Vector Foundations and Products

Overview

This branch develops the Vector I layer of the topic. The main purpose is to make vector algebra meaningful before it is used for lines, planes, angles, and distances.

A vector should be read as a directed movement. Column-vector notation is a compact way of recording that movement component by component.

Caption: The same vector idea can be read as an arrow, a column of components, or an algebraic object.

Scalars, Vectors, and Components

A scalar has magnitude only. A vector has magnitude and direction.

Examples of scalars include distance, speed, mass, and time. Examples of vectors include displacement, velocity, force, and acceleration.

In two dimensions,

$$\mathbf{u}=\left(\begin{array}{c}{u}_1\\ u_2\end{array}\right)$$

means a movement of $u_1$ units in the first coordinate direction and $u_2$ units in the second coordinate direction.

In three dimensions,

$$\mathbf{u}=\left(\begin{array}{c}{u}_1\\ u_2\\ u_3\end{array}\right).$$

Two vectors are equal when they have the same magnitude and direction. They do not need to start at the same point in a diagram.

Magnitude and Unit Vectors

The magnitude of

$$\mathbf{u}=\left(\begin{array}{c}{u}_1\\ u_2\\ u_3\end{array}\right)$$

is

$$|\mathbf{u}|=\sqrt{u_1^2+u_2^2+u_3^2}.$$

If $\mathbf{u}\ne 0$, the unit vector in the direction of $\mathbf{u}$ is

$$\hat{\mathbf{u}}=\frac{\mathbf{u}}{|\mathbf{u}|}.$$

This operation keeps the direction but changes the magnitude to $1$.

Position and Displacement Vectors

A position vector locates a point from the origin. A displacement vector compares two points directly.

If $A$ and $B$ have position vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$, then

$$\boxed{\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}}$$

This is not just a formula. It says:

$$\text{movement from }A\text{ to }B=\text{position of }B-\text{position of }A.$$

Caption: A displacement vector does not depend on where the origin is placed.

Ratio Theorem

Ratio theorem is a position-vector method for locating a point on a line segment or produced line.

Suppose $P$ divides $AB$ internally in the ratio

$$AP:PB=\lambda :\mu .$$

Then

$$\boxed{\overrightarrow{OP}=\frac{\mu \overrightarrow{OA}+\lambda \overrightarrow{OB}}{\lambda +\mu }}$$

The weights are crossed because $P$ is closer to the endpoint with the larger opposite weight.

Caption: If $AP:PB=\lambda :\mu$, the position vector of $P$ is a weighted average of $\overrightarrow{OA}$ and $\overrightarrow{OB}$ using the opposite segment weights.

Midpoint Case

If $P$ is the midpoint of $AB$, then

$$AP:PB=1:1,$$

so

$$\overrightarrow{OP}=\frac{\overrightarrow{OA}+\overrightarrow{OB}}{2}.$$

External Division

If $P$ divides $AB$ externally, the same geometric idea is used but one directed part is taken with the opposite sign. This is why the diagram and the direction of division matter.

Do not use the internal-division formula blindly when $P$ lies outside the segment $AB$.

Vector Addition and Subtraction

Vector addition combines movements.

If

$$\mathbf{a}=\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right),\mathbf{b}=\left(\begin{array}{c}{b}_1\\ b_2\end{array}\right),$$

then

$$\mathbf{a}+\mathbf{b}=\left(\begin{array}{c}{a}_1+b_1\\ a_2+b_2\end{array}\right).$$

Geometrically, this can be read using the triangle law or the parallelogram law.

Caption: Addition follows directed movement; subtraction reverses a vector before adding.

Subtraction is addition of the negative vector:

$$\mathbf{a}-\mathbf{b}=\mathbf{a}+(-\mathbf{b}).$$

Scalar Multiplication and Parallel Vectors

For a scalar $k$,

$$k\mathbf{a}$$

has the same line of direction as $\mathbf{a}$.

• If $k>0$, the direction is unchanged.
• If $k<0$, the direction is reversed.
• If $k=0$, the result is the zero vector.

Two non-zero vectors $\mathbf{a}$ and $\mathbf{b}$ are parallel if

$$\mathbf{a}=k\mathbf{b}$$

for some scalar $k$.

This is often the quickest way to test collinearity in coordinate problems.

Scalar Product

The scalar product of two vectors is defined by

$$\boxed{\mathbf{a}\cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos \theta }$$

where $\theta$ is the angle between the two vectors.

Caption: The scalar product measures the component of one vector in the direction of the other, scaled by the other vector’s magnitude.

For components,

$$\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)\cdot \left(\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right)=a_1{b}_1+a_2{b}_2+a_3{b}_3.$$

The scalar product returns a scalar, not a vector.

What It Measures

The scalar product measures directional agreement.

• If $\mathbf{a}\cdot \mathbf{b}>0$, the angle is acute.
• If $\mathbf{a}\cdot \mathbf{b}=0$, the vectors are perpendicular.
• If $\mathbf{a}\cdot \mathbf{b}<0$, the angle is obtuse.

Finding Angles

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

$$\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}.$$

For acute angles between lines, use an absolute value if direction vectors may point in opposite chosen directions:

$$\cos \theta =\frac{|\mathbf{a}\cdot \mathbf{b}|}{|\mathbf{a}||\mathbf{b}|}.$$

Projections and Resolving

The scalar projection of $\mathbf{a}$ in the direction of a unit vector $\hat{\mathbf{u}}$ is

$$\mathbf{a}\cdot \hat{\mathbf{u}}.$$

The vector projection is

$$(\mathbf{a}\cdot \hat{\mathbf{u}})\hat{\mathbf{u}}.$$

Resolving $\mathbf{a}$ into parts parallel and perpendicular to $\hat{\mathbf{u}}$ gives

$$\mathbf{a}=(\mathbf{a}\cdot \hat{\mathbf{u}})\hat{\mathbf{u}}+\left[\mathbf{a}-(\mathbf{a}\cdot \hat{\mathbf{u}})\hat{\mathbf{u}}\right].$$

Caption: Resolving separates the part along a chosen direction from the part perpendicular to it.

Vector Product

For 3D vectors, the vector product is a vector perpendicular to both input vectors.

Its magnitude is

$$\boxed{|\mathbf{a}\times \mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin \theta }$$

Caption: The vector product creates a normal vector, and its magnitude gives the area of the parallelogram spanned by the two vectors.

For

$$\mathbf{a}=\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right),\mathbf{b}=\left(\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right),$$

the vector product can be calculated by

$$\mathbf{a}\times \mathbf{b}=\left(\begin{array}{c}{a}_2{b}_3-a_3{b}_2\\ a_3{b}_1-a_1{b}_3\\ a_1{b}_2-a_2{b}_1\end{array}\right).$$

This formula is easier to reconstruct if it is read as a component-by-component calculation rather than memorised as three unrelated expressions.

Write

$$\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k},\mathbf{b}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}.$$

Then set up the calculation as

$$\mathbf{a}\times \mathbf{b}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|.$$

This determinant layout is a calculation device. The first-row sign pattern is

$$+,-,+.$$

Expanding along the first row gives

$$\mathbf{a}\times \mathbf{b}=(a_2{b}_3-a_3{b}_2)\mathbf{i}-(a_1{b}_3-a_3{b}_1)\mathbf{j}+(a_1{b}_2-a_2{b}_1)\mathbf{k}.$$

Equivalently,

$$\mathbf{a}\times \mathbf{b}=\left(\begin{array}{c}{a}_2{b}_3-a_3{b}_2\\ a_3{b}_1-a_1{b}_3\\ a_1{b}_2-a_2{b}_1\end{array}\right).$$

The middle sign is the common trap. One safe way to calculate is:

1. For the $\mathbf{i}$ component, cover the first column and calculate $a_2{b}_3-a_3{b}_2$.
2. For the $\mathbf{j}$ component, cover the second column and calculate the negative of $a_1{b}_3-a_3{b}_1$.
3. For the $\mathbf{k}$ component, cover the third column and calculate $a_1{b}_2-a_2{b}_1$.

For example, if

$$\mathbf{a}=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right),\mathbf{b}=\left(\begin{array}{c}4\\ 0\\ 5\end{array}\right),$$

then

$$\mathbf{a}\times \mathbf{b}=\left(\begin{array}{c}2(5)-3(0)\\ 3(4)-1(5)\\ 1(0)-2(4)\end{array}\right)=\left(\begin{array}{c}10\\ 7\\ -8\end{array}\right).$$

You should still check the meaning after calculating: the result should be perpendicular to both original vectors. Here,

$$\left(\begin{array}{c}10\\ 7\\ -8\end{array}\right)\cdot \left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)=10+14-24=0,$$

and

$$\left(\begin{array}{c}10\\ 7\\ -8\end{array}\right)\cdot \left(\begin{array}{c}4\\ 0\\ 5\end{array}\right)=40-40=0.$$

So the calculated vector is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$, as a vector product should be.

Use the vector product when you need:

• a vector perpendicular to two given vectors
• the area of a parallelogram
• a normal vector to a plane
• a test for parallel vectors, since $\mathbf{a}\times \mathbf{b}=0$ for parallel non-zero vectors

Core Examples

Example 1: Displacement from Position Vectors

If

$$\overrightarrow{OA}=\left(\begin{array}{c}3\\ 4\end{array}\right),\overrightarrow{OB}=\left(\begin{array}{c}-5\\ -1\end{array}\right),$$

then

$$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\left(\begin{array}{c}-8\\ -5\end{array}\right).$$

Example 2: Ratio Theorem

Suppose

$$\overrightarrow{OA}=\left(\begin{array}{c}2\\ 1\\ 4\end{array}\right),\overrightarrow{OB}=\left(\begin{array}{c}6\\ 5\\ -2\end{array}\right).$$

If $P$ is the midpoint of $AB$, then

$$\overrightarrow{OP}=\frac{1}{2}\left[\left(\begin{array}{c}2\\ 1\\ 4\end{array}\right)+\left(\begin{array}{c}6\\ 5\\ -2\end{array}\right)\right]=\left(\begin{array}{c}4\\ 3\\ 1\end{array}\right).$$

Example 3: Projection

Let

$$\mathbf{a}=\left(\begin{array}{c}3\\ 4\\ 0\end{array}\right),\mathbf{u}=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right).$$

Since $\mathbf{u}$ is already a unit vector, the projection of $\mathbf{a}$ in the direction of $\mathbf{u}$ is

$$(\mathbf{a}\cdot \mathbf{u})\mathbf{u}=3\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}3\\ 0\\ 0\end{array}\right).$$

The perpendicular component is

$$\left(\begin{array}{c}3\\ 4\\ 0\end{array}\right)-\left(\begin{array}{c}3\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}0\\ 4\\ 0\end{array}\right).$$

Common Errors

• Confusing a vector with the point where an arrow is drawn.
• Treating a position vector and displacement vector as the same object.
• Forgetting that $\overrightarrow{AB}$ means ending position minus starting position.
• Applying the internal Ratio Theorem to an external division problem.
• Assuming vector division exists.
• Treating $\mathbf{a}\cdot \mathbf{b}$ as a vector.
• Forgetting that scalar product measures directional agreement through cosine.
• Using vector product when a scalar projection is needed.
• Forgetting that vector product is a 3D operation in this syllabus context.
• Losing the negative sign in the middle component of a vector product.

Revision Checklist

• Can I explain why a vector is not just a pair or triple of numbers?
• Can I move between arrows and column vectors?
• Can I derive $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$ from a diagram?
• Can I use the Ratio Theorem and identify internal versus external division?
• Can I test parallelism using scalar multiples?
• Can I use scalar product for angles, perpendicularity, and projections?
• Can I use vector product to find a perpendicular vector or area?
• Can I resolve a vector into parallel and perpendicular components?

Related

• Vectors
• Vector Lines, Planes, and 3D Geometry