Applications Using Addition and Subtraction

The development of vector methods in school mathematics is motivated initially by problems involving the vector quantities force, displacement, velocity and the like.

In this chapter, we consider problems of this sort but it is worth mentioning that there are many other objects that can be treated as vectors and the theory of vector spaces has applications in diverse fields. Consider, for example, the computer software that finds the best fitting line or curve to a given set of bivariate data. This is an application of vector mathematics in statistics, where the vectors are the columns of data.

We begin with problems from physics.

Example 1

A small boat sails a distance of $12$12 km on a bearing  $030^\circ$030° and then turns to bearing $250^\circ$250° and travels a further $10$10 km. How far from its starting point is the boat and what should its bearing be to return there?

This problem can be solved using the sine and cosine rules in trigonometry but we will use a vector approach.

The first diagram below represents the problem as it might appear on a map. The second diagram shows the displacements as vectors in coordinate form.

We used some simple trigonometry to obtain the coordinates of the displacement vectors. The resultant vector, in red, should be the sum of the two displacements. Its coordinate representation is

$\left(12\cos60^\circ-10\cos20^\circ,12\sin60^\circ-10\sin20^\circ\right)$(12cos60°−10cos20°,12sin60°−10sin20°)

This simplifies to $\left(-3.40,6.97\right)$(−3.40,6.97). This vector has length $\sqrt{(-3.40)^2+(6.97)^2}\approx7.76$√(−3.40)2+(6.97)2≈7.76. So, the boat has a return journey of $7.76$7.76 km.

The bearing angle of the red arrow is $\arctan\left(\frac{6.97}{-3.4}\right)\approx296^\circ$arctan(6.97−3.4​)≈296°. So, for the return trip, the bearing will need to be $180^\circ$180° away from this, which is $116^\circ$116°.

 

Example 2

A $20$20 kg mass is suspended on a chain $A$A attached to a ceiling timber. This means that there is a tension force of $20\times9.8$20×9.8 N in the chain when it is hanging vertically (if we ignore the weight of the chain). 

Suppose the $20$20 kg mass is pulled horizontally by another chain $B$B until the chain attached to the ceiling is at an angle of $20^\circ$20° from the vertical. Has the tension in chain $A$A increased? 

What are the tension forces in the two chains?

We make use of the fact that the vector with magnitude $T_A$TA​ can be resolved into two orthogonal components. These are shown as green arrows. If the two green arrows acted, they would have the same effect as the red arrow with magnitude $T_A$TA​ alone. In effect, we replace it by the pair of components.

The diagram illustrates the setup. Since $T_A>T_A\cos20^\circ$TA​>TA​cos20°, and this vertical force, magnitude $T_A\cos20^\circ$TA​cos20°,  is balancing the weight force of $196$196 N, we conclude that the tension in chain $A$A is greater than when the weight was hanging vertically.

We form two equations:

$T_A\cos20^\circ+196=0$TA​cos20°+196=0

$T_B+T_A\sin20^\circ=0$TB​+TA​sin20°=0

From the first of these we find $T_A=\frac{-196}{\cos20^\circ}=-208.6$TA​=−196cos20°​=−208.6 N. We have taken the downward weight force to be positive. So, this tension force acts upward.

Then, from the second equation, we obtain $T_B=-(-208.6)\sin20^\circ=71.3$TB​=−(−208.6)sin20°=71.3 N.

 

As an alternative strategy, we might note that in coordinate form the vector with magnitude $T_A$TA​ is $\left(T_A\sin20^\circ,T_A\cos20^\circ\right)$(TA​sin20°,TA​cos20°). The weight vector is $(0,-196)$(0,−196) and the sideways pull $T_B$TB​ is $\left(T_B,0\right)$(TB​,0).

These three vectors should add to the zero vector $(0,0)$(0,0). So, we have $\left(T_A\sin20^\circ+0+T_B,T_A\cos20^\circ-196+0\right)=(0,0)$(TA​sin20°+0+TB​,TA​cos20°−196+0)=(0,0). This vector equation leads to the same two equations as we derived above.

 

Example 3

A river flows at the rate of $0.5$0.5 m/s. A swimmer starting from the river bank aims at the opposite side and maintains a speed of $1.26$1.26 m/s in a direction perpendicular to the flow of the river. What is the swimmer's velocity (magnitude and direction) relative to the river bank?

The swimmer will be swept downstream. The vector representing the swimmer's motion will be the sum of $(0.5,0)$(0.5,0) and $(0,1.26)$(0,1.26). That is, the vector with coordinate form $(0.5,1.26)$(0.5,1.26). The magnitude of this vector is $\sqrt{0.5^2+1.26^2}\approx1.36$√0.52+1.262≈1.36 m/s.

The angle of the motion from the river bank is $\arctan\left(\frac{1.26}{0.5}\right)\approx68.4^\circ$arctan(1.260.5​)≈68.4°.

 

Suppose now that the swimmer wishes to maintain a course directly towards the opposite bank. The swimmer will need to swim upstream at a certain angle to the river bank to achieve this. What is the angle and what will be the swimmer's velocity in the direction directly across the river?

For the moment, we express the vector representing the swimmer's motion relative to the river bank by $(x,y)$(x,y),  where $x=1.26\cos\theta$x=1.26cosθ and $y=1.26\sin\theta$y=1.26sinθ. It must be that the river flow vector and the cross stream vector sum to this vector. So, $(0.5,0)+(0,v)=(x,y)$(0.5,0)+(0,v)=(x,y) where $v$v is the magnitude of the resultant velocity across the stream.

Thus, we can form the equations 

$0.5+0=1.26\cos\theta$0.5+0=1.26cosθ and

$0+v=1.26\sin\theta$0+v=1.26sinθ

From the first equation, we deduce that $\theta=66.6^\circ$θ=66.6°. Then, from the second equation, $v=1.26\times\sin66.6^\circ\approx1.16$v=1.26×sin66.6°≈1.16 m/s.

 

More Worked Examples

Question 1

Question 2

Question 3