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17. Vectors

A vector is a quantity that has a magnitude and a direction (as opposed to a scalar that only has a magnitude). Vectors are used in many areas of science and mathematics. This module introduces vectors and describes how to add, subtract and multiply them.

Do you need this chapter?

Here are some questions on the topics in this chapter to consider. If you can do these you may skip this chapter.

Quiz: Vectors

1.   If [latex]P[/latex] is the point [latex](3, 2, 1)[/latex] and [latex]Q[/latex] is the point [latex](5, 2, 3)[/latex] find the vector [latex]\overrightarrow{PQ}[/latex]

2.   Given that [latex]\vec{a}=2\vec{i}+3\vec{j}-\vec{k}[/latex]  and  [latex]\vec{b}= 5\vec{i}-2\vec{j}+4\vec{k}[/latex] , find:

(a)  [latex]\vec{a}+\vec{b}[/latex]       (b)   [latex]2\vec{a}-3\vec{b}[/latex]     

3.   If  [latex]\vec{p}=(1, 2, 3)[/latex]  and  [latex]\vec{q}=(-3, 1, 3)[/latex]  find (a)  [latex]\vec{p}.\vec{q}[/latex]     (b)  the angle between the two vectors.

4.   If  [latex]\vec{a}=(3, 2, 3)[/latex]  and  [latex]\vec{b}=(2, -1, 1)[/latex]  find  [latex]\vec{a}×\vec{b}[/latex] 

 

Answers

1.   [latex](2, 0, 2)[/latex]

2.   (a)  [latex]7\vec{i}+\vec{j}+3\vec{k}[/latex]     (b)  [latex]-11\vec{i}+12\vec{j}-14\vec{k}[/latex]

3.   (a)   [latex]8[/latex]    (b)  [latex]60.6°[/latex]

4.  [latex](5, 3, -7)[/latex]

Vectors and scalars

One example of a vector is velocity . The velocity of an object is determined by the magnitude (speed) and direction of travel. Other examples of vectors are force, displacement and acceleration.

A scalar is a quantity that has magnitude only. Mass, time and volume are all examples of scalar quantities.

Vectors in three-dimensional space are defined by three mutually perpendicular directions and and can be denoted in a number of ways, most usually with an arrow above  [latex]\vec{a}[/latex]  or  [latex]\vec{b}[/latex]  or  [latex]\vec{c}[/latex]  or  with a tilde below  [latex]\underset{\sim}{u}[/latex]  or   [latex]\underset{\sim}{v}[/latex].

A vector in the opposite direction from  [latex]\vec{a}[/latex]  is denoted by  [latex]-\vec{a}[/latex].

Vectors can be added or subtracted graphically using the triangle rule.

Adding and subtracting vectors

Triangle Rule:

A vector labelled "a"A vector labelled "b".

To add vectors [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex] shown above, place the tail of vector [latex]\vec{b}[/latex] at the head of vector [latex]\vec{a}[/latex] (shown below).

A diagram showing vectors "a" and "b" added head to tail.
Figure 17.1 Addition of two vectors [latex]\vec a[/latex] and [latex]\vec b[/latex] (Copyright © 2025 RMIT University)

The vector sum, [latex]\vec{a}+\vec{b}[/latex], is the vector [latex]\overrightarrow{AC}[/latex] , from the tail of vector [latex]\vec{a}[/latex] to the head of [latex]\vec{b}[/latex].

To subtract [latex]\vec{b}[/latex] from [latex]\vec{a}[/latex], reverse the direction of [latex]\vec{b}[/latex] to give [latex]-\vec{b}[/latex] then add [latex]\vec{a}[/latex] and [latex]-\vec{b}[/latex].

A vector labelled "a".A vector labelled negative "b"

 

[latex]\vec{a}-\vec{b}=\vec{a}+(-\vec{b})[/latex]

A diagram showing vector "a" and vector negative "b" added head to tail
Figure 17.2 Subtraction of two vectors [latex]\vec a - \vec b[/latex] (Copyright © 2025 RMIT University)

Vector [latex]\overrightarrow{AC}[/latex] is equal to the vector [latex]\vec{a}-\vec{b}[/latex].

Components of a vector

The diagram below shows the vector [latex]\overrightarrow{OP}[/latex], where [latex]O[/latex] is the origin and [latex]P[/latex] is the point [latex](1,2,3)[/latex].

A three dimensional graph showing a vector from the origin to the point P(1,2,3)
Figure 17.3 The vector [latex]\overrightarrow{OP}[/latex] has components in the [latex]\vec i,\ \vec j[/latex] and [latex]\vec k[/latex] directions where [latex] \vec i,\ \vec j[/latex] and [latex]\vec k[/latex] are vectors of length one along the [latex] x,\ y [/latex] and [latex]z[/latex] axes respectively. (Copyright © 2025 RMIT University)

[latex]\vec{i}[/latex] , [latex]\vec{j}[/latex] , and [latex]\vec{k}[/latex] are vectors of magnitude one in the direction of the [latex]x[/latex] , [latex]y[/latex], and [latex]z[/latex] axes, respectively.

A vector of magnitude one is called a unit vector.

[latex]1\vec{i}[/latex] is a vector of length [latex]1[/latex] in the direction of the [latex]x[/latex]-axis.

[latex]2\vec{j}[/latex] is a vector of length [latex]2[/latex] in the direction of the [latex]y[/latex]-axis.

[latex]3\vec{k}[/latex] is a vector of length [latex]3[/latex] in the direction of the [latex]z[/latex]-axis.

[latex]\overrightarrow{OP}[/latex] is then the vector [latex]\vec{i}+2\vec{j}+3\vec{k}[/latex].

[latex]1\vec{i}[/latex], [latex]2\vec{j}[/latex] and [latex]3\vec{k}[/latex] are called the components of the vector.

The vector [latex]\overrightarrow{OP}[/latex] is denoted by [latex]\vec{i}+2\vec{j}+3\vec{k}[/latex] but may also be written as [latex]\left(1,2,3\right)[/latex].

In general, the notation [latex](x,y,z)[/latex] is often used to denote the vector [latex]\left(x\vec{i}+y\vec{j}+z\vec{k}\right)[/latex] as well as the co-ordinates of a point [latex]P (x,y,z)[/latex]. The context will determine the correct meaning, as [latex](x,y,z)[/latex] might represent either a point or a vector.

Vectors can be added geometrically (head to tail) but they may also be added or subtracted arithmetically by adding or subtracting their corresponding components.

Examples: Addition and subtraction of vectors

1.  If [latex]\vec{u}=2\vec{i} +3\vec{j} +\vec{k}[/latex] and [latex]\vec{v}=-\vec{i}+2\vec{j}-3\vec{k}[/latex] , find [latex]\vec{u}+\vec{v}.[/latex]

Solution:

[latex]\vec{u}+\vec{v} =(2\vec{i} +3\vec{j} +\vec{k})+(-\vec{i}+2\vec{j}-3\vec{k})[/latex]

[latex]=\vec{i}+5\vec{j}-2\vec{k}.[/latex]

 

2.  If [latex]\vec{a}=(-3,4,2)[/latex] and [latex]\vec{b}=(-1,-2,3)[/latex] , find [latex]\vec{a}+\vec{b}[/latex]  and  [latex]\vec{a}-\vec{b}.[/latex]

Solution:

[latex]\vec{a}+\vec{b} =(-3,4,2)+(-1,-2,3) =(-3+(-1),\,\ 4+(-2),\,\ 2+3) =(-4,2,5)[/latex] and

[latex]\vec{a}-\vec{b} =(-3,4,2)-(-1,-2,3) =(-3-(-1),\,\ 4-(-2),\,\ 2-3) =(-2,6,-1).[/latex]

Now try the following exercise.

Exercises: Addition and subtraction of vectors

Given [latex]\vec{a}=(2,1,1)[/latex] , [latex]\vec{b}=(1,3,-3)[/latex] and [latex]\vec{c}=(0,3,-2)[/latex] find:

1.  [latex]\vec{a}+\vec{b}[/latex]

2.  [latex]\vec{a}+\vec{c}[/latex]    

3.  [latex]\vec{c}-\vec{b}[/latex]

4.  [latex]\vec{a}-\vec{b}[/latex]

 

Answers

1.  [latex]\left(3,4,-2\right)[/latex]

2.  [latex]\left(2,4,-1\right)[/latex]

3.  [latex]\left(-1,0,1\right)[/latex]

4. [latex]\left(1,-2,4\right)[/latex]

 

Directed line segment

The directed line segment, or geometric vector, [latex]\overrightarrow{PQ}[/latex], from a point  [latex]P(x_{1},y_{1},z_{1})[/latex] to a point [latex]Q(x_{2},y_{2},z_{2})[/latex] is found by subtracting the co-ordinates of [latex]P[/latex] (the initial point) from the co-ordinates of [latex]Q[/latex] (the final point).

[latex]\overrightarrow{PQ}=(x_{2}-x_{1})\vec{i}+(y_{2}-y_{1})\vec{j}+(z_{2}-z_{1})\vec{k}[/latex]

Example: Directed line segments

The directed line segment [latex]\overrightarrow{PQ}[/latex] in Fig. 17.4 is represented by the vector [latex]\vec{i}+\vec{j}-2\vec{k}[/latex], or [latex](1,1,-2)[/latex]. Any other directed line segment with the same length and same direction as [latex]\overrightarrow{PQ}[/latex] is also represented by [latex]\vec{i}+\vec{j}-2\vec{k}[/latex] or [latex](1,1,-2)[/latex].

The directed line segment [latex]\overrightarrow{QP}[/latex] has the same length as [latex]\overrightarrow{PQ}[/latex] but is in the opposite direction.

[latex]\overrightarrow{QP} =-\overrightarrow{PQ} =-(\vec{i}+\vec{j}-2\vec{k}) =-\vec{i}-\vec{j}+2\vec{k} =(-1,-1,2).[/latex]

A three dimensional graph from the point P(1,2,3) to the point Q(2,3,1)
Figure 17.4 The directed line segment from point [latex]P[/latex] to point [latex]Q[/latex]. (Copyright © 2025 RMIT University)

Position vectors

The position vector of any point is the directed line segment from the origin [latex]O (0,0,0)[/latex] to that point and is given by the co-ordinates of the point.

For example, the position vector of [latex]P(3,4,1)[/latex] is [latex]3\vec{i}+4\vec{j}+\vec{k}[/latex], or [latex](3,4,1)[/latex].

Exercises: Directed line segments and position vectors

1. Given the points [latex]A(3,0,4)[/latex] , [latex]B(-2,4,3)[/latex], and [latex]C(1,-5,0)[/latex], find:

a)   [latex]\text{(i)}\;\overrightarrow{AB}\quad\text{(ii)} \overrightarrow{AC}\quad \text{(iii)} \overrightarrow{CB}\quad \text{(iv)} \overrightarrow{BC}\quad \text{(v)} \overrightarrow{CA}[/latex]

Compare your answers to (ii) and (v), and also to (iii) and (iv). What do you notice?

(b) What are the position vectors of the points [latex]A,\,B[/latex] and [latex]C[/latex]?

 

Answers

1. [latex]\text{a.)\ (i)} \left(-5,4,-1\right)\quad \text{(ii)} \left(-2,-5,-4\right)\quad \text{(iii)} \left(-3,9,3\right)[/latex]

[latex]\quad \text{(iv)} \left(3,-9,-3\right)\quad \text{(v)} \left(2,5,4\right)[/latex]

They are in opposite directions.

b.)  [latex]\overrightarrow{OA}=3\vec{i}+4\vec{k},\,\  \overrightarrow{OB}=-2\vec{i}+4\vec{j}+3\vec{k}[/latex]  and [latex]\overrightarrow{OC}=\vec{i}-5\vec{j}[/latex].

 

Magnitude of a vector

If  [latex]\vec{a}=a_{1}\vec{i}+a_{2}\vec{j}+a_{3}\vec{k}[/latex]  the length or magnitude of [latex]\vec{a}[/latex] is written as [latex]\left|\vec{a}\right|[/latex] and is evaluated as:

[latex]\left|\vec{a}\right|=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}.[/latex]

Examples: Magnitude of a vector

 

The magnitude of the vector [latex]\vec{i}+2\vec{j}+\vec{k}[/latex] equals [latex]\sqrt{1^{2}+2^{2}+1^{2}}=\sqrt{6}.[/latex]

The length of the vector [latex]2\vec{i}+3\vec{j}-5\vec{k}[/latex] equals [latex]\sqrt{2^{2}+3^{2}+(-5)^{2}}=\sqrt{38}.[/latex]

The length of the vector [latex](6,-1,-2)[/latex] equals [latex]\sqrt{6^{2}+(-1)^{2}+(-2)^{2}}=\sqrt{41}.[/latex]

 

Unit vector

Any vector with a magnitude of one is called a unit vector.

If [latex]\vec{a}[/latex] is any vector, then a unit vector parallel to [latex]\vec{a}[/latex] is written [latex]\hat{a}[/latex] (we call this [latex]a[/latex] “hat”). The “hat” symbolises a unit vector.

The unit vector [latex]\hat{a}[/latex] equals the vector [latex]\vec{a}[/latex] divided by its magnitude [latex]\left|\vec{a}\right|[/latex] . That is

\begin{align*}
\hat{a} &=\dfrac{\vec{a}}{\left|\vec{a}\right|}.
\end{align*}

Rearanging this formula gives

\begin{align*}\vec{a} =\left|\vec{a}\right|\hat{a}. \end{align*}

Of particular importance are the unit vectors [latex]\vec{i}[/latex], [latex]\vec{j}[/latex] and [latex]\vec{k}[/latex] (sometimes written as [latex]\hat{i}[/latex], [latex]\hat{j}[/latex] and [latex]\hat{k}[/latex]), parallel to the [latex]x[/latex], [latex]y[/latex], and [latex]z[/latex] axes, respectively.

Examples: Unit vectors

1. If [latex]\overrightarrow{PQ}[/latex] is the line [latex]2\vec{i}-5\vec{j}+\vec{k}[/latex] find a unit vector parallel to [latex]\overrightarrow{PQ}[/latex].

Solution:

[latex]\left|\overrightarrow{PQ}\right|=\sqrt{2^{2}+\left(-5\right)^{2}+1^{2}}=\sqrt{30}.[/latex]

A unit vector parallel to [latex]\overrightarrow{PQ}[/latex] is

\begin{align*}
\hat{PQ} &=\frac{\overrightarrow{PQ}}{\left|\overrightarrow{PQ}\right|}\\
&=\frac{\left(2\vec{i}-5\vec{j}+\vec{k}\right)}{\sqrt{30}}\\
&=\dfrac{1}{\sqrt{30}}\left(2\vec{i}-5\vec{j}+\vec{k}\right) \\
&=\dfrac{2}{\sqrt{30}}\vec{i}-\dfrac{5}{\sqrt{30}}\vec{j}+\dfrac{1}{\sqrt{30}}\vec{k}.
\end{align*}

Either of the expressions in the last two lines of the previous example is acceptable.

 

2. If [latex]\vec{a}=(1,2,3)[/latex] a unit vector in the direction of [latex]\vec{a}[/latex] is:

\begin{align*}
\hat{a}& =\dfrac{\vec{a}}{\left|\vec{a}\right|} \\
&=\dfrac{\left( \hat i+2\hat j+3\hat k\right)}{\sqrt{1^{2}+2^{2}+3^{2}}}\\
&=\dfrac{\left(\hat i+2\hat j+3\hat k\right)}{\sqrt{14}}\\
&=\frac{1}{\sqrt{14}}\left(\hat i+2\hat j+3\hat k\right).
\end{align*}

Now attempt the questions in the exercise below.

Exercise: Unit vectors

1.  Find the length of the vectors: [latex]\text{(i)}\ (3,-1,-1)\quad \text{(ii)} \ (0,2,4)\quad \text{(iii)} \ (0,-2,0)[/latex]

2.  Given the points [latex]A\,(3,0,4), B\,(0,4,3)[/latex] and [latex]C\,(1,-5,0)[/latex]; find unit vectors parallel to (i) [latex]\overrightarrow{BA}[/latex]   (ii)  [latex]\overrightarrow{CB}[/latex]    (iii)   [latex]\overrightarrow{AC}.[/latex]

 

Answers

1.  i)  [latex]\sqrt{11}[/latex]    ii) [latex]\sqrt{20}=2\sqrt{5}[/latex]   iii)  [latex]2[/latex]

 

2.  i) [latex]\dfrac{(3,-4,1)}{\sqrt{26}}[/latex]     ii) [latex]\dfrac{(-1,9,3)}{\sqrt{91}}[/latex]    iii) [latex]\dfrac{(-2,-5,-4)}{\sqrt{45}}=\dfrac{(-2,-5,-4)}{3\sqrt{5}}[/latex]

 

Multiplication by a scalar

To multiply a vector [latex]\vec{a}=a_{1}\vec{i}+a_{2}\vec{j}+a_{3}\vec{k}[/latex] by a scalar, [latex]m[/latex], multiply each component of [latex]\vec{a}[/latex] by [latex]m[/latex]:

\begin{align*}m\,\vec{a} =m\,a_{1}\vec{i}+m\,a_{2}\vec{j}+m\,a_{3}\vec{k}.\end{align*}

The result is a vector of length [latex]m\times\left|\vec{a}\right|=m\left|\vec{a}\right|.[/latex]

If [latex]m>0[/latex], the resultant vector is in the same direction as [latex]\vec{a}.[/latex]

If [latex]m\lt{0}[/latex], the resultant vector is in the opposite direction to [latex]\vec{a}.[/latex].

Two vectors [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex] are said to be parallel if and only if  [latex]\vec{a}=c\vec{b}[/latex] where [latex]c[/latex] is a real constant.

Examples: Multiplication by a scalar

Multiply [latex]\vec{a}=(3\vec{i}+\vec{j}-2\vec{k})[/latex] by [latex]7[/latex] and show [latex]\left|7\, \vec{a}\right|=7\left|\vec{a}\right| .[/latex]

Solution:

[latex]\begin{align*} 7\,\vec{a} &=7\left(3\vec{i}+\vec{j}-2\vec{k}\right)\\ &=21\vec{i}+7\vec{j}-14\vec{k} \end{align*}[/latex]

so

[latex]\left|7\,\vec{a}\right| =\sqrt{21^{2}+7^{2}+(-14)^{2}} =\sqrt{686}=\sqrt{49\times14} =7\sqrt{14}.[/latex]

Also the magnitude of [latex]\vec{a}[/latex] is [latex]\left|\vec{a}\right| =\sqrt{3^{2}+1^{2}+(-2)^{2}}=\sqrt{14}[/latex] so [latex]7\left|\vec{a}\right| =7\sqrt{14}.[/latex]

Hence [latex]\left|7\,\vec{a}\right| =7\left|\vec{a}\right|=7\sqrt{14}[/latex].

Try the questions below.

Exercises: Multiplication by a scalar

1.  Expand the following: (i) [latex]3\left(\vec{i}+3\vec{j}-5\vec{k}\right)[/latex] (ii) [latex]-4\left(\vec{j}-3\vec{k}\right).[/latex]

2.  If [latex]\vec{a}=\left(2,-2,1\right) ,\,\ \vec{b}=\left(0,1,1\right)[/latex] and [latex]\vec{c}=\left(-1,3,-2\right)[/latex] , find (i) [latex]\left(2\vec{a}+3\vec{b}\right)[/latex]     (ii) [latex]\left(3\vec{a}-2\vec{b}\right)[/latex]     (iii) [latex]\left(2\vec{a}-\vec{b}+2\vec{c}\right)[/latex]     (iv) a unit vector parallel to [latex]2\vec{a}-\vec{b}.[/latex]

3.  Write down a vector three times the length of [latex]\left(6\vec{i}+2\vec{j}-5\vec{k}\right)[/latex] and in the opposite direction.

 

Answers


1.
  i) [latex]\left(3\vec{i}+9\vec{j}-15\vec{k}\right)[/latex]      ii) [latex]\left(-4\vec{j}+12\vec{k}\right)[/latex]

2.  i) [latex]\left(4,-1,5\right)[/latex]   ii) [latex]\left(6,-8,1\right)[/latex]    iii) [latex]\left(2,1,-3\right)[/latex]    iv) [latex]\dfrac{(4,-5,1)}{\sqrt{42}}[/latex]

3.  [latex]\left(-18\vec{i}-6\vec{j}+15\vec{k}\right)[/latex].

 

Multiplication of vectors

There are two ways to multiply two vectors
  1. The scalar product (also known as the dot product) which gives a number.
  2. The vector product (also known as the cross product) which gives a vector.

Scalar product

The scalar, or dot, product of two vectors [latex]\vec{a}=\left(a_{1},a_{2},a_{3}\right)[/latex] and [latex]\vec{b}=\left(b_{1},b_{2},b_{3}\right)[/latex] is a scalar, defined by:

 

\begin{align*}
\vec{a}\cdot\vec{b}&=a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}
\end{align*}

or

\begin{align*}
\vec{a}\cdot\vec{b}&=\left|\vec{a}\right|\left|\vec{b}\right|\cos\theta
\end{align*}

where [latex]\theta[/latex]  is the angle between [latex]\vec{a}[/latex] and [latex]\vec{b}.[/latex]

Properties of the scalar or dot product

1.  If [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex] are non-zero vectors and [latex]\vec{a}[/latex] is perpendicular (at right angles) to [latex]\vec{b}[/latex] then [latex]\vec{a}\cdot\vec{b}=0[/latex], since [latex]\cos\left(\dfrac{\pi}{2}\right)=0[/latex].

2.  If [latex]\vec{a}[/latex] is parallel to [latex]\vec{b}[/latex] then the angle between the vectors is [latex]0[/latex] and [latex]\vec{a}\cdot\vec{b}=\mid\vec{a}\mid\mid\vec{b}\mid[/latex]  as [latex]\cos\left(0\right)=1.[/latex]

3. The dot product does not depend on the order of multiplication:  [latex]\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{a}[/latex]

4. In three dimensions with [latex]\hat{i},\hat{j}[/latex] and [latex]\hat{k}[/latex] unit vectors along the [latex]x[/latex], [latex]y[/latex] and [latex]z[/latex] axes respectively, we have:

[latex]\hat{i}\cdot\hat{j} =\hat{j}\cdot\hat{k}=\hat{k}\cdot\hat{i}=0[/latex]

and

[latex]\hat{i}\cdot\hat{i} =\hat{j}\cdot\hat{j}=\hat{k}\cdot\hat{k}=1.[/latex]

Examples: Dot product of vectors

1. [latex]\left(2\vec{i}+3\vec{j}+4\vec{k}\right)\cdot\left(\vec{i}-2\vec{j}+\vec{k}\right)=\left(2\times1\right)+\left(3\times\left(-2\right)\right)+\left(4\times1\right)=0.[/latex]
2.  [latex]\left(2,-3,-3\right)\cdot\left(1,1,-2\right)=2-3+6=5.[/latex]

 

3.  [latex]\left(5,0,-1\right)\cdot\left(1,4,3\right)=5+0-3=2.[/latex]

 

4. [latex]\left(2\vec{i}+4\vec{k}\right)\cdot\left(-3\vec{i}-2\vec{j}\right)=2×(-3)+0×(-2)+4×0=-6.[/latex]

Complete the exercises below.

Exercises: Dot product of vectors

1.  Calculate the dot product of:
(a) [latex]\left(2,5,-1\right)[/latex] and [latex]\left(4,1,1\right)[/latex]
(b) [latex]3\vec{i}[/latex] and [latex]5\vec{j}[/latex]
(c) [latex]5\vec{k}[/latex] and[latex]\left(\vec{j}+2\vec{k}\right)[/latex]

 

2.  Evaluate:
(a)   [latex]\left(2,0,4\right)\cdot\left(-3,1,3\right)[/latex]
(b)   [latex]\left(0,5,1\right)\cdot\left(4,0,0\right)[/latex]
(c)   [latex]\left(2\vec{i}+3\vec{k}\right)\cdot\left(7\vec{i}+2\vec{j}+4\vec{k}\right)[/latex]

3.
Which of the following vectors are perpendicular to each other?
(a)  [latex]\left(5,2,3\right)[/latex]       (b)  [latex]\left(0,1,-1\right)[/latex]       (c)}   [latex]\left(-2,2,2\right)[/latex]
Answers


1.
(a) [latex]12[/latex]       (b) [latex]0[/latex]     (c) [latex]10[/latex]

2. (a) [latex]6[/latex]      (b) [latex]0[/latex]     (c) [latex]26[/latex]

3.  (a) and (c), (b) and (c)

 

The Angle between two vectors

The angle [latex]\theta,\left(0\leq\theta\leq180°\right)[/latex], between two vectors can be found using the definition of the dot product:
\begin{align*}
\vec{a}\cdot\vec{b}& =\mid\vec{a}\mid\mid\vec{b}\mid\cos\theta .
\end{align*}
Rearranging this formula we get,
\begin{align*}\cos\theta &=\frac{\vec{a}\cdot\vec{b}}{\mid\vec{a}\mid\mid\vec{b}\mid}\end{align*}
and so,
\begin{align*}\theta =\cos^{-1}\left(\frac{\vec{a}\cdot\vec{b}}{\mid\vec{a}\mid\mid\vec{b}\mid}\right).\end{align*}

Examples: Angle between two vectors

 

1. If  [latex]\vec{a}=(2,3,1)[/latex]  and  [latex]\vec{b}=(5,-2,2)[/latex], find the angle [latex]\theta[/latex], between [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex].
  Solution:
  [latex]\theta =\cos^{-1}\left(\dfrac{\vec{a}\cdot\vec{b}}{\mid\vec{a}\mid\mid\vec{b}\mid}\right)[/latex]
[latex]\vec{a}\cdot\vec{b} =\left(2,3,1\right)\cdot\left(5,-2,2\right)=6[/latex]

[latex]\mid\vec{a}\mid =\sqrt{2^{2}+3^{2}+1^{2}}=\sqrt{14},\mid\vec{b}\mid=\sqrt{25+4+4}=\sqrt{33}[/latex]

[latex]\theta =\cos^{-1}\left(\dfrac{6}{\sqrt{33}\times\sqrt{14}}\right)[/latex]

    [latex]=\cos^{-1}\left(0.2791\right)[/latex]

[latex]\theta =73.8^{\circ}[/latex]

The angle between [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex] is [latex]73.8^{\circ}[/latex].

 

2. Find the angle [latex]\theta[/latex] , between [latex]\vec{a}\left(1,0,1\right)[/latex] and [latex]\vec{b}\left(-2,-1,1\right)[/latex].

Solution:

[latex]\vec{a}\cdot\vec{b} =\left(1,0,1\right)\cdot\left(-2,-1,1\right)=-1[/latex]

[latex]\mid\vec{a}\mid =\sqrt{2}[/latex]

[latex]\mid\vec{b}\mid =\sqrt{6}[/latex]

[latex]\theta =\cos^{-1}\left(\dfrac{\vec{a}\cdot\vec{b}}{\mid\vec{a}\mid\mid\vec{b}\mid}\right)[/latex]

[latex]\theta=\cos^{-1}\left(\dfrac{-1}{\sqrt{2}\sqrt{6}}\right)[/latex]

[latex]\theta=\cos^{-1}\left(-0.2887\right)[/latex]

[latex]\theta =106.8^{\circ}[/latex].

The angle between [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex] is [latex]106.8^{\circ}[/latex].

 

Now complete the exercises below.

Exercises: Angle between vectors

1.  Find the angle between the following pairs of vectors:
[latex]\text{(a)}\,\left(1,2,3\right)[/latex]  and  [latex]\left(4,-1,0\right)[/latex]
[latex]\text{(b)}\,\left(2,1,-2\right)[/latex]  and  [latex]\left(1,5,-1\right)[/latex]
[latex]\text{(c)}\,\left(0,5,1\right)[/latex]  and  [latex]\left(2,0,0\right)[/latex]
[latex]\text{(d)}\,\left(1,-2,3\right)[/latex]  and  [latex]\left(-4,1,-3\right)[/latex]
[latex]\text{(e)}\,\left(2,1,-2\right)[/latex]  and  [latex]\left(0,4,0\right)[/latex]

[latex]\text{(f)}\,\left(0,3,0\right)[/latex]  and  [latex]\left(0,1,0\right)[/latex]

2.  If  [latex]\vec{a}=\left(2,2,2\right),\,\vec{b}=\left(3,2,-1\right)[/latex], and  [latex]\vec{c}=\left(-1,4,1\right),[/latex]  show [latex]\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}[/latex]
Answers

1. (a) [latex]86.2°[/latex]       (b) [latex]54.7°[/latex]     (c) [latex]90°[/latex]

(d) [latex]141.8°[/latex]       (e) [latex]70.5°[/latex]     (f) [latex]0°[/latex]

2. [latex]\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}=8[/latex]

 

Vector product

Let the vectors [latex]\vec{a} =a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}[/latex]  and  [latex]\vec{b} =b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}[/latex].
The vector, or cross, product of the two vectors [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex] is the vector,
\begin{align*}
\vec{a}\times\vec{b} &=\left|\vec{a}\right|\left|\vec{b}\right|\sin\left(\theta\right)\,\hat{n}
\end{align*}
where [latex]\left|\vec{a}\right|[/latex] and [latex]\left|\vec{b}\right|[/latex] are the magnitudes of the vectors,  and  [latex]\hat{n}[/latex] is a unit vector that is perpendicular to both [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex] and, [latex]\theta[/latex]  is the angle between the two vectors.
To calculate the cross product, it is more convenient to use the definition that [latex]\vec{a}×\vec{b}[/latex] is equal to the following determinant.

[latex]\vec{a}×\vec{b}=[/latex] [latex]\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3} \end{array}\right|[/latex]

[latex]=\left|\begin{array}{cc} a_{2} & a_{3}\\ b_{2} & b_{3} \end{array}\right|[/latex][latex]\hat{i}[/latex] [latex]-\left|\begin{array}{cc} a_{1} & a_{3}\\ b_{1} & b_{3} \end{array}\right|[/latex][latex]\hat{j}[/latex] [latex]+\left|\begin{array}{cc} a_{1} & a_{2}\\ b_{1} & b_{2} \end{array}\right|[/latex][latex]\hat{k}[/latex]

[latex]=\left(a_{2}  b_{3}-a_{3}b_{2}\right)\hat{i}-\left(a_{1}  b_{3}-a_{3}b_{1}\right)\hat{j}+\left(a_{1}  b_{2}-a_{2}b_{1}\right)\hat{k}.[/latex]
(See the Matrices chapter of this book for information on finding a matrix determinant.)

Properties of the vector or cross product

1. If [latex]\vec{a}[/latex] is parallel to [latex]\vec{b}[/latex] then [latex]\vec{a}\times\vec{b}=\vec{0}[/latex].
This follows from the equation:
\begin{align*}\vec{a}\times\vec{b} =\left|\vec{a}\right|\left|\vec{b}\right|\sin\left(\theta\right)\,\hat{n}.\end{align*}
If the vectors are parallel, then the angle between them is [latex]\theta = 0[/latex], hence [latex]\sin\theta = 0[/latex] and therefore [latex]\vec{a}×\vec{b} = \vec{0}[/latex]. In particular,
\begin{align*}
\hat{i}\times \hat{i} &=\vec{0}.\\
\hat{j} \times \hat{j} &=\vec{0}.\\
\hat{k} \times \hat{k} &= \vec{0}.
\end{align*}
Note that we should write the answer as the zero vector [latex]\vec{0}[/latex] instead of the number [latex]0[/latex] since technically the result of a cross product of vectors must be a vector.

 

2. The order in which you take the cross product is important. In fact,
\begin{align*}\vec{a}\times\vec{b}&=-\vec{b}\times\vec{a}.\end{align*}

 

3. The direction of [latex]\vec{a}\times\vec{b}[/latex] is perpendicular to both [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex] in the direction in which your thumb would point if the fingers of your right hand are curled from [latex]\vec{a}[/latex] to [latex]\vec{b}[/latex]. This is called the right-hand rule. In particular,
\begin{align*}
\hat{i}\times \hat{j} &=\hat{k}.\\
\hat{j} \times \hat{k} &=\hat{i}.\\
\hat{k} \times \hat{i} &= \hat{j}.
\end{align*}

Examples: Cross product of vectors

1. Evaluate [latex]2\hat{i}\times \left( 3 \hat{i}+4 \hat{j} \right).[/latex]

Solution:

\begin{align*}
2\hat{i} \times \left(3\hat{i}+4\hat{j}\right)&=2\hat{i} \times 3\hat{i}+2\hat{i} \times 4\hat{j}\\
&=6\hat{i} \times \hat{i}+8\hat{i} \times \hat{j}\\
&=6(0)+8\hat{k}\\
&=8\hat{k}.
\end{align*}

or

\begin{align*}
2\hat{i}\times \left( 3\hat{i}+4 \hat{j} \right)&=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k}\\
2 & 0 & 0\\
3 & 4 & 0
\end{array}\right|\\
&=\left|\begin{array}{cc}
0 & 0\\
4 & 0
\end{array}\right| \hat{i}
-\left|\begin{array}{cc}
2 & 0\\
3 & 0
\end{array}\right| \hat{j}
+\left|\begin{array}{cc}
2 & 0\\
3 & 4
\end{array} \right| \hat{k}\\
&= 0\hat{i}-0\hat{j}+8\hat{k} \\
&=8\hat{k}.
\end{align*}

2.  Let the vectors  [latex]\vec{a} =2\hat{i}+\hat{j}+3\hat{k}[/latex]  and  [latex]\vec{b} =\hat{i}-2\hat{j}+5\hat{k}[/latex]. Find:

(a)  [latex]\vec{a}×\vec{b}[/latex]
(b) [latex]\vec{b}\times\vec{a}[/latex]

 

Solution (a):
\begin{align*}
\vec{a} \times \vec{b}&=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k}\\
2 & 1 & 3\\
1 & -2 & 5
\end{array}\right| \\
&=2\left|\begin{array}{cc}
1 & 3\\
-2 & 5
\end{array}\right|\hat{i}
-\left|\begin{array}{cc}
1 & 3\\
-2 & 5
\end{array}\right|\hat{j}
+\left|\begin{array}{cc}
2 & 1\\
1 & -2
\end{array}\right| \hat{k}\\
&=(1×5-3×(-2))\hat{i}-(2×5-3×1)\hat{j}+(2×(-2)-1×1)\hat{k} \\
&=11\hat{i}-7\hat{j}-5\hat{k}.
\end{align*}
Solution (b)
\begin{align*}
\vec{b}×\vec{a}&=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k}\\
1 & -2 & 5\\
2 & 1 & 3
\end{array}\right| \\
&=\left|\begin{array}{cc}
-2 & 5\\
1 & 3
\end{array}\right| \hat{i} -\left|\begin{array}{cc}
1 & 3\\
5 & 2
\end{array}\right| \hat{j} +\left|\begin{array}{cc}
1 & -2\\
2 & 1
\end{array}\right| \hat{k}\\
&=(-2×3-5×1)\hat{i}-(1×3-5×2)\hat{j}+(1×1-(-2)×2)\hat{k} \\
&=-11\hat{i}+7\hat{j}+5\hat{k}.
\end{align*}
Note that [latex]\vec{a}\times \vec{b} = -\vec{b}\times \vec{a}[/latex].
3. Find [latex]\vec{a}\times \vec{b}[/latex] if [latex]\vec{a}=2\hat{i}+3\hat{j}+\hat{k}[/latex] and [latex]\vec{b}=5\hat{j}+3\hat{k}[/latex].

Solution:

\begin{align*}
\vec{a} \times \vec{b}&=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k}\\
2 & 3 & 1\\
0 & 5 & 3\end{array}\right| \\
&=\left|\begin{array}{cc}
3 & 1\\
5 & 3
\end{array}\right| \hat{i} -\left| \begin{array}{cc}
2 & 1\\
0 & 3
\end{array}\right| \hat{j} +\left|\begin{array}{cc}
2 & 3\\
0 & 5
\end{array}\right|\hat{k} \\
&=\left(3\times3-5\times1\right)\hat{i} -\left(2\times3-0\times1\right)\hat{j} +\left(2\times5-0\times3\right)\hat{k}\\
&=\left(9-5\right)\hat{i}-\left(6-0\right)\hat{j}+\left(10-0\right)\hat{k} \\
&=4\hat{i}-6\hat{j}+10\hat{k}.
\end{align*}

Try the following exercises.

Exercises: Cross product of vectors

1.  Calculate the following:

a)  [latex]\hat{j}\times\hat{k}[/latex]

b)  [latex]\hat{i}\times4\hat{i}[/latex]

c)   [latex]\left(2\hat{i}+3\hat{j}-\hat{k}\right)\times\left(3\hat{j}+2\hat{k}\right)[/latex]

d)   [latex]3\hat{j}\times5\hat{i}[/latex]

e)   [latex]\left(\hat{i}-3\hat{j}+\hat{k}\right)\times\left(2\hat{i}+\hat{j}-\hat{k}\right)[/latex]

2.   Find a unit vector perpendicular to both [latex]\left(\hat{i}-\hat{k}\right)[/latex] and [latex]\left(\hat{i}+3\hat{j}-2\hat{k}\right)[/latex].

Answers


1.
(a) [latex]\hat{i}[/latex]       (b) [latex]\hat{0}[/latex]     (c) [latex]\ 9\hat{i}-4\hat{j}+6\hat{k}[/latex]

(d) [latex]-15\hat{k}[/latex]       (e) [latex]2\hat{i}+3\hat{j}+7\hat{k}[/latex]

2. [latex]\dfrac{\left(3,1,3\right)}{\sqrt{19}}[/latex]   or [latex]-\dfrac{\left(3,1,3\right)}{\sqrt{19}}=\dfrac{\left(-3,-1,-3\right)}{\sqrt{19}}.[/latex]

 

 

Key takeaways

  • Vectors can be either added geometrically (head to tail) or algebraically by adding the components of the vectors.
  • The vector from the origin to a point [latex]P(x,y,z)[/latex] is called the position vector of [latex]P[/latex] and can be denoted as
    \begin{align*}\overrightarrow{OP}&=x\hat{i}+y\hat{j}+z\hat{k}\end{align*}  or  \begin{align*} \overrightarrow{OP}=(x,y,z).\end{align*}
  • The magnitude of of a vector [latex]\vec{a}=x\hat{i}+y\hat{j}+z\hat{k}[/latex] can be written as [latex]\left|\vec{a}\right|[/latex], where
    \begin{align*} \left|\vec{a}\right|=\sqrt{x^2+y^2+z^2}.\end{align*}
  • If we divide a vector by its magnitude, we get a vector one unit long. We call this a unit vector, [latex]\hat{a}[/latex] where \begin{align*}\hat{a}=\dfrac{\vec{a}}{\left|\vec{a}\right|}.\end{align*}
  • The scalar (or dot) product of two vectors can be found in two ways. If [latex]\vec{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}[/latex] and [latex]\vec{b}=b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}[/latex] then
    \begin{align*}\vec{a} \cdot\vec{b}=\left|\vec{a}\right|\left|\vec{b}\right|\cos{\theta}\end{align*}  and  \begin{align*}\vec{a}\cdot \vec{b}=a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}.\end{align*}
  • If we do not know the angle between two vectors, we can use the two definitions of the dot product to find the angle.
  • The vector (or cross) product of two vectors, [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex] is equal to a vector of magnitude [latex]\left|\vec{a}\right|\left|\vec{b}\right| \left|\sin{\theta}\right|[/latex] at right angles to both [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex] with the direction dictated by a right hand rule.
  • The cross product of two vectors [latex]\vec{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}[/latex] and [latex]\vec{b}=b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}[/latex] can be calculated using the determinant: \begin{align*}\left|\begin{array}{ccc}
    \hat{i} & \hat{j} & \hat{k}\\
    a_{1} & a_{2} & a_{3}\\
    b_{1} & b_{2} & b_{3}
    \end{array}\right|.\end{align*}