Vectors II - Mathematics Form 3 Notes

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Vectors in 3 Dimensions

  • 3 dimensional vectors can be represented on a set of 3 axes at right angles to each other (orthogonal), as shown in the diagram.
  • Note that the z axis is the vertical axis.
  • To get from A to B you would move:
    4 units in the x-direction, (x-component)
    3 units in the y-direction, (y-component)
    2 units in the z-direction. (z-component)

    In component form: 
     vector component form
  • In general: 
    general vector
    general vector graph


Column and Position Vectors

  • In three dimensions, a displacement is represented b a column vector of the form vector pqr where p,q and r are the changes in x,y,z directions respectively.

Example

The displacement from A (3, 1, 4 ) to B ( 7 ,2,6) is represented b the column vector, example column vector

The position vector of A written as OA is position vector OAwhere O is the origin

Addition of vectors in three dimensions is done in the same way as that in two dimensions.

Example

example 2 position vector

Column Vectors in Terms of Unit Vectors

In three dimension the unit vector in the x axis direction is = x axis column vector,that in the dirction of the y axis is = y axis unit vectorwhile that in the direction of z – axis is = z axis unit vector

Diagrammatic Representation of the Vectors.

diagram of unit vectors

Three unit vectors are written as; ijk unit vectors

Example

Express vector unit vectors examplein terms of the unit vector i, j and k

Solution

unit vector solution

Note;

  • The column vector column vector abc can be expressed as ai + bj + ck


Magnitude of a 3 Dimensional Vector.

  • Given the vector AB = xi + yj + 2k,then the magnitude of AB is written as |AB| = √(x2 +y2 +z2)
    magnitude of vector u
  • This is the length of the vector.
  • Use Pythagoras’ Theorem in 3 dimensions.
    AB2 = AR2 + BR2
    = (AP2 + PR2) + BR2
    =(xB − xA)2 + (yB − yA)2 + (zB − zA)2
                    →
    and if
    u = AB then the magnitude of u, |u| = length of AB


Distance Formula for a 3 Dimensionsional Vector

  • Recall that since: =distance formula vectors , then column vector u if then |u| = √(x2 +y+z2)
  • Since x = xB  xA and y = yB yA and zB  zA
    distance formula 3d diagram

Example:

  1. If A is (1, 3, 2) and B is (5, 6, 4)
             →
    Find |AB|
  2. If distance formula vector u. Find |u|

Solution

solution distance vector



Parallel Vectors and Collinearity

Parallel Vectors

  • Two vectors are parallel if one is scalar multiple of the other.i.e vector a is a scalar multiple of b ,i.e. a =kb then the two vectors are parallel.

Note;

  • Scalar multiplication is simply multiplication of a regular number by an entry in the vector

Multiplying by a Scalar

  • A vector can be multiplied by a number (scalar).e.g. multiply a by 3 is written as 3a.
  • Vector 3a has three times the length but is in the same direction as a .In column form, each component will be multiplied by 3.
  • We can also take a common factor out of a vector in component form. If a vector is a scalar multiple of another vector, then the two vectors are parallel, and differ only in magnitude. This is a useful test to see if lines are parallel.

Example

if multiplying by a scalar
parallel vector

Collinear Points

Points are collinear if one straight line passes through all the points. For three points A, B, C - if the line AB is parallel to BC, since B is common to both lines, A, B and C are collinear.

Test for collinearity

Example

A is (0, 1, 2), B is (1, 3, –1 ) and C is (3, 7, –7) Show that A, B and C are collinear.

Solution

collinearity example

 →        →
AB and BC are scalar multiples, so AB is parallel to BC. Since B is a common point, then A, B and C are collinear.

In general the test of collinearity of three points consists of two parts

  • Showing that the column vectors between any two of the points are parallel
  • Showing that they have a point in common.

Example

A (0,3), B (1 ,5) and C ( 4,1 1 ) are three given points. Show that they are collinear.

Solution

solution collinearity

AB and BC are parallel if AB = kBC, where k is a scalar

Therefore AB//BC and point B (1, 5) is common. Therefore A,B,and C are collinear.

Example

Show that the points A (1, 3, 5) ,B(4, 12, 20) and C are collinear.

Solution

Consider vectors AB and AC

collinearity solutin 2
Hence k =2/3
AC =2/3 AB
Therefore AB//AC and the two vectors share a common point A.The three points are thus collinear.

Example

In the figure above OA = a OB = b and OC = 3OB

collinearity example 3

  1. Express AB and AC in terms of a and b
  2. Given that AM = ¾ AB and AN = ½ AC, Express OM and O in terms of a and b
  3. Hence ,show that OM and N are collinear

Solution

  1. AB = OA + OB
    = − a + b
    AC = − a + 3b
  2. OM =OA + AM
    = OA + ¾AB
    = a +¾(− a + b)
    = a − ¾ab
    = ¼a + ¾b
    ON =OA +AN
    =OA + ½AC
    a + ½(− a + 3b)
    a = a − ½a + 3/2b
    = ½ a + 3/2b
  3. OM = kON ¾ b + ¼a = k/2a + 3k/2b
    Comparing the coefficients of a;
    1/4 = k/2
    k = 1/2
    Thus, OM = 1/2 ON.
    Thus two vectors also share a common point ,O .Hence, the points are collinear.


Proportional Division of a Line

Internal Division

  • In the figure below, the line is divided into 7 equal parts
    division of a line
  • The point R lies 4/7 of the ways along PQ if we take the direction from P to Q to be positive, we say R divides PQ internally in the ratio 4 : 3.
  • If Q to P is taken as positive,then R divides QP internally in the ratio 3 : 4 .Hence, QR : RP = 3 : 4 or, 4 QR = 3RP.

External Division

  • In internal division we look at the point within a given interval while in external division we look at points outside a given interval,
  • In the figure below point P is produced on AB
    external division
  • The line AB is divided into three equal parts with BP equal to two of these parts. If the direction from A to B is taken as positive, then the direction from P to B is negative.
  • Thus AP : PB = 5 : -2.In this case we say that P divides AB externally in the ratio 5 : −2 or P divides AB in the ratio 5 : −2.

Points, Ratios and Lines

Finding the ratio in which a point divides a line.

Example:

The points A(2, –3, 4), B(8, 3, 1) and C(12, 7, –1 ) form a straight line. Find the ratio in which B divides AC.

Solution

point ratios

B divides AC in ratio of 3 : 2

 

 

Points Dividing Lines in Given Ratios.

Example:

P divides AB in the ratio 4:3. If A is (2, 1, –3) and B is (16, 15, 11 ), find the co-ordinates of P.
point dividing line

Solution:

→                →       →
AP = 4   so 3AP = 4PB
→     3
PB

3(p a) = 4(b p)
3
p – 3a = 4b – 4p
7p = 4b + 3a
p = 1/7(4b + 3a)
point dividing line example

Points Dividing Lines in Given Ratios Externally.

Example:

Q divides MN externally in the ratio of 3:2. M is (–3, –2, –1 ) and N is (0, –5, 2).Find the co-ordinates of Q.

Note that QN is shown as –2 because the two line segments are MQ and QN, and QN is in the opposite direction to MQ.

 points dividing lines externally



The Ratio Theorem

The figure below shows a point S which divides a line AB in the ratio m : n

ratio theoremdiagram

Taking any point O as origin, we can express s in terms of a and b the positon vectors of a and b respectively.

OS = OA + AS
But AS =   
 AB
               m+n
Therefore, OS = OA +  m   AB
                                m+n
Thus S = a   (−a + b)
                    m+n
= a −   a +  m  b
        m+n     m+n

= (1 − m  )a m b
         m+n       m+n
= (m+n−m)a +  m b
       m+n         m+n
 n   a +  m  b
   m+n     m+n

This is called the ratio theorem. The theorem states that the position vectors s of a point which divides a line AB in the ratio m: n is given by the formula;

S =   n  +  m  b
     m+n      m+n

where a and b are positon vectors of A and B respectively. Note that the sum ofco-ordinates n/m+n and m/m+m= 1

Thus ,in the above example if the ratio m : n = 5 : 3
Then m = 5 and n = 3
OR = 
3  a + b
        5+3    5+3

Thus, r = 3/8a + 3/8b

Example

A point R divides a line QR externally in the ratio 7 : 3. If q and r are position vectors of point Q and R respectively, find the position vector of p in terms of q and r.

Solution

We take any point O as the origin and join it to the points Q, R and P as shown below
ratio theorem example

QP: PR = 7: −3
Substituting m = 7 and n = −3 in the general formulae;
OP =    −
3       7    r
         7+(−3)      7+(−3)

P = −3/4 q + 7/4r

Vectors can be used to determine the ratio in which a point divides two lines if they intersect

Example

In the below OA = a and OB = b. A point P divides OAin the ratio 3:1 and another point O divides AB in the ratio 2 : 5. If OQ meets BP at M
ratio theorem example 2
Determine:

  1. OM : MQ
  2. BM : MP

Let OM : MQ = k : (1 – k) and BM –MP = n : ( 1 – n)

Using the ratio theorem
OQ =
5/7 a + 2/7 b
OM= kOQ
= k
(5/7 a+ 2/7 b)
Also by ratio theorem;
OM = nOP +( 1 – n )OB
But OP = 
3/4 a
Therefore, OM = n (3/4 a) + (1−n)b
Equating the two expressions;
k(5/7 a+ 2/7 b) = n (3/4a) + ( 1−n )b
Comparing the co-efficients
5/7 k = 3/4n………..1
2/7k = 1 − n……..
2
k = 21/25 and n = 10/13
The ratio BM :MP = 10/13 : 3/13
= 10: 3

 



Past KCSE Questions on the Topic

  1. The figure below is a right pyramid with a rectangular base ABCD and VO as the height. The vectors AD= a, AB = b and DV = v
    vectors q1
    1. Express
      1. AV in terms of a and c
      2. BV in terms of a, b and c
    2. M is point on OV such that OM: MV=3:4, Express BM in terms of a, b and cSimplify your answer as far as possible
  2. In triangle OAB, OA = a, OB = b and P lies on AB such that AP: BP = 3.5
    1. Find the terms of a and b the vectors
      1. AB
      2. AP
      3. BP
      4. OP
    2. Point Q is on OP such AQ = −5/8a + 9/40b
      Find the ratio OQ: QP
  3. The figure below shows triangle OAB in which M divides OA in the ratio 2: 3 and N divides OB in the ratio 4:1 AN and BM intersect at X
    vectors q3
    1. Given that OA = a and OB = b, express in terms of a and b:
      1. AN
      2. BM
    2. If AX = sAN and BX = tBM, where s and t are constants, write two expressions for OX in terms of a, b, s and t
      Find the value of s
      Hence write OX in terms of a and b
  4. The position vectors for points P and Q are 4i + 3j and 6j + 6 k respectively. Express vector PQ in terms of unit vectors i, j and k. Hence find the length of PQ, leaving your answer in simplified surd form.
  5. In the figure below, vector OP = p and OR =r. Vector OS = 2r and OQ = 3/2p.
    vectors q5
    1. Express in terms of p and r (i) QR and (ii) PS
    2. The lines QR and PS intersect at K such that QK = mQR and PK = nPS, where m and n are scalars. Find two distinct expressions for OK in terms of p,r,m and n. Hence find the values of m and n.
    3. State the ratio PK: KS
  6. Point T is the midpoint of a straight line AB. Given the position vectors of A and T are i−j + k and 2i+1½ k respectively, find the position vector of B in terms of i, j and k
  7. A point R divides a line PQ internally in the ration 3:4. Another point S, divides the line PR externally in the ration 5:2. Given that PQ = 8 cm, calculate the length of RS, correct to 2 decimal places.
  8. The points P, Q, R and S have position vectors 2p, 3p, r and 3r respectively, relative to an origin O. A point T divides PS internally in the ratio 1:6
    1. Find, in the simplest form, the vectors OT and QT in terms p and r
      1. Show that the points Q, T, and R lie on a straight line
      2. Determine the ratio in which T divides QR
  9. Two points P and Q have coordinates (-2, 3) and (1 , 3) respectively. A translation map point P to P’ (10, 10)
    1. Find the coordinates of Q’ the image of Q under the translation
    2. The position vector of P and Q in (a) above are p and q respectively given that mp – nq = position vector q9
      Find the value of m and n
  10. Given that qi + 1/3j + 2/3 k is a unit vector, find q
  11. In the diagram below, the coordinates of points A and B are (1, 6) and (15, 6) respectively). Point N is on OB such that 3 ON = 2 OB. Line OA is produced to L such that OL = 3 OA
    vectors q11
    1. Find vector LN
    2. Given that a point M is on LN such that LM: MN = 3: 4, find the coordinates of M
    3. If line OM is produced to T such that OM: MT = 6:1
      1. Find the position vector of T
      2. Show that points L, T and B are collinear
  12. In the figure below, OQ = q and OR = r. Point X divides OQ in the ratio 1:2 and Y divides OR in the ratio 3:4 lines XR and YQ intersect at E.
    vectors q12
    1. Express in terms of q and r
      1. XR
      2. YQ
    2. If XE = mXR and YE = nYQ, express OE in terms of:
      1. r, q and m
      2. r, q and n
    3. Using the results in (b) above, find the values of m and n.
  13. Vector q has a magnitude of 7 and is parallel to vector p. Given that p= 3i – j + 1½ k, express vector q in terms of i, j, and k.
  14. In the figure below, OA = 3i + 3j ABD OB = 8i – j. C is a point on AB such that AC:CB 3:2, and D is a point such that OB//CD and 2OB = CD (T17)
    vectors q14
    Determine the vector DA in terms of I and j
  15. In the figure below, KLMN is a trapezium in which KL is parallel to NM and KL = 3NM
    vectors q15
    Given that KN = w, NM = u and ML = v. Show that 2u = v + w
  16. The points P, Q and R lie on a straight line. The position vectors of P and R are 2i + 3j + 13k and 5i – 3j + 4k respectively; Q divides SR internally in the ratio 2: 1 . Find the
    1. Position vector of Q
    2. Distance of Q from the origin
  17. Co-ordinates of points O, P, Q and R are (0, 0), (3, 4), (11 , 6) and (8, 2) respectively. A point T is such that the vector OT, QP and QR satisfy the vector equation OT = QP + ½QT. Find the coordinates of T.
  18. In the figure below OA = a, OB = b, AB = BC and OB: BD = 3:1
    vectors q18
    1. Determine
      1. AB in terms of a and b
      2. CD, in terms of a and b
    2. If CD: DE = 1:k and OA: AE = 1:m determine
      1. DE in terms of a, b and k
      2. The values of k and m
  19. The figure below shows a grid of equally spaced parallel lines
    vectors q19
    AB = a and BC = b
    1. Express
      1. AC in terms of a and b
      2. AD in terms of a and b
    2. Using triangle BEP, express BP in terms of a and b
    3. PR produced meets BA produced at X and PR = 1/9b – 8/3a
      By writing PX as kPR and BX as hBA and using the triangle BPX determine the ratio PR: RX
  20. The position vectors of points x and y are x = 2i + j – 3k and y = 3i + 2j – 2k respectively. Find XY 2. Given that X = 2i + j− 2K, y = -3i + 4j – k and z= 5i + 3j + 2k and that p= 3x – y + 2z, find the magnitude of vector p to 3 significant figures.
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