Exercise 9.2

Vectors in 2-D

Imagine we have the points O, A, B, C, D & E as shown in the diagram below:

There are a few things we may want to find out about the diagram, such as:

• What are the coordinates of the point 'B'?

• What is the length: |OA|?

• What are the coordinates of 'D'?

• Are DE and AB parallel?

• What is the angle 'θ' between OA and AB?

• What is the equation of the line AB?

• Where does this line cut the 'x' and 'y' axes?

• Does the point F(20,14) lie on the line DE produced ('produced' means 'extended' in this context)?

• Where do the lines AB and CD meet? 

All of these questions could happily be answered using our existing techniques; but we will instead learn to use 'Vector Techniques' to answer them. The reason being that these vector techniques can also be applied to 3-D problems, which we will encounter later...

Vector Notation

You are familiar with writing the coordinates of 'A' as (4,3)

In vector notation, we may write the coordinates of 'A' as 4i + 3j
where 'i' stands for 'units along the x axis'
and 'j' stands for 'units along the y axis'

This is called the position vector of 'A' (i.e. the coordinates of 'A')

The vector which takes us from 'A' to 'B' is shown on the diagram. This is called a relative vector 'AB'. It is usual to write a relative vector as a column vector, as shown on the diagram:

		┌   ┐
	AB =    │ 5 │		or    AB = 5i + 12j
		│12 │
		└   ┘
 

It tells us that, to get from 'A' to 'B', we have to move '5 units along x' and '12 units along y'
 

	Simply subtract:	Coordinates  -	Coordinates 
				  of 'Q'	  of 'P'
 

e.g. Find the relative vector from AC:

	AC =	  C   -	  A
		┌   ┐	┌   ┐	┌   ┐
	AC =	│11 │ -	│ 4 │ =	│ 7 │	(or: 7i - 3j)
		│ 0 │	│ 3 │	│-3 │
		└   ┘	└   ┘	└   ┘
 
 

e.g. Find the relative vector OA:

	OA =	  A   -	  O
		┌   ┐	┌   ┐	┌   ┐
	OA =	│ 4 │ -	│ 0 │ =	│ 4 │	(or: 4i + 3j)
		│ 3 │	│ 0 │	│ 3 │
		└   ┘	└   ┘	└   ┘
 

So the relative vector OA is the same as the position vector 'A'
That is always true:
 

 
e.g. 'E' has coordinates (17.5, 9). Find the vector OE:

		┌      ┐
	OE =	│ 17.5 │	or       OE = 17.5i + 9j
		│  9   │
		└      ┘
 

Again showing that there is little difference between the coordinates of a point, it's position vector and the relative vector from the origin to the point!

Adding Vectors

Relative Vectors can be added together to form a route to a point
In the diagram, we are show that the vector which "takes us from 'A' to 'B' is":

		┌   ┐
	AB =	│ 5 │
		│12 │
		└   ┘
 

So, if we start from 'A' and then 'add' the vector which takes us from 'A to B'; then we end up at 'B':

	OB =	  OA  +	  AB
		┌   ┐	┌   ┐	┌   ┐
	OB =	│ 4 │ +	│ 5 │ =	│ 9 │		(or: 9i + 15j)
		│ 3 │	│12 │	│15 │
		└   ┘	└   ┘	└   ┘
 

So, the coordinates of 'B' are: (9, 15)
 

e.g. The point 'P' has coordinates (7, -2) and the vector PQ is -2i + 3j. Find the coordinates of Q:

	OQ =	  OP  +	  PQ
		┌   ┐	┌   ┐	┌   ┐
	OQ =	│ 7 │ +	│-2 │ =	│ 5 │		(or: 5i + j)
		│-2 │	│ 3 │	│ 1 │
		└   ┘	└   ┘	└   ┘
 

So, the coordinates of 'Q' are (5, 1)

The length of a Vector

		┌   ┐
	PQ =    │ p │		or	    PQ = pi + qj
		│ q │
		└   ┘
 

e.g. Find the length of the vector OA in the diagram:

		┌   ┐
	OA =	│ 3 │		 |OA| = √ 3² + 4²
		│ 4 │		 
		└   ┘		      = 5 units	
 
 

e.g. Find the length of the vector AB in the diagram:

		┌   ┐
	AB =	│ 5 │		 |OA| = √ 5² + 12²
		│12 │		 
		└   ┘		      = 13 units	
 
 

e.g. Find the length of the vector OB in the diagram:

	Earlier, we found the coordinates of 'B' were (9, 15)
 
		┌   ┐
	OB =	│ 9 │		 |OB| = √ 9² + 15²
		│15 │		 
		└   ┘		      = 17.49 units	
 
 

Equivalent Vectors

In the diagram; the vector OA is labelled 'a' (it is common to name relative vectors using a small letter, underlined).

But you'll notice that the vector CD is also labelled 'a'. This means that the vector CD is both the same length and the same direction as the vector OA:

		┌   ┐
	CD =	│ 4 │
		│ 3 │
		└   ┘
 
 

e.g. Since we now know CD, find the coordinates of 'D':

	OD =	  OC  +	  CD
		┌   ┐	┌   ┐	┌   ┐
	OD =	│11 │ +	│ 4 │ =	│15 │		(or: 15i + 3j)
		│ 0 │	│ 3 │	│ 3 │
		└   ┘	└   ┘	└   ┘
 

So, the coordinates of 'D' are: (15, 3)
 

e.g. Hence find the relative vector DE:

	DE =	   E  -	  D
		┌   ┐	┌   ┐	┌   ┐
	AC =	│17½│ -	│15 │ =	│2.5│		(or: 2.5i - 6j)
		│ 9 │	│ 3 │	│ 6 │
		└   ┘	└   ┘	└   ┘
 
 

Parallel Vectors

Parallel vectors have the same direction, but different lengths.
It is fairly easy to spot parallel vectors when they are written in the column vector format; because there will be a 'constant' that one vector can be multiplied to get the other:

		  AC  		 AB
		┌   ┐	 	┌   ┐	
		│2.5│  ——×2——>	│ 5 │ 		So: 2AC = AB
		│ 6 │  ——×2——>	│12 │	
		└   ┘		└   ┘	
 

Which tells us that they ARE parallel; but also that AB is twice as long as AC
 

e.g. Are the vectors: PQ = 6i + 4j and RS = 3i + 2j parallel?

		  PQ  		 RS
		┌   ┐		┌   ┐	
		│ 6 │  ——×½——>	│ 3 │ 		So: ½PQ = RS
		│ 4 │  ——×½——>	│ 2 │	
		└   ┘		└   ┘	
 
 

e.g. The points L, M, N and P have coordinates; (5,3), (8,5),  (4,8) & (-2,4) respectively. Describe the shape LMNP:

Start with a quick sketch of the shape on a grid...

So, the sides of the shape are LM, MN, NP and PL

	LM =	  M   -	  L
		┌   ┐	┌   ┐	┌   ┐
	LM =	│ 8 │ -	│ 5 │ =	│ 3 │  ═══════╗
		│ 5 │	│ 3 │	│ 2 │  ═══════╣
		└   ┘	└   ┘	└   ┘         ║
				              ║
	MN =	  N   -	  M	              ║
		┌   ┐	┌   ┐	┌   ┐         ║
	MN =	│ 4 │ -	│ 8 │ =	│-4 │         ║	
		│ 8 │	│ 5 │	│ 3 │         ║×-2
		└   ┘	└   ┘	└   ┘         ║
				              ║
	NP =	  P   -	  N	              ║
		┌   ┐	┌   ┐	┌   ┐         ║
	NP =	│-2 │ -	│ 4 │ =	│-6 │  ◄══════╣
		│ 4 │	│ 8 │	│-4 │  ◄══════╝
		└   ┘	└   ┘	└   ┘
 
	PL =	  L   -	  P
		┌   ┐	┌   ┐	┌   ┐
	PL =	│ 5 │ -	│-2 │ =	│ 7 │	
		│ 3 │	│ 4 │	│ 7 │
		└   ┘	└   ┘	└   ┘
 
The first thing we notice is that the vectors: LM and NP are parallel
No other vectors are parallel and a quadrilateral is only one pair of parallel sides is a: TRAPEZIUM
 
 

e.g. 'F' is a point along DE produced such that F is four times as far from D as E is. Find the coordinates of 'F':

Since we are told that 'F' is along the same line
And is four times as far:
 
		  DF  =   4  ×	 DE
		┌   ┐		┌   ┐	
		│...│ =	  4  ×	│2.5│ 	
		│...│   	│ 6 │	
		└   ┘		└   ┘	
 
And so, by finding the vector OF, we can find the coordinates of 'F':
 
	OF =	  OD  +	  DF
		┌   ┐	┌   ┐	┌   ┐
	OF =	│15 │ +	│10 │ =	│25 │		(or: 25i + 27j)
		│ 3 │	│24 │	│27 │
		└   ┘	└   ┘	└   ┘
 
 

e.g. 'R' is a point along AB such that it's distance from 'B' is three times it's distance from A.
Find the coordinates of 'R'.

Another way of saying this is to say:	The vector AR is ¼ of the vector AB
 
						  AR  =   ¼  ×	 AB
						┌   ┐		┌   ┐	
						│...│ =	  ¼  ×	│ 5 │ 	
						│...│   	│12 │	
						└   ┘		└   ┘	
 
And so, by finding the vector OR, we can find the coordinates of 'R':
 
	OR =	  OA  +	  AR
		┌   ┐	┌   ┐	┌    ┐
	OR =	│ 4 │ +	│5/4│ =	│5.25│		(or: 5.25i + 6j)
		│ 3 │	│ 3 │	│ 6  │
		└   ┘	└   ┘	└    ┘
 
 

The Vector Equation of a Line

In order to find the vector equation of a line, we need to know the position vector of any point on the line and the direction vector of the line (i.e. any relative vector that lies along the line or is parallel to the line):

			 r   =    OP    +    λ  PQ 
 

We will deal with an example to show how easy this is to do, before we go about explaining what it means...
 

e.g. Find the vector equation of the line AB in the diagram:

Since we know the vector position vector of 'A' and we know the relative vector 'AB' (which is a vector along the line)

		r  =	 OA      +  λ    AB
			┌   ┐		┌   ┐	
		r =	│ 4 │    +  λ	│ 5 │	 	or: r = (4i + 3j) + λ(5i + 12j)
			│ 3 │		│12 │
			└   ┘		└   ┘
			╘═╦═╛		╘═╦═╛
      the position vector ╝		  ╚ the dirn vector
   of a point on the line       	    parallel to the line
  

So, this is the equation of the line "though 4i +3j" and "parallel to the vector 5i + 12j"

To understand this equation, we need to start by understanding what 'r' and 'λ' stand for:

1. 'λ' is a scalar (that means "an ordinary number, not a vector"). It is the variable in the equation; so putting in different values of 'λ' generates the different 'points' on the line...

2. 'r' is the position vector of any point on the line AB

For example: If we put in λ=0 into the equation:

		┌   ┐		┌   ┐		┌   ┐
	r =	│ 4 │    +  0	│ 5 │	  =	│ 4 │  		(or:  4i + 3j)
		│ 3 │		│12 │		│ 3 │
		└   ┘		└   ┘		└   ┘
 
	Which is the position vector of the point 'A' on the line AB
 

If we put in λ=1 into the equation:

		┌   ┐		┌   ┐		┌   ┐
	r =	│ 4 │    +  1	│ 5 │	  =	│ 9 │  		(or:  9i + 15j)
		│ 3 │		│12 │		│15 │
		└   ┘		└   ┘		└   ┘
 
	Which is the position vector of the point 'B' on the line AB
 

If we put in λ=½ into the equation:

		┌   ┐		┌   ┐		┌   ┐
	r =	│ 4 │   +  ½	│ 5 │	  =	│6.5│  		(or:  6.5i + 9j)
		│ 3 │		│12 │		│ 9 │
		└   ┘		└   ┘		└   ┘
 
	Which is the position vector of 'M', the midpoint of AB
 

If we put in λ=¼ into the equation:

		┌   ┐		┌   ┐		┌    ┐
	r =	│ 4 │   +  ¼	│ 5 │	  =	│5.25│		(or:  5.25i + 6j)
		│ 3 │		│12 │		│ 6  │
		└   ┘		└   ┘		└    ┘
 
	Which is the position vector of the point 'R' on the line AB
 

So, because we used the point 'A' and vector 'AB' to form the equation of the line; when we put in 'λ=0' we get the point 'A' and when be put in 'λ=1' we get the point 'B' and when we put in 'λ=1/2' we get the point which is halfway between 'A' and 'B'
[Kinda reminds you of the Grand Old Duke of York - when they were up they were up, when they were down they were down and when they were halfway up...]

Actually; we could have found the equation of the line AB in a different way:

We could have started with the point 'B' and the vector AB:

	r  =	 OB     +   η    AB
		┌   ┐		┌   ┐	
	r =	│ 9 │   +   η	│ 5 │	(or: r = (9i + 15j) + η(5i + 12j)
		│15 │		│12 │	
		└   ┘		└   ┘	
 
Note: Since I have written the equation a different way, I have used a different variable 'η'
      so we don't get mixed up...
 

Although this is still the equation of the line AB; this time when we put in 'η=0' we get the point 'B'; when we put in 'η=-1' we get the point 'A' and when we put in 'η=1' we get the point with coordinates (14, 27); which is further along the line AB...

Alternatively; we could have used the two points A(4, 3) and R(5.25, 6) to find the line AB:

We know the vector position vectors of 'A' & 'R'
 
So we can find the  relative vector 'AR':		AR =	  R   -	  A
								┌   ┐	┌   ┐	┌   ┐
							AR =	│5¼ │ -	│ 4 │ =	│1¼ │	
								│ 6 │	│ 3 │	│ 3 │
								└   ┘	└   ┘	└   ┘
 
Now we can find the equation of the line AR (which is really the same line as AB):
 
	r  =	  OA    +   κ    AR
		┌   ┐		┌   ┐	
	r  =	│ 4 │   +   κ	│1¼ │	
		│ 3 │		│ 3 │	
		└   ┘		└   ┘	
 

So, the question is posed; how can we tell that these three equations all represent the same line?
We'll answer that in a little while
 

e.g. Find the equation of the line CD:

Since we know the vector position vector of 'C' And we know the relative vector 'CD' (is the same as OA)

		r  =	 OC      +  μ    CD
			┌   ┐		┌   ┐	
		r =	│11 │    +  μ	│ 4 │		 or: r = (11i + 0j) + μ(4i + 3j)
			│ 0 │		│ 3 │
			└   ┘		└   ┘
			╘═╦═╛		╘═╦═╛
      the position vector ╝		  ╚ the dirn vector
   of a point on the line       	    parallel to the line
 
 

e.g. Find the equation of the line through 9i + 2j, parallel to 5i + 2j:

We know the vector position vector of a point on the line '9i + 2j' And we know a vector parallel to the line '5i + 2j;

		r  =	 OA      +  λ    AB
			┌   ┐		┌   ┐	
		r =	│ 9 │    +  λ	│ 5 │		 or: r = (9i + 2j) + λ(5i + 2j)
			│ 2 │		│ 2 │
			└   ┘		└   ┘
			╘═╦═╛		╘═╦═╛
      the position vector ╝		  ╚ the dirn vector
   of a point on the line       	    parallel to the line
  
 

Finding a Particular Point on a Line

1. You may be asked to find a point for a particular value of 'λ':

e.g. Find the point on r = (4i + 3j) + λ(5i + 12j) where λ=4:

		┌   ┐		┌   ┐		┌   ┐
	r =	│ 4 │    +   4	│ 5 │	  = 	│24 │  =  24i + 51j
		│ 3 │		│12 │		│51 │
		└   ┘		└   ┘		└   ┘
 
 

2. You may be asked to find a point given a particular 'x' or 'y' coordinate:

e.g. Find the point on r = (4i + 3j) + λ(5i + 12j) where x=-1:

Since we don't know the 'y' coordinate of the point where x=-1 we can call the point (-1, ?):
  
	┌   ┐		┌   ┐		┌   ┐
	│ 4 │   +   λ	│ 5 │	  = 	│-1 │  ————► 4 +  5λ = -1  ————►  λ = -1 ═╗
	│ 3 │		│12 │		│ ? │  ————► 3 + 12λ =  ?  ◄══════════════╝
	└   ┘		└   ┘		└   ┘
 
	Solving the TOP ROW gives: λ = -1
 
	Substituting into the BOTTOM ROW gives: y = -9
 
	So the point is (-1, -9)
 
 

e.g. Find the point on r = (2i - 8j) + λ(3i + 5j) where y=2:

Since we don't know the 'x' coordinate of the point where y=2 we can call the point (?, 2):
 
	┌   ┐		┌   ┐		┌   ┐
	│ 2 │   +   λ	│ 3 │	  = 	│ ? │  ————►  3 + 3λ = ? ◄════════════╗
	│-8 │		│ 5 │		│ 2 │  ————► -8 + 5λ = 2 ————► λ = 0 ═╝
	└   ┘		└   ┘		└   ┘
 
	Solving the BOTTOM gives: λ = 2
 
	Substituting into the BOTTOM ROW gives: x = 9
 
	So the point is (9, 2)
 
 

e.g. Does the point (7, 11) lie on r = (3i + 3j) + λ(2i + 5j):

Pretend we don't know the 'y' coordinate of the point where x=7 we can call the point (7, ?):
 
	┌   ┐		┌   ┐		┌   ┐
	│ 3 │   +   λ	│ 2 │	  = 	│ 7 │  ————►  3 + 2λ = 7 ————►  λ = 2 ═╗
	│ 3 │		│ 5 │		│ ? │  ————►  3 + 5λ = ? ◄═════════════╝
	└   ┘		└   ┘		└   ┘
 
	Solving the TOP ROW gives: λ = 2
 
	Substituting into the BOTTOM ROW gives: x = 13
 
	So the point where x is '7' is (7, 13), NOT (7, 11)
  
	So (7, 11) is not on the line!!
 
 

Finding where a line intercepts the axes

1. To find where a line intercepts the x axis; we put 'y=0'

e.g. Find where r = (3i + 4j) + λ(2i - 8j) intercepts the x axis:

We put y=0, so we can call the point (?, 0):
 
	┌   ┐		┌   ┐		┌   ┐
	│ 3 │   +   λ	│ 2 │	  = 	│ ? │  ————► 3 + 3λ = ? ◄════════════╗
	│ 4 │		│-8 │		│ 0 │  ————► 4 - 8λ = 0 ————► λ = ½ ═╝
	└   ┘		└   ┘		└   ┘
 
	Solving the BOTTOM ROW gives: λ = 1/2
 
	Substituting into the TOP ROW gives: x = 4
 
	So the point is (4, 0)
 
 

2. To find where a line intercepts the y axis; we put 'x=0'

e.g. Find where r = (i + 5j) + λ(2i - 6j) intercepts the y axis:

We put x=0, so we can call the point (0, ?):

	┌   ┐		┌   ┐		┌   ┐
	│ 1 │   +   λ	│ 2 │	  = 	│ 0 │  ————► 1 + 2λ = 0 ————►  λ = -½ ═╗
	│ 5 │		│-6 │		│ ? │  ————► 5 - 6λ = ? ◄══════════════╝
	└   ┘		└   ┘		└   ┘

	Solving the TOP ROW gives: λ = -1/2
 
	Substituting into the BOTTOM ROW gives: y = 8

	So the point is (0, 8)
 
 

Identifying Equivalent Equations

Before, we learnt that there are many different ways of writing the vector equation of the line AB

We posed the question;
"How do we tell that:  r = (9i + 15j) + λ(5i + 12j) and r = (4i + 3j) + κ(1¼i + 3j)
represent the same line?"

1. The direction vectors must be parallel:

	   d.v. of eqn 2    d.v. of eqn 1
		┌   ┐		┌   ┐	
		│1¼ │  ——×4——>	│ 5 │ 	
		│ 3 │  ——×4——>	│12 │	
		└   ┘		└   ┘	
 

2. We must prove that the 'point' given on the 2nd equation is also a point that lies on the 1st equation:

			┌   ┐
So, we know that	│ 4 │	lies on the second line
			│ 3 │
			└   ┘
 
We have to check it this point also lies on the first line:
 
Ignore the 'y' coordinate and call the point (4, ?)
 
	┌   ┐		┌   ┐		┌   ┐
	│ 9 │   +   λ	│ 5 │	  = 	│ 4 │  ————►  9 +  5λ = 4 ————►  λ = -1 ═╗
	│15 │		│12 │		│ ? │  ————► 15 + 12λ = ? ◄══════════════╝
	└   ┘		└   ┘		└   ┘
 
	Solving the TOP ROW gives: λ = -1
 
	Substituting into the BOTTOM ROW gives: y = 3
 
	So the point where x is 4 is: (4, 3)
 

Which shows that both lines pass through the point (4, 3) and since we have already shown that the direction vectors are parallel, we can say that these two equations represent the SAME line...

Finding where two Lines Intersect

To find where two lines intersect:

1. Make them equation to each other

2. Read the "Top Line" of this vector equation and call this equation (1)

3. Read the "Bottom Line" of this vector equation and call this equation (2)

4. Solve simultaneously to find 'λ' and 'µ'

5. Put the value of 'λ' back into it's equation to find the point of intersection

e.g. Find where the lines AB [ r = (4i + 3j) + λ(5i + 12j) ] and CD [ r = (11i) + µ(4i + 3j) ] intersect:

	┌   ┐		┌   ┐		┌   ┐		┌   ┐
	│ 4 │   +   λ	│ 5 │	  = 	│11 │   +   μ	│ 4 │  ——�E#9658; 4+ 5λ = 11+4μ ——�E#9658; 4+5λ = 11+ 5μ
	│ 3 │		│12 │		│ 0 │		│ 3 │  ——�E#9658; 3+12λ =    3μ ——�E#9658; 1+4λ =      μ
	└   ┘		└   ┘		└   ┘		└   ┘                     Multiply bottom eqn
							                          by 5 and subtract�E/font>
	Solving equations simultaneously gives: λ=-1 and μ=-3
 
	Putting λ = -1 into line (1):
 
		┌   ┐		┌   ┐		┌   ┐
	r =	│ 4 │    +   -1	│ 5 │	  = 	│-1 │  =  -i - 9j
		│ 3 │		│12 │		│-9 │
		└   ┘		└   ┘		└   ┘
 

		┌   ┐		┌   ┐	
	  r =	│ 4 │    +   λ	│ 5 │	  
	  │	│ 3 │		│12 │	
	  ▼	└   ┘		└   ┘
	┌   ┐	┌   ┐		┌   ┐	
	│ x │=	│ 4 │    +   λ	│ 5 │	<=  x = 4 +  5λ   
	│ y │	│ 3 │		│12 │	<=  y = 3 + 12λ
	└   ┘	└   ┘		└   ┘

					So: λ = x - 4	      and:	 λ = y - 3
						  5  			      12   
 
					=>		x - 4  =  y - 3
							  5        12   
 

	=>		x - 5  =  y - 2
			  7         3   
 

	=>		x - 2  =  y + 3
			  4         5   
 

	=>		x - 2  =  y - 1
			  5         -1   
 

	=>		x - 0  =  y - 2
			  -1        3   
 

	=>		x - 5  =  y - 2
			  7         0   
 
But this one doesn't seem to make sense, because we can't "divide by 0"
  
In fact, the only way it comes close to making sense is if y=2
(then the top of the fraction would also be zero)
 
	=> In fact the equation is: y = 2
 
 

	OB =	  OA  +	  AB
		┌   ┐	┌   ┐	┌   ┐
	OB =	│ 4 │ +	│ 3 │ =	│.. │	(or: ..i + ..j)
		│ 1 │	│ 9 │	│.. │
		└   ┘	└   ┘	└   ┘
 

		  AB  		 AP
		┌   ┐		┌   ┐	
		│ 3 │  ——×⅓——>	│.. │ 		
		│ 9 │  ——×⅓——>	│.. │	
		└   ┘		└   ┘	
 

		┌   ┐	
	i.e.	│ 3 │ 			
		│ 9 │   	
		└   ┘	
 	

		┌   ┐		
		│ 1 │ 		
		│ 3 │   
		└   ┘	
 

			┌   ┐		┌   ┐	
		r =	│ 4 │    +  λ	│ 1 │	 	or: r = (4i + j) + λ(i + 3j)
			│ 1 │		│ 3 │
			└   ┘		└   ┘
			╘═╦═╛		╘═╦═╛
      the position vector ╝		  ╚ the dirn vector
   of a point on the line       	    parallel to the line
 

			┌   ┐		┌   ┐	
		r =	│ 7 │    +  λ	│ 1 │	 	or: r = (7i + 10j) + λ(i + 3j)
			│10 │		│ 3 │		
			└   ┘		└   ┘		
      

			┌   ┐		┌   ┐	
		r =	│ 5 │    +  λ	│ 1 │	 	or: r = (5i + 4j) + λ(i + 3j)
			│ 4 │		│ 3 │	
			└   ┘		└   ┘	
 	  

	┌   ┐		┌   ┐		┌   ┐
	│ 4 │   +   λ	│ 1 │	  = 	│ 0 │  ————►  4 +  λ = 0  ————►  λ = -4 ═╗
	│ 1 │		│ 3 │		│ ? │  ————►  1 + 3λ = ?  ◄══════════════╝
	└   ┘		└   ┘		└   ┘
 
	Solving the TOP ROW gives: λ = -4
 
	Substituting into the BOTTOM ROW gives: y = ..
 
	So the point where x is '0' is (0, �E
 
	Therefore (0, -11) [is/is not]? on the line!!
 

		┌   ┐	
	OA =	│ 4 │			(or: 4i + 3j)
		│ 3 │	
		└   ┘	
 

		  OA  		 OP
		┌   ┐		┌   ┐	
		│ 4 │  —×2/3�Egt;	│8/3│ 		
		│ 3 │  —×2/3�Egt;	│.. │	
		└   ┘		└   ┘	
 

		┌   ┐	
	a  =	│12 │
		│ 0 │	
		└   ┘	
 

	AB =	  B   -	  A
		┌   ┐	┌   ┐	┌   ┐
	AB =	│ 2 │ -	│12 │ =	│.. │
		│10 │	│ 0 │	│.. │
		└   ┘	└   ┘	└   ┘
 

		  AB  		 AX
		┌   ┐		┌   ┐	
		│.. │  —×1/5�Egt;	│.. │ 		
		│.. │  —×1/5�Egt;	│.. │	
		└   ┘		└   ┘	
 

	OX =	  OA  +	  AX
		┌   ┐	┌   ┐	┌   ┐
	OX =	│12 │+	│.. │=	│.. │
		│ 0 │	│.. │	│.. │
		└   ┘	└   ┘	└   ┘
 

		┌   ┐	
	b  =	│ 2 │
		│10 │	
		└   ┘	
 

		  b  		 AY
		┌   ┐		┌   ┐	
		│ 2 │  ——×n——>	│2n │ 		
		│10 │  ——×n——>	│10n│	
		└   ┘		└   ┘	
 

	BY =	  BA  +	  AY
		┌   ┐	┌   ┐	┌	┐
	BY =	│.. │ +	│2n │ =	│2n+10	│
		│.. │	│10n│	│10n-10	│
		└   ┘	└   ┘	└   	┘
 

		 OX  		 AY
		┌   ┐		┌	┐	
		│10 │  ——×k——>	│2n+10	│ 		
		│ 2 │  ——×k——>	│10n-10	│	
		└   ┘		└   	┘	
 

	AB =	  B  -	  A
		┌   ┐	┌   ┐	┌   ┐
	AB =	│-1 │-	│-2 │=	│.. │
		│ 5 │	│ 3 │	│.. │
		└   ┘	└   ┘	└   ┘
 

	┌   ┐		┌   ┐	
   r = 	│-2 │   +   λ	│.. │	  
	│ 3 │		│.. │		
	└   ┘		└   ┘
 

	┌   ┐		┌   ┐	
   r = 	│-1 │   +   λ	│.. │	  
	│ 5 │		│.. │		
	└   ┘		└   ┘
 

	┌   ┐		┌   ┐	
   r = 	│ 0 │   +   λ	│ 1 │	  
	│ 1 │		│ 2 │		
	└   ┘		└   ┘
 

Putting in λ = 2:
					  A
	┌   ┐		┌   ┐		┌   ┐
   r =	│ 0 │   +   2	│ 1 │	  = 	│ 2 │  			So A = 2i + 5j 
	│ 1 │		│ 2 │		│ 5 │  
	└   ┘		└   ┘		└   ┘
 
Putting in λ = 5 ...
 

	┌   ┐		┌   ┐		┌   ┐
	│ 0 │   +   λ	│.. │	  = 	│ ? │  ————► 0 +  λ = ? ◄════════════╗
	│ 4 │		│.. │		│ 0 │  ————► 4 - 2λ = 0 ————► λ = �E═╝
	└   ┘		└   ┘		└   ┘
 
	Solving the BOTTOM ROW gives: λ = �E
 
	Substituting into the TOP ROW gives: x = �E

	So the point is (�E 0)
 

	┌   ┐		┌   ┐		┌   ┐
	│ 0 │   +   λ	│ 2 │	  = 	│ 6 │  ————►  0 + 2λ = 6 ————► λ = �E═╗
	│-5 │		│ 1 │		│ ? │  ————► -5 +  λ = ? ◄════════════╝
	└   ┘		└   ┘		└   ┘
 
	Solving the TOP ROW gives: λ = �E
 
	Substituting into the BOTTOM ROW gives: y = �E
 
	Which shows that (6, 2) [is/is not]? on the line
 

	=>		x - 0  =  y - 3
			  4         2  

		┌   ┐		┌   ┐	
	r =	│ 0 │    +  λ	│ 4 │	 		or:   r = (3j) + λ(4i + 2j)
		│ 3 │		│ 2 │		
		└   ┘		└   ┘		
      

	=>		x - -3  =  y - 2
			   1         4   
 

	=>		x - A  =  y - 3
			  B         0   
 

		┌   ┐		┌   ┐	
	r =	│ 1 │    +  λ	│ 1 │	 		or:  	 r = ...
		│ 3 │		│ 0 │		
		└   ┘		└   ┘