The Concept of a Vector and Linear Operations on Vectors

Vector as a Directed Line Segment

A geometric vector or simply a vector is a mathematical object having magnitude and direction. A vector is represented by a directed line segment.

Let A and B be two points in space. The directed line segment from A to B is denoted as AB or $$\overrightarrow{AB}$$, or by one bold letter like a or b. The point A is called the initial point, and the point B is called the terminal point of the line segment.

The length (or magnitude) of the vector AB or a is denoted by $$\Vert\mathbf{AB}\Vert$$ or $$\Vert\mathbf{a}\Vert$$, respectively.

A vector is called the zero vector if its initial and terminal points are the same. The zero vector has no direction and has a length equal to zero.

Collinear and Equal Vectors

Vectors are said to be collinear if they lie either on the same line or on parallel lines.

Two vectors are called equal if they are collinear, have the same length and the same direction. All zero vectors are considered equal. Figure 2 shows equal and unequal vectors a and b.

It follows from the definition of equality of vectors that the point of application of a vector can be chosen arbitrarily, that is, we do not distinguish between two equal vectors having different points of application and resulting from one another by parallel translation.

Linear Operations on Vectors

Linear operations on vectors include vector addition and multiplication of vectors by real numbers.

The sum a + b of two vectors a and b is the third vector that begins at the initial point (at the tail) of the first vector and goes to the terminal point (to the head) of the second vector, provided that vector b is attached to the head of vector a. This rule for adding two vectors is usually called the triangle rule. This name is explained by the fact that, in accordance with the indicated rule, the vectors a, b, and a + b form a triangle.

Addition of vectors has the same basic properties as addition of real numbers:

Commutative Law

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

The commutativity property is illustrated by the parallelogram rule for the addition of vectors.

Indeed, in parallelogram ABCD we have

$\mathbf{AB} = \mathbf{DC} = \mathbf{a},\;\mathbf{BC} = \mathbf{AD} = \mathbf{b}.$

Then the diagonal AC is given by

$\mathbf{AC} = \mathbf{AB} + \mathbf{BC} = \mathbf{AD} + \mathbf{DC},$

or

$\mathbf{AC} = \mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}.$

Associative Law

$\left({\mathbf{a} + \mathbf{b}}\right) + \mathbf{c} = \mathbf{a} + \left({\mathbf{b} + \mathbf{c}}\right)$

Existence of a Zero Vector

$\mathbf{a} + \mathbf{0} = \mathbf{a}$

Existence of an Opposite Vector

$\mathbf{a} + \left({-\mathbf{a}}\right) = \mathbf{0}$

The vector $$\left({-\mathbf{a}}\right)$$ is opposite to vector $$\mathbf{a}$$ if it has the same length but the opposite direction.

Subtraction of Vectors

The difference between vector $$\mathbf{a}$$ and vector $$\mathbf{b}$$, written $$\mathbf{a - b},$$ is such a vector $$\mathbf{c}$$ which, in sum with the vector $$\mathbf{b}$$, gives the vector $$\mathbf{a}.$$ So if $$\mathbf{c = a - b},$$ then based on the vector addition properties

$\mathbf{c + b} = \left({\mathbf{a - b}}\right) + \mathbf{b} = \mathbf{a} + \left({\mathbf{-b + b}}\right) = \mathbf{a + 0} = \mathbf{a}.$

To get the difference between vectors $$\mathbf{a}$$ and $$\mathbf{b}$$, we bring these vectors to a common initial point and connect the head of vector $$\mathbf{b}$$ with the head of vector $$\mathbf{a}.$$

Scalar Multiplication of Vectors

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

The product $$\lambda \mathbf{a}$$ of the vector $$\mathbf{a}$$ by the real number $$\lambda$$ is the vector $$\mathbf{b}$$ that is collinear to the vector $$\mathbf{a}$$ and has a length equal to $$\Vert\lambda\mathbf{a}\Vert.$$ The direction of the vector $$\mathbf{b}$$ coincides with the direction of vector $$\mathbf{a}$$ if $$\lambda \gt 0$$ and is opposite to the direction of $$\mathbf{a}$$ in the case of $$\lambda \lt 0.$$

When $$\lambda = 0$$ or $$\mathbf{a = 0},$$ the product $$\lambda\mathbf{a}$$ is the zero vector whose direction is undefined.

When a vector $$\mathbf{a}$$ is multiplied by a number $$\lambda$$, the vector $$\mathbf{a}$$ stretches $$\lambda$$ times if $$\lambda \gt 1$$ and shrinks $$\lambda$$ times if $$\lambda \lt 1.$$

The operation of multiplying a vector by a number has the following properties:

Distributive Law with Respect to Vector Addition

$\lambda\left({\mathbf{a + b}}\right) = \lambda\mathbf{a} + \lambda\mathbf{b}$

Distributive Law with Respect to Number Addition

$\left({\lambda + \mu}\right)\mathbf{a} = \lambda\mathbf{a} + \mu\mathbf{a}$

Associative Property of Numerical Factors

$\lambda\left({\mu\mathbf{a}}\right) = \left({\lambda\mu}\right)\mathbf{a}$

Identity Element of Scalar Multiplication

$1\cdot\mathbf{a} = \mathbf{a}$

These properties are of fundamental importance because they allow us to perform operations on vectors in the same way as we do with numbers.

Solved Problems

Click or tap a problem to see the solution.

Example 1

Vectors a, b are the diagonals of a parallelogram ABCD. Express the vectors AB, BC, CD, DA in terms of vectors a and b.

Example 2

Medians AD, BE and CF are drawn in triangle ABC. Find the sum of vectors AD, BE and CF.

Example 3

In triangle ABC, the line AM is the bisector of angle BAC and point M lies on the side BC. Find the vector $$\mathbf{AM = m}$$ if $$\mathbf{AB = c},$$ $$\mathbf{AC = b}.$$

Example 4

Find the value of parameter t at which vectors u and v are collinear if $$\mathbf{u} = \left({2 + t}\right)\mathbf{a} + \left({6 - t}\right)\mathbf{b} + 2\mathbf{c},$$ $$\mathbf{v} = t\mathbf{a} + 2\mathbf{b} + \left({t - 1}\right)\mathbf{c}.$$

Example 1.

Vectors a, b are the diagonals of a parallelogram ABCD. Express the vectors AB, BC, CD, DA in terms of vectors a and b.

Solution.

Using the linear properties of vectors, we get:

$\mathbf{AB} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} = \frac{1}{2}\left({\mathbf{a + b}}\right);$
$\mathbf{BC} = \frac{1}{2}\mathbf{a} - \frac{1}{2}\mathbf{b} = \frac{1}{2}\left({\mathbf{a - b}}\right);$
$\mathbf{CD} = -\mathbf{AB} = - \frac{1}{2}\left({\mathbf{a + b}}\right);$
$\mathbf{DA} = -\mathbf{BC} = \frac{1}{2}\left({\mathbf{b - a}}\right);$

Check that the sum of all 4 vectors is equal to the zero vector:

$\mathbf{AB + BC + CD + DA} = \frac{1}{2}\left({\mathbf{a + b}}\right) + \frac{1}{2}\left({\mathbf{a - b}}\right) - \frac{1}{2}\left({\mathbf{a + b}}\right) + \frac{1}{2}\left({\mathbf{b - a}}\right) = \frac{1}{2}\left({\mathbf{\cancel{a} + \cancel{b} + \cancel{a} - \cancel{b} - \cancel{a} - \cancel{b} + \cancel{b} - \cancel{a}}}\right) = \mathbf{0}.$

Example 2.

Medians AD, BE and CF are drawn in triangle ABC. Find the sum of vectors AD, BE and CF.

Solution.

Points F, D, E the midpoints of the corresponding sides of the triangle. Therefore

$\mathbf{AD} = \mathbf{AB} + \mathbf{BD} = \mathbf{AB} + \frac{1}{2}\mathbf{BC},$
$\mathbf{BE} = \mathbf{BC} + \mathbf{CE} = \mathbf{BC} + \frac{1}{2}\mathbf{CA},$
$\mathbf{CF} = \mathbf{CA} + \mathbf{AF} = \mathbf{CA} + \frac{1}{2}\mathbf{AB}.$

Notice that the sum of vectors lying on the sides of the triangle is equal to zero:

$\mathbf{AB + BC + CA} = \mathbf{0}.$

Now it's easy to find that the sum of vectors representing the medians:

$\mathbf{AD + BE + CF} = \mathbf{AB} + \frac{1}{2}\mathbf{BC} + \mathbf{BC} + \frac{1}{2}\mathbf{CA} + \mathbf{CA} + \frac{1}{2}\mathbf{AB} = \frac{3}{2}\left({\mathbf{AB + BC + CA}}\right) = \frac{3}{2}\cdot\mathbf{0} = \mathbf{0}.$

Example 3.

In triangle ABC, the line AM is the bisector of angle BAC and point M lies on the side BC. Find the vector $$\mathbf{AM = m}$$ if $$\mathbf{AB = c},$$ $$\mathbf{AC = b}.$$

Solution.

Remember the angle bisector theorem which states that an angle bisector divides the opposite side of a triangle into two segments that are proportional to the lengths of the other two sides. This means that

$\frac{\Vert{BM}\Vert}{\Vert{MC}\Vert} = \frac{\Vert{\mathbf{c}}\Vert}{\Vert{\mathbf{b}}\Vert},$

or

$\frac{\Vert{BM}\Vert}{\Vert{BC}\Vert} = \frac{\Vert{\mathbf{c}}\Vert}{\Vert{\mathbf{b}}\Vert + \Vert{\mathbf{c}}\Vert}, \Rightarrow {\Vert{BM}\Vert} = {\Vert{BC}\Vert}\frac{\Vert{\mathbf{c}}\Vert}{\Vert{\mathbf{b}}\Vert + \Vert{\mathbf{c}}\Vert}.$

Using the linear properties of vectors, we get

$\mathbf{m} = \mathbf{c} + \mathbf{BM} = \mathbf{c} + \mathbf{BC}\frac{\Vert{\mathbf{c}}\Vert}{\Vert{\mathbf{b}}\Vert + \Vert{\mathbf{c}}\Vert}.$

The vector $$\mathbf{BC}$$ can be expressed in terms of $$\mathbf{b}$$ and $$\mathbf{c}:$$

$\mathbf{BC = b - c}.$

Therefore

$\mathbf{m} = \mathbf{c} + \left({\mathbf{b - c}}\right)\frac{\Vert{\mathbf{c}}\Vert}{\Vert{\mathbf{b}}\Vert + \Vert{\mathbf{c}}\Vert} = \frac{\mathbf{c}\left({\Vert{\mathbf{b}}\Vert + \Vert{\mathbf{c}}\Vert}\right) + \left({\mathbf{b - c}}\right)\Vert{\mathbf{c}}\Vert}{{\Vert{\mathbf{b}}\Vert + \Vert{\mathbf{c}}\Vert}} = \frac{\mathbf{c}\Vert{\mathbf{b}}\Vert + \cancel{\mathbf{c}\Vert{\mathbf{c}}\Vert} + \mathbf{b}\Vert{\mathbf{c}}\Vert - \cancel{\mathbf{c}\Vert{\mathbf{c}}\Vert}}{\Vert{\mathbf{b}}\Vert + \Vert{\mathbf{c}}\Vert} = \frac{\mathbf{b}\Vert{\mathbf{c}}\Vert + \mathbf{c}\Vert{\mathbf{b}}\Vert }{\Vert{\mathbf{b}}\Vert + \Vert{\mathbf{c}}\Vert}.$

Example 4.

Find the value of parameter t at which vectors u and v are collinear if $$\mathbf{u} = \left({2 + t}\right)\mathbf{a} + \left({6 - t}\right)\mathbf{b} + 2\mathbf{c},$$ $$\mathbf{v} = t\mathbf{a} + 2\mathbf{b} + \left({t - 1}\right)\mathbf{c}.$$

Solution.

Collinear vectors satisfy the relation

$\mathbf{u} = \lambda\mathbf{v}.$

In that case we should have

$\frac{2 + t}{t} = \frac{6 - t}{2} = \frac{2}{t - 1} = \lambda.$

Let's check is there a value $$t$$ that satisfies the given equations. Find $$t$$ and $$\lambda$$ from the first equation:

$\frac{2 + t}{t} = \frac{6 - t}{2}, \Rightarrow 2\left({2 + t}\right) = t\left({6 - t}\right), \Rightarrow 4 + 2t = 6t - t^2, \Rightarrow t^2 - 4t + 4 = 0, \Rightarrow \left({t - 2}\right)^2 = 0, \Rightarrow t = 2.$

Then $$\lambda = 2.$$ Check if the $$3\text{rd}$$ proportion is correct:

$\frac{2}{t - 1} = \frac{2}{2 - 1} = 2.$

Thus at $$t = 2,$$ the vectors u and v are collinear.