If you asked for directions and simply got the answer "\(400\, m\)", this would not be helpful at all. What way should you go? Should you go \(400\, m\) left, right, forwards, backwards? For all you know, you might have to dig a hole \(400\, m\) deep underground, or fly \(400\, m\) into the air. This answer is not helpful because they have only provided a scalar quantity, a simple distance. If they had given you a distance with a direction, you would know exactly what to do. A distance with a direction is a vector quantity.
Definition of Vector Calculus
Vectors are mathematical objects representing movements or points in more than one dimension.
A vector is a mathematical object that has both direction and magnitude. A 2-dimensional vector can be written as:
\[ \begin{bmatrix} x \\ y\end{bmatrix} = x \vec{i} + y \vec{j}. \]
The left is called a column vector, and the second is called component form.
\( \vec{i}\) and \( \vec{j} \) are known as the standard unit vectors. These can be written as:
\[ \vec{i} = \begin{bmatrix} 1 \\ 0\end{bmatrix}, \: \vec{j} = \begin{bmatrix} 0\\1 \end{bmatrix}. \]
On a computer, vectors are often lowercase letters and written in bold. When handwritten, it is standard to underline them \((\underline{v})\), overline them \((\overline{v})\) or draw an arrow above them \((\overrightarrow{v})\). If you are specifically talking about a vector between two points, say point \(A\) and point \(B,\) this vector is normally written as the two points with an arrow above them, \( \overrightarrow{AB}.\)
Vectors can be thought of as arrows, pointing from one place to another. If the vector in 2d space is \( 3\vec{i} + 2 \vec{j},\) and the vector begins at the origin, it will point to \( (3, 2) \) on the \((x,y)\) plane.
A vector can be thought of as an arrow, pointing from one place to another.
The vector above could represent the movement of 3 units in the \(x\) direction and 2 units in the \(y\) direction, or it could represent the point \( (3, 2) \) in the \( (x,y)\) plane. Because of this, we distinguish vectors into direction vectors and position vectors.
The direction vector \( a \vec{i} + b \vec{j}\) is a vector representing a movement of \(a\) in the positive \(x\) direction and \(b\) in the positive \(y\) direction.
The position vector \( a \vec{i} + b \vec{j}.\) represents the point \( (a, b) \) in 2D space.
If you apply a direction vector from the origin, you will get to the corresponding position vector.
It is important to be able to write column vectors in component form and vice-versa. Let's look at some examples of this.
Vectors \(\vec{u}\) and \(\vec{v}\) are given below.
\[ \begin{align} \vec{u} & = \begin{bmatrix} 3 \\ -1 \end{bmatrix} \\ \vec{v} & = 3 \vec{i} + \vec{j}. \end{align} \]
Write
- Vector \(\vec{u}\) in component form,
- Vector \(\vec{v} \) in column vector form.
Solution
1. To write a column vector in component form, you write the number in the first position as the coefficient of \( \vec{i} \) and the number in the second position as the coefficient of \( \vec{j} \).
\[ \vec{u} = 3 \vec{v} - \vec{j}. \]
2. To write a vector in component form as a column vector, simply put the coefficients of each unit vector into their position in the column vector, remembering: the coefficient of \(\vec{i} \) goes in the first position, and the coefficient of \(\vec{j} \) goes in the second position. This gives:
\[ \vec{v} = \begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix}. \]
Vector Addition Calculus
Just as with normal numbers, vectors can be added, subtracted, and multiplied. Let's first look at the addition and subtraction of vectors.
Addition and Subtraction of Vectors
When vectors are added, it is essentially like lining the arrows of the direction vectors tip to tip.
When vectors are added, it is essentially like placing the arrows of the direction vectors tip to tip.
Above are two vectors \(\vec{v}\) and \(\vec{u}\), being added together. As you can see, the sum of these two vectors is the same as just stacking the vectors tip to tip. This makes sense when you think about a vector as a form of movement. If you first walk 3 steps right and 2 steps forwards, have a break, and then walk another step right and another two steps forward, you have moved to the same position as if you had just walked 4 steps forward and 4 steps right in one go.
Vectors can be subtracted visually by joining the arrows up tip to tip but drawing the vector that is being subtracted backward.
Similarly to vector addition, a vector \(\vec{v}\) can be subtracted from a vector \(\vec{u}\) by putting them tip to tip, but with vector \(\vec{v}\) facing in the opposite direction.
Numerically, vectors can be added or subtracted by adding or subtracting the individual components. In component form, this makes visual sense, as it looks exactly the same as collecting like terms when doing algebra.
Find
- \( (\vec{i} + 2 \vec{j}) + (4\vec{i} - 3\vec{j}), \)
- \( ( 3\vec{i} - \vec{j}) - (\vec{i} - 2 \vec{j}). \)
Solution
1. You can add up all the terms as if the unit vectors were any other type of algebraic quantity,
\[ (\vec{i} + 2 \vec{j}) + (4\vec{i} - 3\vec{j}) = 5 \vec{i} - \vec{j}. \]
2. Here you can expand the bracket normally as if they were other algebraic quantities, and then simplify it:
\[ \begin{align} ( 3\vec{i} - \vec{j}) - (\vec{i} - 2 \vec{j}) & = 3 \vec{i} - \vec{j} - \vec{i} + 2 \vec{j}\\ & = 2 \vec{i} + \vec{j} \end{align} \]
For column vectors, just add or subtract the numbers that are in the same position in each vector together.
Find
- \[ \begin{bmatrix} 3 \\ -1 \end{bmatrix} + \begin{bmatrix} 2 \\ -4 \end{bmatrix}. \]
- \[ \begin{bmatrix} 1 \\ -1 \end{bmatrix} - \begin{bmatrix} 4 \\ 1 \end{bmatrix}. \]
Solution
1. The number in the first position of our new vector will be the number in the first position of our first vector (3) added to the number in the first position of our second vector (2), so it will be 5. Do the same for the other row to get
\[ \begin{bmatrix} 3 \\ -1 \end{bmatrix} + \begin{bmatrix} 2 \\ -4 \end{bmatrix} = \begin{bmatrix} 5 \\ -5 \end{bmatrix}. \]
2. Do exactly the same as for question one, but this time subtracting the second number instead of adding it to the first corresponding number:
\[ \begin{bmatrix} 1 \\ -1 \end{bmatrix} - \begin{bmatrix} 4 \\ 1 \end{bmatrix} = \begin{bmatrix} -3 \\ -2 \end{bmatrix}. \]
Vector Calculus Properties
There are many important properties of vectors within calculus, but first you must learn about the three methods of multiplication that exists for vectors.
Multiplication of a Vector by a Scalar
Vectors can be multiplied by real numbers. The real numbers here are called "scalers", because they scale the vector to a different size.
If a vector is multiplied by 3, it is essentially the same as stacking 3 of that vector tip to tip. If the scalar is negative, the output of the multiplication will be facing in the opposite direction to the original vector. This reflects the multiplication of real numbers, as when you multiply \(x\) and \(y\) together, it is the same as adding \(x\) to itself \(y\) times.
Vectors can be multiplied by a scalar \(a\) visually by drawing the arrow tip to tip with itself a times.
Numerically, the multiplication of a vector by a scalar is done by multiplying each component in the vector by that scalar. For a column vector, this just means multiplying each entry in the vector by the scalar. For a vector in normal vector form, this looks just like expanding a bracket in any other equation.
This form of multiplication allows us to determine when two vectors are parallel.
A vector \(\vec{v}\) is parallel to another vector \(\vec{u}\) if and only if there is a scalar \(a\) such that \( \vec{v} = a \vec{u}.\)
Don't confuse the multiplication of a vector by a scalar with the scalar multiple of vectors. The scalar multiple of vectors is a way of multiplying vectors together, giving a scalar as the output. For more information, see Scalar Products.
Let's see some examples of multiplying a vector by a scalar.
The vectors \(\vec{v}\) and \(\vec{u}\) are given below,
\[ \begin{align} \vec{u} & = \begin{bmatrix} 2 \\ -1 \end{bmatrix} \\ \vec{v} & = 4 \vec{i} - \vec{j}. \end{align} \]
Find
- \[ 3 \vec{u}. \]
- \[ -\frac{1}{2} \vec{v}. \]
Solution
1. Multiply each component of the column vector by 3:
\[ \begin{align} 3 \vec{u} & = 3 \begin{bmatrix} 2 \\ -1 \end{bmatrix} \\ & = \begin{bmatrix} 6 \\ -3 \end{bmatrix}. \end{align} \]
2. Multiply each component of the column vector by \(-\frac{1}{2}\):
\[ \begin{align} -\frac{1}{2} \vec{v} & = -\frac{1}{2} (4\vec{i}-\vec{j}) \\&= -2 \vec{i}+\frac{1}{2}\vec{j} .\end{align} \]
The Dot Product and Cross Product
The dot product and scaler product are two different ways of multiplying two vectors together. The dot product gives a scaler as the output, while the cross product gives another vector as an output.
Given two three dimensional vectors \( \vec{a} = a_1 \vec{i} + a_2 \vec{j} + a_3 \vec{k} \) and \( \vec{b} = b_1 \vec{i} + b_2 \vec{j} + b_3 \vec{k} \), the following are true:
The dot product or scalar product of a 2D vector is
\[ \vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3. \]
The cross product or vector product of a 3D vector is
\[ \vec{a} \times \vec{b} = \begin{bmatrix} a_2 b_3 - a_3 b_2 \\ a_1 b_3 - a_3 b_1 \\ a_1 b_2 - a_2 b_1 \end{bmatrix}. \]
The dot product can be thought of as representing how much two vectors 'overlap'. This means that if two vectors are parallel and pointing in the same direction, the dot product will be maximized, but if the two vectors are orthogonal, the dot product will be 0.
The scalar product of two vectors gives a third vector that is perpendicular to both vectors, and is 0 when the vectors are perpendicular. In 2D, the dot product is considered the standard product, and the cross product does not exist. For AP, you do not need to work with vectors in more that 2 dimensions, so only the dot product is required.
Let's look at an example of finding the dot product
Given the following vectors:
\[ \vec{a} = \begin{bmatrix} 4 \\ -2 \\ 3 \end{bmatrix}, \hspace{1cm} \vec{b} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}, \]
find \( \vec{a} \cdot \vec{b}. \)
SolutionMultiply together the component pairs, and add them all up.
\[ \begin{align} \vec{a} \cdot \vec{b} & = 4 \cdot 1 + (-2) \cdot 0 + 3 \cdot (-1) \\ & = 4 + 0 -3 \\ & = 1 . \end{align} \]
Properties of Vector Addition and Multiplication
Just as there are properties of regular arithmetic, such as associativity, distributivity and others, the same exist for vector arithmetic. For any vectors \(\vec{p}, \vec{q}, \vec{r}\) and scalars \( a, b\) the following properties hold:
Commutativity: \[ \vec{p} + \vec{q} = \vec{q} + \vec{p} .\]
Associativity of addition: \[ (\vec{p} + \vec{q}) + \vec{r} = \vec{p} + (\vec{p} \vec{r}). \]
Distributivity of vectors: \[ a (\vec{p} + \vec{q}) = a \vec{p} + a \vec{q} .\]
Distributivity of scalars: \[ (a + b) \vec{p} = a \vec{p} + b \vec{p}. \]
Associativity of scalars: \[ a (b \vec{p}) = (ab) \vec{p}. \]
Commutativity of dot product: \[ \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}. \]
Distributivity of dot product over addition: \[\vec{a} \cdot ( \vec{b} + \vec{c} ) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}. \]
Anticommutativity of cross product: \[ \vec{a} \times \vec{b} = - \vec{b} \times \vec{a}. \]
Distributivity of cross product over addition: \[ \vec{a} \times (\vec{b} + \vec{c} ) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}. \]
Scalar triple product: \[\begin{align} \vec{a} \cdot ( \vec{b} \times \vec{c} ) &= \vec{b} \cdot ( \vec{c} \times \vec{a}) \\ &= \vec{c} \cdot (\vec{a} \times \vec{b}). \end{align}\]
Important Formulas for Vectors
There are many important formulas about vectors within mathematics. Here you will look at the most essential formulas.
Head Minus Tail Rule
An important equation in Vector mathematics is the head minus tail rule. This is a rule for calculating the vector between two points, if you have vectors going from another point to each of those two points.
The vector between two points can be found using the head minus tail rule.
Given vectors \(\vec{OA}, \vec{OB},\) the vector \(\vec{AB}\) is:\[ \vec{AB} = \vec{OB} - \vec{OA} .\]
If the point \(O\) is the origin, this simplifies to:
\[ \vec{AB} = \vec{B} = \vec{A}, \]
where \(\vec{A}, \vec{B}\) are the position vectors of points \(A\) and \(B\) respectively.
Magnitude of a Vector
As stated in the definition of a vector, a vector has both direction and magnitude. Magnitude is the length of the vector, and can be calculated using the Pythagorean Theorem.
The magnitude of a vector is its length. For the vector,
\[ \vec{v} = \begin{bmatrix} x \\ y \end{bmatrix} = x \vec{i} + y \vec{j}, \]
the magnitude of \( \vec{v} \) is
\[ | \vec{v} | = \sqrt{ x^2 + y^2 }. \]
The magnitude of a vector can also be called the Euclidean norm (often just norm) or the modulus of the vector.
The magnitude of the vector represents the distance you would be travelling if you walked directly along the line. This distance is often known colloquially as "as the crow flies". Since this is a distance, the magnitude of a vector is always positive, assuming all of its components are real numbers.
The magnitude, or Euclidean norm, is not the only way of calculating the length of a vector, but it is the most common. In some scenarios, it makes more sense to use a different way of defining length.
For example, if you wish to get the Eurostar train from London to Amsterdam, you must first go from London to Brussels, and then from Brussels to Amsterdam. This is a much greater distance than the Euclidean distance from London to Amsterdam, so a different norm must be used when working out train travel distances.
The notation for the magnitude of a vector looks just like the absolute value of a real or complex number, and this is no coincidence. Just as the magnitude here represents the distance "as the crow flies" of the vector, the absolute value or modulus of a real or complex number is how far away that number is from the origin.
Let's look at some examples of calculating the magnitude of a vector.
Calculate the magnitude of the following vectors:
- \[ \vec{u} = 3 \vec{i} - 4 \vec{j} \]
- \[ \vec{v} = \begin{bmatrix} 2 \\ 4 \end{bmatrix}. \]
Solution
1. You need to take the sum of the squares of the coefficients of the unit vectors, and then square root this answer. This will be
\[ \begin{align} | \vec{u} | & = \sqrt{3^2 + (-4)^2} \\ & = \sqrt{9 + 16} \\ & = \sqrt{25} \\ & = 5. \end{align} \]
2. This time, the components will be the values in the column vector. The magnitude of \( \vec{v} \) will be
\[ \begin{align} | \vec{v} | & = \sqrt{ 2^2 + (-4)^2} \\ & = \sqrt{ 4 + 16} \\ & = \sqrt{20}. \end{align} \]
The Direction Angle of a Vector
In two dimensions, a vector can be determined using just the magnitude and the angle of the vector from the positive \(x\)-axis.
The angle of a vector is always measured between it and the positive x-axis.
The formula for the angle \(\theta\) between a vector \(\vec{u} = x\vec{i} + y \vec{j}\) and the positive \(x\)-axis is:
\[\theta = \tan^{-1}{\frac{y}{x}}. \]
The angle should be between \(0^\circ\) and \(360^\circ,\) so you may have to add your answer to \(360^\circ\) if your calculator gives you a negative answer.
Let's look at an example of calculating the direction angle of a vector.
Find the direction angle of \(6 \vec{i} - 7 \vec{j}. \)
Solution
Using the formula, with \(6\) in place of \(x\) and \(-7\) in place of \(y\) gets:
\[ \theta = \tan^{-1}{\frac{-7}{6}} = -49.40^\circ \]
to 2 decimal places. This angle is not between \(0^\circ\) and \(360^\circ,\) so you must add \(360^\circ\) to it:
\[ \theta = -49.40^\circ + 360^\circ = 310.80^\circ\]
to 2 decimal places. This is the size of the direction angle.
Normal Vectors
An important type of vector in mathematics is the unit vector. You have already met some unit vectors in this article, the standard unit vectors, \( \vec{i}\) and \( \vec{j}.\)
A unit vector is a vector with a magnitude equal to \(1\).
The normalized vector of a vector \( \vec{v} \) is the unit vector pointing in the same direction as vector \( \vec{v}. \) This is denoted by \( \hat{v} \), and is often referred to as "v hat". \( \hat{v} \) can be calculated by:
\[ \hat{v} = \frac{1}{|\vec{v}|} \vec{v}. \]
The normalized version of a vector is the vector that is parallel to the original vector, but with magnitude \(1\).
Let's look at normalizing some vectors.
Normalize the following vectors:
- \[\vec{v} = \begin{bmatrix} 3 \\ -4 \end{bmatrix} \]
- \[\vec{u} = 5 \vec{i} -2 \vec{j}. \]
Solution
1. First, calculate the magnitude of the vector:
\[ \begin{align} | \vec{v} | & = \left| \begin{bmatrix} 3 \\ -4 \end{bmatrix} \right| \\ & = \sqrt{3^2 + (-4)^2 } \\ & = \sqrt{9 + 16 + } \\ & = \sqrt{25} = 5. \end{align} \]
Now, multiply the vector by \( \frac{1}{|\vec{v}|} \) to get \( \hat{v} \). Remember that to multiply a vector by a scalar, you multiply each of the entries in the vector by the scalar. This gives you
\[ \begin{align} \hat{v} & = \frac{1}{|\vec{v}|} \vec{v} \\ & = \frac{1}{5} \begin{bmatrix} 3 \\ -4 \end{bmatrix} \\ & = \begin{bmatrix} \frac{3}{5} \\ \frac{-4}{5} \\ \frac{1}{5} \end{bmatrix} \end{align}\]
This is the final normalized vector.
2. Again, the first step is to calculate the magnitude of the vector:
\[ \begin{align} | \vec{v} | & = | 5 \vec{i} -2 \vec{j}| \\ & = \sqrt{5^2 + (-2)^2} \\ & = \sqrt{25 + 4} \\ & = \sqrt{29}. \end{align} \]Now, multiply \(\vec{u} \) by \( \frac{1}{| \vec{u} |} \). Since it is in unit vector form, this is just like expanding the brackets:
\[ \begin{align} \hat{u} & = \frac{1}{|\vec{u}|} \vec{u} \\ & = \frac{1}{\sqrt{29}} (5 \vec{i} -2 \vec{j}) \\ & = \frac{5}{\sqrt{29}} \vec{i} -\frac{2}{\sqrt{29}} \vec{j}. \end{align} \]
This is the normalized vector.
Cross and Dot Product Formulas
The cross and dot product both have formulas that allow you to calculate the angle \(\theta\) between two vectors.
The angle between two vectors.
Given two vectors \(\vec{a}\) and \(\vec{b},\) these formulas are:
\[ \begin{align} \vec{a} \cdot \vec{b} &= | \vec{a} | | \vec{b} | \cos{\theta} \\ | \vec{a} \times \vec{b} | &= |\vec{a} | | \vec{b} | \sin{\theta}. \end{align} \]
Projectile Motion using Vector Calculus
Vector-valued functions are types of functions that take a scalar \(t\) as input, and output a vector. In physics, it is standard to define position as a vector-valued function \(\vec{s}(t),\) where the input \(t\) is time and the output is the position vector at time \(t.\) This makes for a useful way of defining the way that projectiles move in 2D or 3D space.
The velocity of the particle can then be calculated by differentiating the given position function \( \vec{s}(t), \) and the acceleration will be the second derivative of the position function, and hence the derivative of the velocity function. For more information, see Vector-Valued Functions and Vector-valued motion - position, speed, acceleration.
Applications of Vector Calculus in Real Life
Many areas of Physics require a good knowledge of vector calculus, including mechanics, quantum physics and general relativity. Beyond physics, vector calculus is also an essential part of modern computer programming, including graphics and machine learning. In fact, gradient descent, an incredibly important part of machine learning, is the method of calculating the steepest descent to find the local minimum of a function. Within machine learning, this is essential as it allows for the minimization of loss functions or error within the algorithm. For more applications see Vector-valued motion - position, speed, acceleration.
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