VECTOR ANALYSIS 2
VECTOR ANALYSIS 2
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2 = 2 + 2 – 2 cos
Therefore
2 = 2 + 2 – 2 cos
02. USED TO FIND THE PROJECTION OF ONE VECTOR ONTO ANOTHER VECTOR
– Suppose the projection of a onto b
i.e
Cos Q =
Cos Q =
proj = cos Q ———– i
Also
. = cos Q
= cos Q ———–ii
Equalizing i and ii as follows;
proj = cos Q =
proj =
similary
proj =
Where;
Proj = projection of onto
Proj = projection of onto
VECTOR PROJECTION
This is given by
V proj =.
And
V. proj =.
Where;
V. proj = vector projection of onto
V. proj = vector projection of onto
03. TO FIND THE WORKDONE
– Consider the diagram below
Force applied (F) = component of tone
cos Q =
Cos Q =
F = Cos Q
Also
Distance d =
d = ….ii
Hence
Work done = Force applied (F) x distance (d)
W.D = F x d
W. D = cos Q x
= cos Q
= .
W.D =
Note
i) Force F in the direction of vector
Force applied = F.
ii) Distance in the direction of vector
Displacement = d.
Individual
i) = 2i + + 2
= i + +
W.D =
04. TO PROVE COMPOUND ANGLE FORMULA OF COSINE
i.e cos (A + B) = cos A cos B – sin A sin B
– consider the vector diagram below.
Diagram
= (cos A) + (sin A)
= (cos B) – (sin B)
Hence
= cos (A + B)
but
. =
= cos A cos B + -sin A sin 0
= cos A cos B – sin A sin B
Also
=
=
=
= 1
=
=
=
= 1
Therefore
Cos A cos B – sin A sin B = (1)(1) cos (A + B)
Cos (A + B) = cos A cos B – sin A sin B
Proved
Pg. 2 drawing
= (cos A) + (sin A) j
= (cos B) I + (sin B)
Hence
. = cos (A – B)
But
. =
. = cos A cos B + sin A sin B
Also
=
=
=
= 1
=
=
=
= 1
There
Cos A cos B + sin A sin B = (1) (1) cos (A – B)
Cos (A – B) = cos A cos B + sin A sin B
Proved
06. TO PROVE THAT AN INSCRIBED ANGLE SUBTENDING A SEMI – CIRCLE IS A RIGHT ANGLE
– Consider the vector diagram below.
Pg. 2 drawing
To prove that
< SR = 900
. = 0
-b + + – = 0
– + =
= +
+ + -= 0
+ =
= –
Hence
. = ( + ) ( – )
. = 2 – (2
. = 2 – 2
. = 2 – 2
But
= = radius, r
. = 0
Proved
QUESTION
17. Find the projection of + 2 – 3 onto + 2 + 2
18. Find the vector projection of onto. If = 2 +2 + and = 3+ + 2
19. Find the work done of the force of (2i + 3i + k) n s pulling a load (3i + j k) m
20. Find the work done of the force of (2i + 3j + k) N is pulling a load a distance of 2m in the direction of 2m in the direction of
= 3 + 2 + 2
21. Find a vector which has magnitude of 14 in the direction of 2 + 3 +
CROSS (VECTOR) PRODUCT (X or )
x = sin Ø.
Where
– is the unit vector perpendicular to both vector and
= x
=
Hence
x = sin Ø.
1 =
= sin Ø
Therefore
= sin Ø
OR
= sin Ø
Where
Ø – is the angle between the vector and
Again
Suppose the vector
Hence
+ =
+ = – j +
Note
i) If you cross two vectors, the product is also the vector.
ii) Cross (vector) product uses the knowledge of determinant of 3 x 3 matrix.
iii) From the definition.
= sin
Individual
= (1, 0, 0)
j = (0, 1, 0)
k = (0, 0, 1)
= (1) (1) (0)
= 0
= 0
Hence
i x i = j x = x = 0
iv) From the definition
x = k
Generally
Consider the component vector
For anticlockwise (+ve)
Pg. 4 drawing
i) x =
ii) x =
iii) x =
For clockwise (-ve)
i) x =
ii) x = –
iii) x = –
THEOREM
From the definition
–
= 0
Is the condition for collinear (parallel) vectors
a) ≠
= –
Questions
22. If = 2 + 6 + 3 and = + 2 + 2. Find the angle between and
23. Determine a unit vector perpendicular to = 2 – 6 – 3 and = 4 + 3 –
24. If = 2 + j + 2k and = 3 + 2 +
Find â‹€
Box product
-This involves both cross and dot product
Suppose. x then start with cross (x) followed by DOT (.)
. x =. x)
-This is sometimes called scalar triple product
Note
-If scalar triple product (box product) of three vectors. and = 0
-Then the vector , and are said to be COMPLANAR
Question
25. If = 2 + + 2
= 2 + and
= 3 + 2 + k
Find
a) . x c
b) x.
APPLICATION OF CROSS PRODUCT
USED TO PROVE SINE RULE
– Consider the diagram below
+ + – = 0
+ =
Cross by on both sides of
x + x = x
0 + x = x
x + x
Crossing by on both sides of eqn …. 1 above
+ =
x + + = x
x + = x
x = x c
– ( x = x
x = – x )
x = x
Equation i and ii as follows
x = x = x
Sin. = sin. = sin.
Sin = Sin = sin
Dividing the whole eqn by
= =
= =
Sine rule
USED TO PROVED COMPOUND ANGLE FORMULA OF SINE
Sin (A + B) = sin A cos B + sin B cos A
Consider the vector diagram below
Pg. drawing
= (cos A) + (sin A) + o
= (cos A) + (sin A) + o
Hence
= sin (A + B)
But
=
= i + k
= i (o) – j (o) + k (-cos A sin B – sin A cos B)
= -k [sin A cos B + cos A sin B]
= 2
= 2
= sin A cos B + cos A sin B
Also
=
= 2
2 = 2 B
= 1
Therefore
Sin A cos B + cos A sin B = (1) (1) sin (A+B)
Sin (A + B) = Sin A cos B + cos A sin B
Proved
USED TO DETERMINE/ TO PROVE COMPOUND AND FORMULA OF SINE
Sin (A – B) = sin A cos B – Cos A sin B
– Consider the vector diagram below
Pg.6 drawing
= (cos A) + (sin A) + o
= (cos B) + (sin B) + o
= = sin (A – B)
but
=
= i – j + k
= 2
= 2
= sin A cos B – cos A sin B
Also
= + 0
=
=
= 1
=
=
=
= 1
Therefore
Sin A cos B – cosA sin B = (1) (1) sin (A – B)
Sin (A – B) = sin A cos B – cos A sin B
Proved
USED TO FIND THE AREA OF THE TRIANGLE
– Consider the triangle ABC below
Pg. 7 drawing
Area (A) = ½ x base (b) x height (h)
A = ½ bh ……i
Also
Sin =
Sin =
h = Sin ….ii
And
b = …iii
Substitute …ii and …iii into 1 as follows
A = ½ sin
A = ½
USED TO FIND THE AREA OF PARALLELOGRAM
– Consider the parallelogram below
Pg. 7 drawing
Area = length (l) x height (h)
= A = Lh …i
L = …ii
Also
Sin Q =
h = Sin Q….ii
Substitute ii and iii into —1 as follows
Area (A) = sin Q
A =
Generally
Area (A) =
=
=
=
Where
= –
= –
= –
= –
USED TO FIND THE VOLUME OF PARALLELOPIPED
– Consider the diagram below
Pg. 7 drawing
– Suppose P Q R and S are the vertices of the parallelepiped, hence the volume (v) of the parallelepiped is given by:
Volume (v) = base area (A) x height (h)
V = x
V =
V =
V =
Again, for the sides with position vectors, and
Volume (v) = x
=
=
USED TO FIND THE VOLUME OF A TETRAHEDRON
– Consider the tetrahedron with vertices P, Q, R and S
Pg. 8 drawing
Volume (v) = 1/3 x base area x altitude
V = 1/3 x ½ x
V = 1/6 x
V =
Therefore
Volume = 1/6
USED TO FIND THE VECTOR PERPENDICULAR TO THE PLANE
– Consider the diagram below
Pg. 8 drawing
Where
= is the vector perpendicular or normal to the plane
= x
Question
26. Find the area of the triangle ABC whose vertices are A (2, 1, 1) B (3, 2, 1) and C (-2, -4, -1)
27. The position vector of the points A, B and C are (2, 4, 3), (6, 3, -4) and (7, 5, -5) respectively. Find the angle between and and hence the area of the triangle ABC
28. Find the area of the parallelogram whose vertices P, Q and R are (2, 1, 1), (3, 2, 1) (2, 4, 1)
29. Find the volume of the parallelepiped the edges are A (1, 0, 2) B (2, -1, 3) C (4, 1, 3) and D (1, -1, 1)
30. Find the volume of the tetrahedron whose sides are = 2 + , = – 3 ` + , and – + 2
COLLINEAR AND COPLANAR VECTORS
1. COLLINEAR VECTOR
These are vectors having the same slope (re direction).
Pg. 10 drawing
= μ
= t
Where
Æ›, μ and t are scalar
Again and to be collinear x = 0
2. COPLANAR VECTOR
These are vectors which lie on the same plane
Eg. Pg. 10 drawing
For the vectors, and to be coplanar
= 0
= 0
= 0
Generally
= = = 0
Question
31. Given that
= 3 + 4
32. Find the value are collinear vectors 2 – + , + 2+ 3 and 3 + + 5 are coplanar.
33. Find unit vector in the direction of = 6 + 3 + and state its length
LINEAR COMBINATION OF VECTORS
Suppose that are vectors and, β and γ are real numbers (sealers). Then a vector = + β + γ is a linear combination of vectors
NB
To solve vectors means to put the vectors into linear form
Question
34. If = + , = – and = 3 – 4 resolve into vectors parallel to
35. Express the vector r = 10 – 3- as a linear function of given that
= 2 – +
= 3 + 2 – and
= – + 3 – 2
Note: Required to be placed in a right position
Subtopic: Dot Product
Proving cosine rule using dot product
Consider the triangle ABC above
Subtopic: Cross Product
Proof of sine rule Consider the ∆ABC with sides A,B and C respectively
Construct a line AH which lies on BC and perpendicular.
Equating area
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