Reviewing length of a vector
Clement Radcliffe,Contributor

At the outset let us review the Home Work given last week.

Home Work

The diagram below shows vector b and vector c.

Express in the form (p q)

(i) b

(ii) c + b

(iv) b - c

Solution

(i) b = (3 6)

(ii) c = (-4 -7)

(iii) b + c = (3 6) + (-4 -2) = (-1 4)

(iv) b - c = (3 6) - (-4 -2) = ( (3 + 4) (6 + 2) ) = (7 8)

We will now continue vector by reviewing LENGTH OF A VECTOR

The vectorAB = (2 3) may be illustrated on the CASTESIAN DIAGRAM as follows:

As ACB is a right angled triangle
Then using Pythagoras' Theorem
AB² = AB² + CB²

... AB² = 2² + 3² = 4 + 9 = 13
... AB = square root of 13

It follows the form vectorAB = (x y), then the length of AB² = x² + y²

NB. the length of AB may be expressed as -AB- or the modulus of AB.
While this aspect of vector presented above is relatively simple the points noted are sometimes missed by students of the detriment. Please note them well.

Practice example

In the diagram above, A and B are points such that vectorOA = a and vector OB = b. The poinp p (not shown) is such that vector OP = 1/2a + b

(i) Write vectorOP in the form (x y)

(ii) Determine the length of OP

SOLUTION

(i) From the diagram, the coordinates of A = (6 , 8) and B(5 , 11)

It was illustrated in last week's lesson that if the coordinates of A is (6 , 8), then the position vectorOA = (6 8)

... position vectors

a = (6 8) and b = (5 11)

Since vectorOP = 1/2 a + b

then vectorOP = 1/2 (6 8) + (5 11) = (3 4) + (5 11) = (8 15)

... vectorOP = (8 15)

I do hope that you realize that the coordinates of the point P are (8, 15)

(ii) Using the formula for length -:

OP² = x² + y² [using Pythagoras' theorem]
= 8² + 15² = 64 + 225 = 289
... length of OP = square root of 289 = 17

I do hope that you had no difficulty in understanding the above. If this is the case, let us attempt another example.

EXAMPLE

Using the graph

(i) Express each of the position vectors vectorOA and vectorOB in the form (x y)

(ii) Determine the vectors:

(a) 3 OA, (b) -2OB, hence determine (c) 3 vectorOA - 2 vectorOB

(iii) If vectorOA + vectorOB = c, show that [c] = square root of 34

SOLUTION

(i) The coordinates of A and B respectively are (3, 4) and (2, 1)

... the position vectors are vectorOA = (3 4) and vectorOB (2 -1)

Note that the coordinates of A and B were used to determine the position vectors vectorOA and vectorOB. You could also have read off the components directly fromt he graph. In this case joining OA and OB should make it easier.

----> ------> ( ) ( )

(ii) (a) Given tha OA = (3 4) then 3OA = (3 x 3) (3 x 4) = ( 9 12)

... Answer is (9 12)

(b) Given that vectorOB = (2 -1) then vector-2OB = (-2 x 2) (-2 x -1) = (-4 2)

... Answer is (-4 2)

(c) vector3OA - vector2OB = (9 12) + (-4 2) = (5 14)

... Answer is (5 14)

• The hence in the question indicates that the answers in (a) and (b) should be used to solve the part (c). While other methods may be used, the method you are directed to use is usally the simplest approach. Always obey the instructions.
• A common error is to subtract the two vectors found above, instead of adding. this is justified as follows:

vector3OA - vector2OB = vector3OA + vector(-2OB)

You therefore add both answers.

(iii) Since vectorOA + vectorOB = c

... c = (3 4) + (2 -1) = (5 3)

As [c] is the length of vector c,

then [c] = suqare root of (5² + 3²) = square root of (25 + 9) = square root of 34

We will begin the review of Matrices Next Week. Have a productive week.

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