11.03 Working with vectors in i, j notation

Vectors in $i$i, $j$j notation

It is often convenient to use a special pair of vectors as the basis for the vectors in the plane. These are two vectors, called $\mathbf{i}$i and $\mathbf{j}$j, that are at right-angles to one another and that have unit length. In terms of the Cartesian coordinate system, the vector $\mathbf{i}$i points in the direction usually designated the $x$x-axis, and the vector $\mathbf{j}$j points in the direction usually designated the $y$y-axis. 

This convention makes it very easy to switch between a representation of a vector as a coordinate pair and an equivalent representation as a linear combination of basis vectors (which we will look at in more detail later).

Worked examples

Question 1

The coordinate pair $(4,3)$(4,3) represents the same vector as the sum $4\mathbf{i}+3\mathbf{j}$4i+3j.

 

Care is needed if the vector under consideration does not have its tail at the origin of the coordinate system. Remember that an arrow translated so that its length and direction do not change, is considered to be the same vector.

Question 2

The vector represented by the green arrow is clearly the same vector as in the previous diagram. You should look at this diagram carefully to make sure that you understand why the two vectors are the same.

Practice questions

Question 3

Question 4

Question 5

Addition and subtraction

The vectors $\mathbf{i}$i and $\mathbf{j}$j introduced earlier form what is called an orthonormal basis for a two-dimensional or plane vector space. These two fundamental or basis vectors have unit length and are at right-angles to one another.

Any vector in the plane can be expressed as a linear combination of the basis vectors $\mathbf{i}$i and $\mathbf{j}$j. We can add or subtract vectors expressed in this way by simply collecting like terms.

Worked examples

Question 6

Suppose $\mathbf{w}=2\mathbf{i}+6\mathbf{j}$w=2i+6j and $\mathbf{u}=5\mathbf{i}+\mathbf{j}$u=5i+j.

Then, we have $\mathbf{w}+\mathbf{u}=(2+5)\mathbf{i}+(6+1)\mathbf{j}=7\mathbf{i}+7\mathbf{j}$w+u=(2+5)i+(6+1)j=7i+7j.

 

Question 7

Suppose $\mathbf{w}=6\mathbf{i}+2\mathbf{j}$w=6i+2j and $\mathbf{u}=5\mathbf{i}-3\mathbf{j}$u=5i−3j.

Then, $\mathbf{w}-\mathbf{u}=(6-5)\mathbf{i}+(2-(-3))\mathbf{j}=\mathbf{i}+5\mathbf{j}$w−u=(6−5)i+(2−(−3))j=i+5j.

 

Vector addition and subtraction expressed this way conforms to one of the two distributive laws in vector arithmetic. For any vector $\mathbf{v}$v, it is an axiom that $a\mathbf{v}+b\mathbf{v}=(a+b)\mathbf{v}$av+bv=(a+b)v. In the examples above we have merely used $\mathbf{i}$i and $\mathbf{j}$j instead of $\mathbf{v}$v.

 

In the exercises, you are asked to verify the associative property for the scalars in vector addition. It is not difficult to see that this follows from the associative law in the addition of the real numbers. 

That is, we know that for real numbers $a$a, $b$b and $c$c, it is true that $(a+b)+c=a+(b+c)$(a+b)+c=a+(b+c). It follows that $(a+b)\mathbf{v}+c\mathbf{v}=a\mathbf{v}+(b+c)\mathbf{v}=(a+b+c)\mathbf{v}$(a+b)v+cv=av+(b+c)v=(a+b+c)v.

Practice questions

Question 8

Question 9

Question 10

Magnitude and direction

Roughly speaking, vectors are quantities with both magnitude and direction. If we wish to determine whether two vectors are equal, we need to check whether they have the same magnitude and whether they have the same direction.

Two vectors are considered to have the same direction if they are scalar multiples of one another.

Worked examples

Question 11

Two vectors, given in coordinate form, are $(-3,17)$(−3,17) and $(6,-34)$(6,−34).  

If plotted on the Coordinate Plane, it is apparent that the two are pointing in the same direction although with opposite sense. And one has twice the length of the other.  We could take out a scalar multiple from the second vector and write them as $(-3,17)$(−3,17) and $-2(-3,17)$−2(−3,17). Clearly, they only differ by the scalar multiple $-2$−2.

We say these two vectors are parallel.

Question 12

We are given two line segments. One joins the points $(2,4)$(2,4) and $(1,5)$(1,5). The other connects $(0,1)$(0,1) and $(-2,3)$(−2,3). Are the line segments parallel?

In terms of the basis vectors $\mathbf{i}$i and $\mathbf{j}$j, the line segments are equivalent to the vectors $\mathbf{i}-2\mathbf{i}+5\mathbf{j}-4\mathbf{j}=-\mathbf{i}+\mathbf{j}$i−2i+5j−4j=−i+j

and

$-2\mathbf{i}-0\mathbf{i}+3\mathbf{j}-\mathbf{j}=-2\mathbf{i}+2\mathbf{j}=2\left(-\mathbf{i}+\mathbf{j}\right)$−2i−0i+3j−j=−2i+2j=2(−i+j).

Again, the second vector is a multiple of the first. So the line segments are parallel.

 

The vector $(-2,3)$(−2,3) or $-2\mathbf{i}+3\mathbf{j}$−2i+3j has magnitude $\sqrt{(-2)^2+3^2}=\sqrt{13}$√(−2)2+32=√13. This is just an application of Pythagoras's theorem.

The same is true in three (or more) dimensions. The magnitude of a vector $(a,b,c)$(a,b,c) is written with vertical lines: $|(a,b,c)|$|(a,b,c)| and is given by $\sqrt{a^2+b^2+c^2}$√a2+b2+c2.

Thus, if we can first express a vector in its coordinate form, we can calculate its magnitude.

Worked examples

Question 13

A line segment joins the points $(2,4)$(2,4) and $(1,5)$(1,5). What is the magnitude of the vector that represents this line segment?

We saw above that the vector can be represented as $-\mathbf{i}+\mathbf{j}$−i+j or just $(-1,1)$(−1,1). So, its magnitude is $\sqrt{(-1)^2+1^2}=\sqrt{2}$√(−1)2+12=√2.

 

Unit vectors

A unit vector has magnitude $1$1. Given any vector, we may wish to express it as a multiple of the corresponding unit vector.

The notation used for unit vectors uses the caret symbol. That is, if $\mathbf{v}$v is a vector, its corresponding unit vector is $\mathbf{\hat{v}}$^v.

If we know the magnitude of a vector we can calculate the unit vector by just multiplying the vector by the reciprocal of its magnitude. That is,

$\mathbf{\hat{v}}=\frac{1}{|\mathbf{v}|}\mathbf{v}$^v=1|v|​v

This is true because we can express vector $\mathbf{v}$v as a multiple of its corresponding unit vector. That is, $\mathbf{v}=|\mathbf{v}|\mathbf{\hat{v}}$v=|v|^v which we rearrange to give the above formula. (Observe that the magnitude of a vector is a scalar.)

Worked example

Question 14

What is the unit vector for $\mathbf{w}=4\mathbf{i}+5\mathbf{j}$w=4i+5j and how would $\mathbf{w}$w be expressed in terms of $\mathbf{\hat{w}}$^w?

We have, $|\mathbf{w}|=\sqrt{4^2+5^2}=\sqrt{41}$|w|=√42+52=√41. So, $\mathbf{\hat{w}}=\frac{1}{\sqrt{41}}\mathbf{w}$^w=1√41​w

 or

$\mathbf{w}=\sqrt{41}\mathbf{\hat{w}}$w=√41^w

We are willing to believe that the expression on the right has the correct magnitude, $\sqrt{41}$√41, because in general. if $\mathbf{v}=a\mathbf{u}$v=au, then

$|\mathbf{v}|$|v|	$=$=	$|a\mathbf{u}|$|au|
	$=$=	$|a||\mathbf{u}|$|a||u|

We leave it as a challenge to verify that this last statement is true.

Practice questions

Question 15

Question 16

Scalar multiplication

How are the magnitude and the direction of a vector affected by multiplication by a scalar?

It seems reasonable to say that the direction of a vector is unaffected by scalar multiplication. In the plane, we can represent any vector by its orthonormal components $\mathbf{i}$i and $\mathbf{j}$j. For a vector $\mathbf{v}=a\mathbf{i}+b\mathbf{j}$v=ai+bj in the Coordinate Plane, we can think of its slope $\frac{b}{a}$ba​. Suppose now that the vector is multiplied by a scalar $k$k. We need to find the slope of the new vector $k\mathbf{v}$kv.

Since $k\mathbf{v}=k\left(a\mathbf{i}+b\mathbf{j}\right)$kv=k(ai+bj), we have

$k\mathbf{v}=ka\mathbf{i}+kb\mathbf{j}$kv=kai+kbj and its slope is $\frac{kb}{ka}=\frac{b}{a}$kbka​=ba​ as before.

We have used this idea to define what is meant by vectors having the same direction: the directions are the same if the vectors are scalar multiples of one another.

However, multiplication by a scalar does affect the magnitude of a vector.

Consider again the vector $\mathbf{w}=k\mathbf{v}$w=kv. We wish to determine the magnitude $|\mathbf{w}|$|w|, 

Beginning with the form $\mathbf{w}=k\left(a\mathbf{i}+b\mathbf{j}\right)$w=k(ai+bj), we have

$|\mathbf{w}|$|w|	$=$=	$|ka\mathbf{i}+kb\mathbf{j}|$|kai+kbj|
	$=$=	$\sqrt{(ka)^2+(kb)^2}$√(ka)2+(kb)2
	$=$=	$\sqrt{k^2a^2+k^2b^2}$√k2a2+k2b2
	$=$=	$\sqrt{k^2\left(a^2+b^2\right)}$√k2(a2+b2)
	$=$=	$k\sqrt{a^2+b^2}$k√a2+b2
	$=$=	$|k||\mathbf{v}|$|k||v|

We used the absolute value symbol $|k|$|k| because $\sqrt{k^2}$√k2 is positive whatever the sign of $k$k. 

The general rule is,

$\left|k\mathbf{v}\right|=\left|k\right|\left|\mathbf{v}\right|$|kv|=|k||v|

 

How do the magnitudes of sums and differences of vectors compare with the magnitudes of the separate vectors?

It turns out that $\left|\mathbf{u}+\mathbf{v}\right|$|u+v| is always less than or equal to $\left|\mathbf{u}\right|+\left|\mathbf{v}\right|$|u|+|v|. This is called the triangle inequality.

It is easy to see that this is true with a diagram showing the addition of two arrow vectors $\mathbf{u}$u and $\mathbf{v}$v.

It can be shown in a more general way with the help of some algebra after an operation called the scalar product has been introduced. We assume it is true, for the moment.

It follows that

$|\mathbf{u}|$|u|	$=$=	$|\mathbf{u-v+v}|$|u−v+v|
	$\le$≤	$|\mathbf{u-v}|+|\mathbf{v}|$|u−v|+|v|
$\therefore\ |\mathbf{u-v}|$∴ |u−v|	$\ge$≥	$|\mathbf{u}|-|\mathbf{v}|$|u|−|v|

Also, 

$|\mathbf{v}|$|v|	$=$=	$|\mathbf{v-u+u}|$|v−u+u|
	$\le$≤	$|\mathbf{v-u}|+|\mathbf{u}|$|v−u|+|u|
$\therefore\ |\mathbf{v-u}|$∴ |v−u|	$\ge$≥	$|\mathbf{v}|-|\mathbf{u}|$|v|−|u|
$\therefore\ |\mathbf{u-v}|$∴ |u−v|	$\ge$≥	$|\mathbf{v}|-|\mathbf{u}|$|v|−|u|

On comparing these two expressions for $|\mathbf{u-v}|$|u−v|, we conclude that

$|\mathbf{u-v}|\ge\left||\mathbf{u}|-|\mathbf{v}|\right|$|u−v|≥||u|−|v||

 

For the exercises following this chapter, you will also need to recall the fact that there is only one way to express a vector in the plane as a linear combination of two independent vectors. That is, if $\mathbf{w}=a\mathbf{u}+b\mathbf{v}$w=au+bv where $\mathbf{u}$u and $\mathbf{v}$v are independent, then the scalars $a$a and $b$b are determined.

This is useful in solving certain equations.

Worked examples

Question 17

If $\mathbf{a}=\mathbf{i}+2\mathbf{j}$a=i+2j and $\mathbf{b}=5\mathbf{i}-3\mathbf{j}$b=5i−3j, find $|\mathbf{a}+\mathbf{b}|$|a+b|. Compare this magnitude with the sum $|\mathbf{a}|+|\mathbf{b}|$|a|+|b|.

$|\mathbf{a}+\mathbf{b}|$|a+b|	$=$=	$\left|\mathbf{i}+2\mathbf{j}+5\mathbf{i}-3\mathbf{j}\right|$|i+2j+5i−3j|
	$=$=	$\left|6\mathbf{i}-\mathbf{j}\right|$|6i−j|
	$=$=	$\sqrt{6^2+(-1)^2}$√62+(−1)2
	$=$=	$\sqrt{37}$√37

On the other hand,

$|\mathbf{a}|+|\mathbf{b}|$|a|+|b|	$=$=	$|\mathbf{i}+2\mathbf{j}|+|5\mathbf{i}-3\mathbf{j}|$|i+2j|+|5i−3j|
	$=$=	$\sqrt{5}+\sqrt{34}$√5+√34

With a calculator, we can confirm that $\sqrt{37}\le\sqrt{5}+\sqrt{34}$√37≤√5+√34, as expected.

Question 18

Suppose $\mathbf{u}=a\mathbf{i}-3\mathbf{j}$u=ai−3j and $\mathbf{v}=4\mathbf{i}+b\mathbf{j}$v=4i+bj. The sum of these two vectors is the zero vector. Find the scalars $a$a and $b$b.

The sum is $\mathbf{u}+\mathbf{v}=a\mathbf{i}-3\mathbf{j}+4\mathbf{i}+b\mathbf{j}=(a+4)\mathbf{i}+(-3+b)\mathbf{j}$u+v=ai−3j+4i+bj=(a+4)i+(−3+b)j.

But, this is equal to the zero vector, $0\mathbf{i}+0\mathbf{j}$0i+0j. Hence, $a+4=0$a+4=0 and $-3+b=0$−3+b=0. We deduce that $a=-4$a=−4 and $b=3$b=3.

 

Practice question

Question 19

Finding unit vectors

Given any non-zero vector, there is always a vector with unit length that has the same direction as the given vector.

A vector with unit length is called a unit vector and it is notated $\mathbf{\hat{v}}$^v. How do we calculate the unit vector?

Since the vector $\mathbf{v}$v and the corresponding unit vector $\mathbf{\hat{v}}$^v have the same direction, it must be that $\mathbf{v}$v is a scalar multiple of $\mathbf{\hat{v}}$^v. We can write $\mathbf{v}=k\mathbf{\hat{v}}$v=k^v. 

Then, 

$\left|\mathbf{v}\right|$|v|	$=$=	$\left|k\mathbf{\hat{v}}\right|$|k^v|
	$=$=	$\left|k\right|\left|\mathbf{\hat{v}}\right|$|k||^v|
	$=$=	$|k|\times1$|k|×1

Thus, the scalar we require is $|k|=\left|\mathbf{v}\right|$|k|=|v|. Therefore, $\mathbf{v}=\left|\mathbf{v}\right|\mathbf{\hat{v}}$v=|v|^v and so, 

$\mathbf{\hat{v}}=\frac{1}{\left|\mathbf{v}\right|}\mathbf{v}$^v=1|v|​v

 

Worked example

Question 20

A vector $\mathbf{u}$u is given by the coordinate pair $\left(6,\sqrt{3}\right)$(6,√3). Find the corresponding unit vector.

We have $\left|\mathbf{u}\right|=\sqrt{6^2+(\sqrt{3})^2}=\sqrt{39}$|u|=√62+(√3)2=√39.

Therefore, $\mathbf{\hat{u}}=\frac{1}{\sqrt{39}}\left(6,\sqrt{3}\right)$^u=1√39​(6,√3).

In terms of the $\mathbf{i},\mathbf{j}$i,j basis notation, this is $\mathbf{\hat{u}}=\frac{6}{\sqrt{39}}\mathbf{i}+\frac{1}{\sqrt{13}}\mathbf{j}$^u=6√39​i+1√13​j.

It is easy to check that this is indeed a unit vector.

Practice questions

Question 21

Question 22

Applications of vectors

Vectors have many real world applications including physics and engineering.

Momentum

Any object that is moving has momentum.  Momentum is a measurement of a moving body and is a product of its mass and velocity.  

Think about these two scenarios

An elephant is skateboarding down a hill.  His large mass and high velocity create a large momentum.  

 

Alternatively, a grasshopper sits on the same skateboard.  His smaller mass creates a smaller momentum. 

Formally, we define the momentum (p) of a moving object to be, the product of its mass (m) and the velocity (v).  

• Mass is a scalar quantity.
• Velocity is a vector quantity.
• So momentum is the result of a scalar multiplication (mass) of the vector (velocity). 

We can use proportional reasoning to further understand the relationship between momentum and the mass and velocity. For example, if a velocity is held constant, but the mass of an object is tripled, then this will result in a tripling of the momentum.  

 

Work

Work is related to force and distance traveled by the formula

$W=\mathbf{F\cdot s}$W=F·s

where $W$W is the work done when a force given by vector $\mathbf{F}$F moves an object through a displacement vector $\mathbf{s}$s. As expected, the two vectors give a scalar value to the quantity work. The formula uses a special vector operation, the dot- or scalar product.

If $\mathbf{F}=\left(f_1,f_2,f_3\right)$F=(f1​,f2​,f3​) and $\mathbf{s}=\left(s_1,s_2,s_3\right)$s=(s1​,s2​,s3​), then $\mathbf{F\cdot s}=f_1s_1+f_2s_2+f_3s_3$F·s=f1​s1​+f2​s2​+f3​s3​.

 

Vector resolutes

We have seen that any vector in the plane can be uniquely expressed as a linear combination of two independent vectors. 

Conversely, a vector can be resolved into component vectors. This is a pair of independent vectors that when added have the effect of the original vector. 

Usually, the independent vectors chosen are the orthogonal unit vectors notated $\mathbf{i}$i and $\mathbf{j}$j. These are understood to align with the horizontal and vertical axes, respectively, of a Cartesian coordinate system.

We use ordinary trigonometric relations, sine and cosine, when resolving vectors in this way.

 

Worked example

Question 23

A vector has a direction angle of $27^\circ$27° and can be resolved into horizontal and vertical components such that the horizontal component has magnitude $14$14 N. What is the vertical component?

We understand the direction angle to be the angle measured counterclockwise from the positive horizontal axis.

The vector $\mathbf{v}$v, colored green, has magnitude related to the given information by $\cos27^\circ=\frac{14}{\mathbf{|v|}}$cos27°=14|v|​.  Thus, $\mathbf{|v|}=\frac{14}{\cos27^\circ}$|v|=14cos27°​.

Then by Pythagoras, the vertical component has magnitude $\sqrt{\left(\frac{14}{\cos27^\circ}\right)^2-14^2}$√(14cos27°​)2−142. This is $\approx7.1$≈7.1 N.

 

Question 24

A projectile is launched with a velocity of $750$750 km/h at an angle of elevation of $48^\circ$48°. What are the horizontal and vertical components of the velocity?

Horizontally: $750\cos48^\circ\approx501.8$750cos48°≈501.8 km\h

Vertically: $750\sin48^\circ\approx557.4$750sin48°≈557.4 km\h

 

If the projectile is launched over level ground, for how long will it fly before hitting the ground and how far will it travel horizontally?

We can use the formula from Physics that relates the initial velocity $u$u and the final velocity $v$v under a constant acceleration $a$a acting for time $t$t. That is, $v=u+at$v=u+at. 

The projectile will have a vertical velocity of $-u$−u at the moment when it hits the ground.

We know the acceleration downwards due to gravity: $-9.8$−9.8 m/s. And, we need to convert the velocities to m/s rather than km/h.

So, using the formula and the previous result, we have $-557.4\times\frac{1000}{3600}=557.4\times\frac{5}{18}-9.8t$−557.4×10003600​=557.4×518​−9.8t. This gives $t=31.6$t=31.6. That is, $31.6$31.6 seconds before impact.

Finally, the distance traveled horizontally is the horizontal velocity multiplied by the time in flight. That is, $501.8\times\frac{5}{18}\times31.6=4404.7$501.8×518​×31.6=4404.7 m.

 

Question 25

We can use a procedure developed from the scalar product to find the area of a triangle formed by a pair of vectors located at the origin.

From geometry, we know that the area $A$A of the triangle formed by the line segments defined by two vectors $\mathbf{a}$a and $\mathbf{b}$b is $A=\frac{1}{2}|\mathbf{a}||\mathbf{b}|\sin\theta$A=12​|a||b|sinθ.

We also have an expression for $\cos\theta$cosθ in terms of the scalar product. That is, $\cos\theta=\frac{\mathbf{a\cdot b}}{|\mathbf{a}||\mathbf{b}|}$cosθ=a·b|a||b|​. So, we can make a useful substitution if we first express the area formula in terms of $\cos\theta$cosθ.

Thus, $A=\frac{1}{2}|\mathbf{a}||\mathbf{b}|\sqrt{1-\cos^2\theta}$A=12​|a||b|√1−cos2θ, and therefore

$A$A	$=$=	$\frac{1}{2}|\mathbf{a}||\mathbf{b}|\sqrt{1-\left(\frac{\mathbf{a\cdot b}}{|\mathbf{a}||\mathbf{b}|}\right)^2}$12​|a||b|√1−(a·b|a||b|​)2
	$=$=	$\frac{1}{2}\sqrt{|\mathbf{a}|^2|\mathbf{b}|^2-\left(\mathbf{a\cdot b}\right)^2}$12​√|a|2|b|2−(a·b)2

This can be simplified if we write the vectors in coordinate form.

$A$A	$=$=	$\frac{1}{2}\sqrt{\left(a_1^2+a_2^2\right)\left(b_1^2+b_2^2\right)-\left(a_1b_1+a_2b_2\right)^2}$12​√(a21​+a22​)(b21​+b22​)−(a1​b1​+a2​b2​)2
	$=$=	$\frac{1}{2}\sqrt{a_1^2b_1^2+a_1^2b_2^2+a_2^2b_1^2+a_2^2b_2^2-a_1^2b_1^2-2a_1b_1a_2b_2-a_2^2b_2^2}$12​√a21​b21​+a21​b22​+a22​b21​+a22​b22​−a21​b21​−2a1​b1​a2​b2​−a22​b22​
	$=$=	$\frac{1}{2}\sqrt{a_1^2b_2^2-2a_1b_1a_2b_2+a_2^2b_1^2}$12​√a21​b22​−2a1​b1​a2​b2​+a22​b21​
	$=$=	$\frac{1}{2}\sqrt{\left(a_1b_2-a_2b_1\right)^2}$12​√(a1​b2​−a2​b1​)2
	$=$=	$\frac{1}{2}\left(a_1b_2-a_2b_1\right)$12​(a1​b2​−a2​b1​)

The expression $a_1b_2-a_2b_1$a1​b2​−a2​b1​ happens to be the determinant of a matrix formed by letting the coordinates of the vectors be the matrix columns. 

This idea extends to the areas of triangles formed by vectors in $3$3-dimensional space where it involves an operation called the vector cross product which can be defined for $3$3-dimensional vectors. Again, the formula involves determinants.

Practice questions

Question 26

Question 27

Question 28