Reviewing vectors
Clement Radcliffe,Contributor

As we continue the review of vectors, we will begin with the solution to last week's homework.

By reviewing the graph you should have determined the following:

b (3 2)

c (6 -3)

d (-3 4)

Please plot the following:-

x = (2 3)

y = (3 6)

Continuing Vectors

The arithmetic operations may be applied to vectors as follows:

• -1 x vectorAB = VectorBA

I do hope that you will realize that the negative sign reverses the direction of the vector.

• If vectorAB = (x₁ y₁) then k vectorAB = ( (k x₁) (k y₁) )

where k is a constant

• If vectorPQ = (x₁ y₁) and vectorRS = (x₂ y₂) then vectorPQ + vectorRS
  = ( (x₁ + x₂) (y₁ + y₂) )
• If vectorPQ - vectorRS may be evaluated as vectorPQ + 9-1) x vectorRS. I hope the implication is clear to you.

(You simply reverse the sign in vectorRS and add both vectors).
From the above, therefore, vectorPQ - vectorRS = ( (x₁ - x₂) (y₁ - y₂) )

Let us attempt the following examples which will clearly illustrate the above. It is advisable that you evaluate the answers and check them against mine.

Examples: Given the vector b = (2 6) and c = (-4 -2)

Determine the following vectors: -b, -c, 2b, -3c, 2b - 3c

Answers

1. -b = (-2 -6)

2. -c = (4 2)

3. 2b = (4 12)

4. -3c = (12 6)

5. 2b - 3c = (4 12) + (12 6) = ( (4 + 12) (12 + 6) ) = (16 18)

In number (3) above could you use the Cartesian diagram to investigate the relationship between the vectors b and 2b?

NB: Did you notice that 2b is twice the length of b and in the smae direction? I am sure that you will find it interesting to attempt all the questions in the Cartesian diagram and endeavour to determine the various relationships.

If you have mastered the above we will go on to a Special Vector - THE POSITION VECTOR.

POINTS TO NOTE

• The position vector begins at the origin.
• It is denoted by p or alternatively vectorOP where o is the origin.
• Given the point P(2, 3) then the position vectorOP = (2 3)

Let us do the following together.

The points A(1 , 2) B(5 , 2) c(6 , 4) and D(2 , 4) are the vertices of a quadrilateral ABCD.

(a) Express in the form = (x y)

(i) The position vectors vectorOA, vectorOB, vectorOC and vectorOD when o is the origin (0,0)

Solution

(a) (i) Given A(1 , 2), then the position vector vectorOA = (1 2)

Similarly, vectorOB = (5 2), vectorOC = (6 4), vectorOD = (2 4)

REMINDERS

Given the points A(x₁ y₁) and B(x₂, y₂) then vectorAB = ( (x₂ - x₁) (y₂ - y₁) )

This part of course, is based on the fact that in the vector triangle OAB

vectorOA + vectorAB = vectorOB

... vectorAB = vectorOB - vectorOA

From the above example, given the points A(x₁ , y₁) and B(x₂, y₂).

... vectorOA = (x₁ , y₁) and vectorOB = (x₂ y₂)

... vectorAB = (x₂ y₂) - (x₁ , y₁)

... vectorAB = ( (x₂ - x₁) (y₂ - y₁) )

Substituting from above

... vectorAB = ( (5 - 1) (2 - 2) ) = (4 0)

ALTERNATIVELY

Since vectorAB = vectorOB - vectorOA

Given the position vectors, vectorOA = (1 2) and vectorOB = (5 2)

... vectorAB = (5 2) - (1 2) = (4 0)

Home Work

The diagram above shows vector b and vector c

Express in the form (p q)

(i) b

(ii) c

(iii) c + b

(iv) b - c

Clement Radcliffe is the principal of Glenmuir High School in May Pen.