $\hat{v} = \frac{\vec{v}}{||\vec{v}||}$.

  • Distance between two points: Calculated using the magnitudes of displacement vectors. If the points are A and B, the distance is $||\vec{AB}|| = \sqrt{(x<em>2 - x</em>1)^2 + (y<em>2 - y</em>1)^2 + (z<em>2 - z</em>1)^2}$.

  • Collinearity: Points are collinear if they lie on the same straight line. Vectors connecting these points are parallel, meaning one is a scalar multiple of the other.

• Geometrical Applications:

  • Use of the ratio theorem: This theorem is used to find the position vector of a point that divides a line segment in a given ratio. If point P divides line segment AB in the ratio m:n, then $\vec{OP} = \frac{n\vec{OA} + m\vec{OB}}{m+n}$.

Scalar and Vector Products in Vectors

• Concepts and Properties:

  • Scalar product (dot product): The scalar product of two vectors $\vec{a}$ and $\vec{b}$ is defined as $\vec{a} \cdot \vec{b} = ||\vec{a}|| ||\vec{b}|| \cos\theta$, where $\theta$ is the angle between the vectors. The dot product results in a scalar value.

  • Vector product (cross product): The vector product of two vectors $\vec{a}$ and $\vec{b}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$. Its magnitude is given by $||\vec{a} \times \vec{b}|| = ||\vec{a}|| ||\vec{b}|| \sin\theta$, where $\theta$ is the angle between the vectors. The cross product results in a vector.

• Calculations:

  • Magnitude of a vector: As mentioned earlier, for a vector $\vec{v} = \begin{pmatrix} x \ y \ z \end{pmatrix}$, the magnitude is $||\vec{v}|| = \sqrt{x^2 + y^2 + z^2}$.

  • Angle between two vectors: This can be found using the dot product formula: $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{||\vec{a}|| ||\vec{b}||}$. The angle $\theta$ can then be found using the inverse cosine function.

• Geometrical Meanings:

  • $a \cdot \hat{n}$: Represents the component of vector $\vec{a}$ in the direction of the unit vector $\hat{n}$. This is the projection of $\vec{a}$ onto $\hat{n}$.

  • $a \times \hat{n}$: Results in a vector perpendicular to both $\vec{a}$ and $\hat{n}$. The magnitude of this vector is the area of the parallelogram formed by $\vec{a}$ and $\hat{n}$.
    

    *Note: Triple products $a \cdot (b \times c)$ and $a \times (b \times c)$ are excluded.

Three-Dimensional Vector Geometry

• Equations:

  • Vector equations of lines: A line can be represented as $\vec{r} = \vec{a} + t\vec{d}$, where $\vec{r}$ is the position vector of any point on the line, $\vec{a}$ is the position vector of a known point on the line, $\vec{d}$ is the direction vector of the line, and $t$ is a scalar parameter.

  • Cartesian equations of lines: These can be derived from the vector equation by expressing the components in terms of x, y, and z. For example, $\frac{x - x<em>0}{a} = \frac{y - y</em>0}{b} = \frac{z - z<em>0}{c}$, where $(x</em>0, y<em>0, z</em>0)$ is a point on the line and $(a, b, c)$ is the direction vector.

  • Vector equations of planes: A plane can be represented as $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$, where $\vec{r}$ is the position vector of any point on the plane, $\vec{a}$ is the position vector of a known point on the plane, and $\vec{n}$ is a normal vector to the plane.

  • Cartesian equations of planes: These can be derived from the vector equation as $ax + by + cz = d$, where $(a, b, c)$ is the normal vector to the plane, and $d$ is a constant.

• Calculations:

  • Foot of the perpendicular from a point to a line or a plane: This involves finding the point on the line or plane that is closest to the given point, which requires using vector projections and solving systems of equations.

  • Distance from a point to a line: Calculated using the formula $d = \frac{||(\vec{a} - \vec{p}) \times \vec{v}||}{||\vec{v}||}$, where $\vec{a}$ is the position vector of the point, $\vec{p}$ is the position vector of a point on the line, and $\vec{v}$ is the direction vector of the line.

  • Distance from a point to a plane: Calculated using the formula $d = \frac{|(\vec{a} - \vec{p}) \cdot \vec{n}|}{||\vec{n}||}$, where $\vec{a}$ is the position vector of the point, $\vec{p}$ is the position vector of a point on the plane, and $\vec{n}$ is the normal vector to the plane.

  • Angle between two lines: Found using the direction vectors of the lines. If $\vec{d<em>1}$ and $\vec{d</em>2}$ are the direction vectors, then $\cos\theta = \frac{\vec{d<em>1} \cdot \vec{d</em>2}}{||\vec{d<em>1}|| ||\vec{d</em>2}||}$.

  • Angle between a line and a plane: Found using the direction vector of the line and the normal vector of the plane. If $\vec{d}$ is the direction vector of the line and $\vec{n}$ is the normal vector of the plane, then $\sin\theta = \frac{|\vec{d} \cdot \vec{n}|}{||\vec{d}|| ||\vec{n}||}$.

  • Angle between two planes: Found using the normal vectors of the planes. If $\vec{n<em>1}$ and $\vec{n</em>2}$ are the normal vectors, then $\cos\theta = \frac{\vec{n<em>1} \cdot \vec{n</em>2}}{||\vec{n<em>1}|| ||\vec{n</em>2}||}$.

• Relationships:

  • Relationships between two lines (coplanar or skew): Coplanar lines lie in the same plane, while skew lines do not. Coplanar lines can be parallel or intersecting. The condition for coplanarity can be determined by checking if the scalar triple product of the vectors connecting points on the lines is zero.

  • Relationships between a line and a plane: A line can be parallel to the plane, perpendicular to the plane, or lie in the plane. These relationships can be determined by examining the dot product of the direction vector of the line and the normal vector of the plane.

  • Relationships between two planes: Two planes can be parallel, perpendicular, or intersecting. The angle between the planes can be found using their normal vectors.
    

    *Note: Finding the shortest distance between two skew lines and finding an equation for the common perpendicular to two skew lines are excluded.

References

• Websites:

  • http://www.h2maths.site

  • https://www.mathsisfun.com/algebra/vectors.html

• Books:

  • Ho Soo Thong, Tay Yong Chiang & Koh Khee Meng, “College Mathematics Syllabus C Volume 1”, Pan Pacific Publications.

  • Pure Mathematics by Alan Sherlock, Elizabeth Roebuck, Timothy Heneage, Shirley Beck: Chapter 16

  • Pure Mathematics 2 by L Bostock, S Chandler: Chapter 5

  • My First Step in using TI-84 Plus CE for H1 and H2 Math by Attal Lam Chapter 1(a): Vector Algebra and Scalar Product

Introduction

• Pre-requisite: GCE “O” Level Elementary Mathematics – 2-dimensional vectors

• Scalar Quantity: Defined by magnitude alone (e.g., length, distance, speed). Scalars are fully described by a numerical value and do not have a direction.

• Vector Quantity: Has both magnitude and direction (e.g., force, displacement, velocity). Vectors require both a numerical value and a direction to be fully defined.

• Geometrical Representation: A vector a (directed from P to Q) is represented geometrically by a directed line $\vec{PQ}$. The length of the line represents the magnitude of the vector, and the arrow indicates its direction.

• Notation: $\vec{PQ} = a$ or $\vec{a}$ (in the direction from P to Q).

2-Dimensional / 3-Dimensional Vectors

• 2D Vectors:

  • Examples: $\begin{pmatrix} 1 \ 1 \ \end{pmatrix}$, $\begin{pmatrix} -4 \ 2 \ \end{pmatrix}$. These vectors lie in a two-dimensional plane and are defined by two components.

• 3D Vectors:

  • Examples: $\begin{pmatrix} 1 \ 1 \ 4 \ \end{pmatrix}$, $\begin{pmatrix} -4 \ 2 \ 0 \ \end{pmatrix}$. These vectors exist in a three-dimensional space and are defined by three components.

• Column vector form. Vectors are often represented in column form to facilitate matrix operations and calculations.

Geometrical Representation of 2D Vectors

• 2D vectors can be represented on the Cartesian plane.

  • The $i$ value corresponds to the x-coordinate.

  • The $j$ value corresponds to the y-coordinate.

  • The zero vector corresponds to the origin $(0, 0)$, denoted by $0$ or $\vec{0}$. It has no magnitude or direction.

  • $i = \begin{pmatrix} 1 \ 0 \ \end{pmatrix}$, $j = \begin{pmatrix} 0 \ 1 \ \end{pmatrix}$. These are the standard unit vectors along the x and y axes, respectively.

• Example:

  • $\vec{OA} = 5i + 2j = \begin{pmatrix} 5 \ 2 \ \end{pmatrix}$

• Magnitude/Modulus of $\vec{OA}$:

  • $|\vec{OA}| = \sqrt{5^2 + 2^2} = \sqrt{29}$ (by Pythagoras theorem). The magnitude represents the length of the vector from the origin to point A.

• General Magnitude/Modulus of a Vector $\vec{PQ} = \begin{pmatrix} x \ y \ \end{pmatrix} = xi + yj$:

  • $|\vec{PQ}| = \sqrt{x^2 + y^2}$

Geometrical Representation of 3D Vectors

• 3D vectors can be represented on the Cartesian space.

  • The $i$ value corresponds to the x-coordinate.

  • The $j$ value corresponds to the y-coordinate.

  • The $k$ value corresponds to the z-coordinate.

  • $0$ or $\vec{0}$ corresponds to the origin $(0, 0, 0)$.

  • $i = \begin{pmatrix} 1 \ 0 \ 0 \ \end{pmatrix}$, $j = \begin{pmatrix} 0 \ 1 \ 0 \ \end{pmatrix}$, $k = \begin{pmatrix} 0 \ 0 \ 1 \ \end{pmatrix}$. These are the standard unit vectors along the x, y, and z axes, respectively.

• Example:

  • $\vec{OB} = 10i + 4j + 5k = \begin{pmatrix} 10 \ 4 \ 5 \ \end{pmatrix}$

• Magnitude/Modulus of $\vec{OB}$:

  • $|\vec{OB}| = \sqrt{10^2 + 4^2 + 5^2} = \sqrt{141}$

• General Magnitude/Modulus of a Vector $\vec{PQ} = \begin{pmatrix} x \ y \ z \ \end{pmatrix} = xi + yj + zk$ :

  • $|\vec{PQ}| = \sqrt{x^2 + y^2 + z^2}$
    

    *Recall : In Chp 0A, $| x |$ is the distance of the real number x from zero on a number line.

Example 1

Find the magnitude of the following vectors, in exact form:

• (i) $\vec{OA} = 3i + 4j$

• (ii) $\vec{OB} = 6i + 3j + 4k$

• (iii) $\vec{OC} = \begin{pmatrix} 1 \ 1 \ -2 \ \end{pmatrix}$

Solution

• (i) $|\vec{OA}| = \sqrt{3^2 + 4^2} = 5$

• (ii) $|\vec{OB}| = \sqrt{6^2 + 3^2 + 4^2} = \sqrt{61}$

• (iii) $|\vec{OC}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6}$

General Types of Vectors

• Equal Vectors

• Negative Vectors

• Parallel Vectors

• Unit Vectors

• Position Vectors

• Displacement Vectors

Equal Vectors

Two vectors are equal if and only if they have the same magnitude and direction. Equal vectors are indistinguishable in terms of their effect.

• i.e. $\vec{a} = \vec{b} \Leftrightarrow |\vec{a}| = |\vec{b}|$ AND $\vec{a}$ and $\vec{b}$ are in the same direction

Negative Vectors

If two vectors $\vec{a}$ and $\vec{b}$ have the same magnitude but are parallel in opposite directions, then $\vec{a} = -\vec{b}$ or $-\vec{a} = \vec{b}$. Negative vectors effectively reverse the direction of a vector while keeping its magnitude constant.

• From the diagram, we can see that the directed line segments $\vec{PQ}$ and $\vec{QP}$ have the same magnitude but are parallel in opposite directions.

• $\therefore \vec{PQ} = -\vec{QP}$

Given that ABCD is a parallelogram, state all the pairs of equal vectors.

• $\vec{AB} = \vec{DC}$ and $\vec{AD} = \vec{BC}$

Parallel (//) Vectors

If $\vec{a}$ and $\vec{b}$ are non-zero vectors, then

$\vec{a} // \vec{b} \Leftrightarrow \vec{a} = \lambda \vec{b}$ for some $\lambda \in \mathbb{R}$, where $\lambda \neq 0$.

In other words, if $\vec{a}$ and $\vec{b}$ are any two vectors and $\vec{a} = \lambda \vec{b}$ for some $\lambda \in \mathbb{R}$, \
eq 0, then there are 2 possibilities .

• Either $\vec{a}$ and $\vec{b}$ are parallel,

• $\vec{a} = \vec{b} = 0$

Note: To prove ABCD is a parallelogram, we need to prove either $\vec{AB} = \vec{DC}$ OR $\vec{AD} = \vec{BC}$. This is because opposite sides of a parallelogram are equal and parallel.

Example 2

Determine if the following pairs of vectors are parallel

• (i) $4i + 4j – 2k$ and $-2i – 2j + k$

• (ii) $3i – 5k$ and $\begin{pmatrix} 0 \ 3 \ -5 \ \end{pmatrix}$.

Solution

• (i) Since $\begin{pmatrix} 4 \ 4 \ -2 \ \end{pmatrix} = -2 \begin{pmatrix} -2 \ -2 \ 1 \ \end{pmatrix}$
  $\Rightarrow$ they are parallel.

• (ii) There exist no unique $\lambda$ such that $\begin{pmatrix} 3 \ 0 \ -5 \ \end{pmatrix} = \lambda \begin{pmatrix} 0 \ 3 \ -5 \ \end{pmatrix}$
  $\Rightarrow$ they are not parallel.

Example 3

Given that $2i + hj + 4k$ is parallel to $5i – 4j + pk$, find the values of $h$ and $p$.

Solution

Since $\begin{pmatrix} 2 \ h \ 4 \ \end{pmatrix}$ is parallel to $\begin{pmatrix} 5 \ -4 \ p \ \end{pmatrix}$,

$\begin{pmatrix} 2 \ h \ 4 \ \end{pmatrix} = \lambda \begin{pmatrix} 5 \ -4 \ p \ \end{pmatrix}$

for some $\lambda \in \mathbb{R}$ where $\lambda \neq 0$.

$\begin{aligned}2 &amp;= 5\lambda \ h &amp;= -4\lambda \ 4 &amp;= p\lambda \end{aligned}$

From (1), $\lambda = \frac{2}{5}$

From (2), $h = -\frac{8}{5}$

From (3), $p = \frac{10}{5} = 2$

Unit Vector

The unit vector of $\vec{a}$, denoted by $\hat{a}$, is the vector in the direction of $\vec{a}$ with magnitude 1, i.e. $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$. Unit vectors are useful for specifying direction without magnitude.

Example 4

Given $\vec{a} = i + 2j -k$, find $\hat{a}$.

Solution

$|\vec{a}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6}$

Therefore, $\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{6}} \begin{pmatrix} 1 \ 2 \ -1 \ \end{pmatrix}$

Example 5

Find the vector of length 6 in the direction of $2i + 2j + k$.

Solution

Let $\vec{a}$ be the required vector.

Then, $\hat{a} = \frac{2i + 2j + k}{\sqrt{2^2 + 2^2 + 1^2}} = \frac{1}{3} \begin{pmatrix} 2 \ 2 \ 1 \ \end{pmatrix}$

$\vec{a} = 6\hat{a} = 6 \cdot \frac{1}{3} \begin{pmatrix} 2 \ 2 \ 1 \ \end{pmatrix} = \begin{pmatrix} 4 \ 4 \ 2 \ \end{pmatrix}$

Position Vectors

To specify the position of a point P, we introduce a point of reference, which is the fixed point O (origin).

The position vector of a point P relative to the point O is indicated by the directed line segment $\vec{OP}$. Position vectors are essential for describing spatial relationships.

The vector $\vec{OP}$ is called the position vector of P relative to O.

Displacement Vectors

The displacement vector $\vec{v}$ with initial point $(x<em>1, y</em>1, z<em>1)$ and ending point $(x</em>2, y<em>2, z</em>2)$ is given by

$\vec{v} = \begin{pmatrix} x<em>2 - x</em>1 \ y<em>2 - y</em>1 \ z<em>2 - z</em>1 \ \end{pmatrix}$.

E.g. Given the points A ($-4 , 1, 3$) and B($-2, -5, 2$), then

the position vector $\vec{OA} = \begin{pmatrix} -4 \ 1 \ 3 \ \end{pmatrix}$ and the position vector $\vec{OB} = \begin{pmatrix} -2 \ -5 \ 2 \ \end{pmatrix}$.

The displacement vector $\vec{AB} = \begin{pmatrix} 2 \ -6 \ -1 \ \end{pmatrix}$ and the displacement vector $\vec{BA} = \begin{pmatrix} -2 \ 6 \ 1 \ \end{pmatrix}$.

The distance between A and B = $\sqrt{2^2 + (-6)^2 + (-1)^2} = \sqrt{41}$

Vector Algebra

Vector Addition and Subtraction

If $\vec{a} = a<em>1i + a</em>2j + a<em>3k = \begin{pmatrix} a</em>1 \ a<em>2 \ a</em>3 \ \end{pmatrix}$ and $\vec{b} = b<em>1i + b</em>2j + b<em>3k = \begin{pmatrix} b</em>1 \ b<em>2 \ b</em>3 \ \end{pmatrix}$, then

$\vec{a} \pm \vec{b} = (a<em>1 \pm b</em>1)i + (a<em>2 \pm b</em>2)j + (a<em>3 \pm b</em>3)k = \begin{pmatrix} a<em>1 \pm b</em>1 \ a<em>2 \pm b</em>2 \ a<em>3 \pm b</em>3 \ \end{pmatrix}$.

Note: These results can be extended to more than 2 vectors.

Example 6

Given that $\vec{a} = 2i + 3j + 2k$, $\vec{b} = 4i – 3j + k$ and $\vec{c} = 6i – 11j$. Find

• (i) $\vec{a} + \vec{b}$,

• (ii) $\vec{a} – \vec{b} + \vec{c}$.

Solution

• (i) $\vec{a} + \vec{b} = \begin{pmatrix} 2 \ 3 \ 2 \ \end{pmatrix} + \begin{pmatrix} 4 \ -3 \ 1 \ \end{pmatrix} = \begin{pmatrix} 6 \ 0 \ 3 \ \end{pmatrix}$

• (ii) $\vec{a} – \vec{b} + \vec{c} = \begin{pmatrix} 2 \ 3 \ 2 \ \end{pmatrix} - \begin{pmatrix} 4 \ -3 \ 1 \ \end{pmatrix} + \begin{pmatrix} 6 \ -11 \ 0 \ \end{pmatrix} = \begin{pmatrix} 4 \ -5 \ 1 \ \end{pmatrix}$

Geometrical Representation of Vector Addition and Subtraction

• The vector $\vec{OP}$ is defined as the sum of $\vec{a}$ and $\vec{b}$ and is written as $\vec{a} + \vec{b}$ or $\vec{b} + \vec{a}$. Vector addition is commutative, meaning the order does not affect the result.

• The vector $\vec{OQ}$ is defined as the difference of $\vec{a}$ and $\vec{b}$ and is written as $\vec{a} - \vec{b}$ or $-\vec{b} + \vec{a}$. Vector subtraction is not commutative.

• The vector $\vec{OR}$ is defined as the difference of $\vec{a}$ and $\vec{b}$ and is written as $-\vec{a} + \vec{b}$ or $\vec{b} - \vec{a}$.

• Similarly, for any vector $\vec{AB}$, it can be expressed as follows:-
  $\vec{AB} = \vec{AO} + \vec{OB} = -\vec{OA} + \vec{OB} = \vec{OB} - \vec{OA}$

• $\vec{AB}$ is called displacement vector.

Multiplication of a Vector by a Scalar

If $\vec{a} = a<em>1i + a</em>2j + a<em>3k = \begin{pmatrix} a</em>1 \ a<em>2 \ a</em>3 \ \end{pmatrix}$, then

$\lambda \vec{a} = \lambda a<em>1i + \lambda a</em>2j + \lambda a<em>3k = \begin{pmatrix} \lambda a</em>1 \ \lambda a<em>2 \ \lambda a</em>3 \ \end{pmatrix}$.

Example 7

Given that $\vec{a} = 2i + 3j + 2k$ and $\vec{b} = 4i – 3j + k$. Find

• (i) $\vec{a} + 2\vec{b}$

• (ii) $3\vec{a} – \vec{b}$

Solution

• (i) $\vec{a} + 2\vec{b} = \begin{pmatrix} 2 \ 3 \ 2 \ \end{pmatrix} + 2\begin{pmatrix} 4 \ -3 \ 1 \ \end{pmatrix} = \begin{pmatrix} 10 \ -3 \ 4 \ \end{pmatrix}$

• (ii) $3\vec{a} – \vec{b} = 3\begin{pmatrix} 2 \ 3 \ 2 \ \end{pmatrix} - \begin{pmatrix} 4 \ -3 \ 1 \ \end{pmatrix} = \begin{pmatrix} 2 \ 12 \ 5 \ \end{pmatrix}$

Geometrical Representation of Multiplication of a Vector by a Scalar

For any $\lambda \in \mathbb{R}$, $|\lambda \vec{a}| = |\lambda| |\vec{a}|$. Scalar multiplication changes the magnitude of the vector by a factor of |$\lambda$|.

Important results of Vector Operations

• (i) $\vec{a} + \vec{b} = \vec{b} + \vec{a}$

• (ii) $(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})$

• (iii) $\lambda (\vec{a}) = \lambda \vec{a}$, $\lambda \in \mathbb{R}$

• (iv) $\lambda (\mu \vec{a}) = (\lambda \mu) \vec{a}$, $\lambda, \mu \in \mathbb{R}$

• (v) $(\lambda + \mu)\vec{a} = \lambda \vec{a} + \mu \vec{a}$

Collinear Points

If 3 points A, B and C are collinear, i.e. A, B, and C all lie on a straight line, then

$\vec{AB} = \lambda \vec{BC}$, for some $\lambda \in \mathbb{R}$, $\lambda \neq 0$. Collinear points are fundamental in geometry and linear algebra.

Example 8

The position vectors of points A, B and C relative to a fixed point O are $\vec{a} = 4i – 9j – k$, $\vec{b} = i + 3j + 5k$ and $\vec{c} = 2i – j + 3k$. Prove that A, B and C are collinear.

Solution

$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 1 \ 3 \ 5 \ \end{pmatrix} - \begin{pmatrix} 4 \ -9 \ -1 \ \end{pmatrix} = \begin{pmatrix} -3 \ 12 \ 6 \ \end{pmatrix}$

$\vec{AC} = \vec{OC} - \vec{OA} = \begin{pmatrix} 2 \ -1 \ 3 \ \end{pmatrix} - \begin{pmatrix} 4 \ -9 \ -1 \ \end{pmatrix} = \begin{pmatrix} -2 \ 8 \ 4 \ \end{pmatrix}$

Since $\vec{AB} = \frac{3}{2} \vec{AC}$

A, B and C are collinear.

Ratio Theorem

Consider a point P, which divides the line segment AB in the ratio $\lambda : \mu$

The position vector of P, $\vec{OP} = \vec{OA} + \vec{AP}$ write $\vec{AP}$ in terms of $\vec{AB}$

$$\vec{OP} = \vec{OA}