me 2304: 3d geometry & vector calculus dr. faraz junejo
TRANSCRIPT
ME 2304: 3D Geometry & Vector Calculus
Dr. Faraz Junejo
Multiplication of a Vector
By a scalar b = m a
a
Magnitude m a
Direction along a
b
We can increase or decrease the magnitude of a vector by multiplying the vector by a scalar
Scalar Multiplication
In the examples, vector B (2 units) is half the size of vector A (which is 4 units) . We can write:
B = 0.5 A This is an example of a scalar multiple. We have
multiplied the vector A by the scalar 0.5.
ExerciseGiven a = (15,-6,24) , b = (5,-2, 8) and c = (-15/2, 3, -12).a. Show that a & b have the same direction.b. Show that a & c have opposite direction
By inspecting a & b, it can be seen that: a = 3b or b = 1/3 a
Since the scalar 3 or 1/3 is positive, this implies a & b have the same direction.
Whereas, by inspecting a & c, it can be seen that: a = -2c or c = -1/2 a
Since the scalar -2 or -1/2 is negative, this implies a & c have opposite directions.
Multiplication of Vectors(contd)By a vector
two types
Dot (Scalar) Product
Cross (Vector) Product
Scalar Product
a
b
multiplication SCALAR
a . b
DOT PRODUCT
Dot Product (contd)
a . b
a
b
= a b cos 360 -
In words, a.b is the length of a times the length of b times the cosine of the angle between a and b
Dot Product (contd)Magnitude of either one ofthe vectors
Component of the other along the direction of the first
X
a
b
bcos a
b acos
Dot Product: SummaryIf we have any two vectors, say a and b, the dot product of a and b is given by:
cosbaba
Where, |a| and |b| are the magnitudes of a and b respectively, and θ is the angle between the 2 vectors.
The dot product of the vectors a and b is also known as the scalar product since it always returns a scalar value.
W = F . d
Work done on a particle by a
constant force F during a
displacement d is
Examples in Physics
Exercise
A small cart weighing 100 lb is pushed up an incline that makes an angle of 30 Degrees with the horizontal. Find the work done against gravity in pushing the cart a distance of 80ft.
80 ft
Θ=30
R (a1,a2)
T xP
Q (0,-100)
y
The vector PQ represents the force of gravity acting vertically downward with a magnitude of 100 lb.
The corresponding vector PQ is (0i-100j)
From triangle PTR, we conclude that 403080a , 34030cos80 21 Sina
jiThus 40340PR , Therefore, the work done against gravity is PQ.PR =
4000 ft-lb
Find the dot product of force vectors F1= 4N and F2= 6N acting at 40 Degrees to each other ?
F1=4N
F2=6N
θ
cos2121 FFFF
40)6)(4(21 CosFF
NFF 38.1821
Scalar Product: Applications
To test whether two vectors are perpendicular.
If the dot product of two non zero vectors is zero, the vectors are perpendicular.
0143
)k1j2i()k1j2i(3
ˆˆˆˆˆˆ
Finding the angle between two vectors.
ba
ba
cos
Scalar Product: Applications
cosbaba
ba
ba
1cos
kj2i2a ˆˆˆ
k2j3i6b ˆˆˆ
79)21
4(cos
7.3
4cos 1
What is the angle between a and b ?
39122 222 a
749236 222 b
42612 ba
Find the dot product of vectors P and Q, if they are acting at right angle to each other and
Exercises
,5,7 QP
Answer: 0
Find P.Q , if P= 6i+5j and Q= 2i-8jAnswer: -28
Find the angle between P=3i-5j and Q=4i+6j. Draw a rough sketch to verify your answer.
Answer: 115.3 Degrees
If, the work done by F in acting from P to q is
60 and ,3,40 mDNF
Answer: 60 Joules
b
Scalar Product: Applications
Q
P
XY
a
Finding the projection of one vector on another vector.
Projection of a on b is XY The projection of a onto b will be
given by:
b
b cosaaprojb
In summary, the proja b has length
cosa , and directionb
b
It is called the scalar component of a in the direction of b
Vector Projection (contd)
baba cos
Therefore, scalar component of a in the direction of b
b
baa
cos
kj2i2a ˆˆˆ
k2j3i6b ˆˆˆ
Vector Projection (contd)What is the scalar component of a in the direction of b ?
Scalar component of a on b is
7
4
b
ba
Since, projection of a onto b is:
b
b cosaaprojb
we know,
b
baa
cos
Therefore,
bb
ba
b
b
b
baaprojb
2
Similarly, projection of b onto a is given by:
aa
abbproja
2
b onto a of Projection
• If b is a vector not equal to zero, then any vector “a” can be projected onto “b” (as mentioned earlier) as well as onto a vector
• It can be seen from figure shown below that using vector addition it can be written:
b magnitude of b
a
b
b
aojbPr
aojb
Pr
aProjaaProj
aProjaProja
bb
bb
Find the vector projection of b = (6i+3j+2k) onto a=i-2j-2k and the scalar component of b in the direction of a
)22()2()2()1(
4662
2222 kjia
a
abbproja
kjikjibproja 9
8
9
8
9
4)22(
9
4
Now the scalar component of b in the direction of a will be given by:
3
4
3
466cos
a
abb
Cross Product
a
b a b
CROSS PRODUCT
multiplication VECTOR
Vector Product (contd)
a
b
Magnitude : Area of the parallelogram generated by a and b.
Vector Product (contd)
a
b
Magnitude : sinabah
sinbh
Vector Product (contd)
a
b
ba
sinbh
Direction : Perpendicular to both a and b.
Right-Hand Rule
The order of vector multiplicationis important.
Geometrical Interpretation
a
b
b sin
sin|ba| ab
A = a b
Area of the parallelogram formed by a being the base and height being b sin
Cross Product: SummarySuppose we have 2 vectors a and b. These 2 vectors lie on a plane and the unit vector n is normal (at right angles) to that plane
The cross product (also known as the vector product) of a and b is given by:
sinbaba
The right hand side represents a vector at right angles to the plane containing vectors A and B.
Cross Product: Properties
00sin|ba| ab
and a and b are not null vectors, then a is parallel to b.
If 0ba
Cross Product: Determinant
321
321
kji
bbb
aaaba
• The cross product can be expressed as
• Expanding the determinants gives
ˆ ˆ ˆ
ˆ ˆ ˆy z x yx zx y z
y z x yx zx y z
A A A AA AA A A
B B B BB BB B B
i j k
A B i j k
ˆ ˆ ˆy z z y x z z x x y y xA B A B A B A B A B A B A B i j k
Vector Products Using Determinants
ExampleA simple cross product
132
215
ˆˆˆ
kji
k
j
i
ˆ)]1.2(3.5[
ˆ)]1.5(2.2[
ˆ]2.3)1.1[(
-5
)ˆˆ3ˆ2()ˆ2ˆˆ5( kjikji
9
17
kji ˆ17ˆ9ˆ5
CHECK
)( ba
is perpendicular to ba
&
Example (Contd)
)ˆˆ3ˆ2()ˆ2ˆˆ5( kjikji
017.29.15.5
)ˆ17ˆ9ˆ5).(ˆ2ˆˆ5(
kjikji
ExampleFind A x B and B x A, if
134
112
kji
-2i-6j+10K
)34(B and ),22( kjikjiA
B x A = 2i+6j-10k, which implies B x A = - (A x B)
Finding a unit vector perpendicular to a plane.
Find a unit vector perpendicular to the plane containing two vectors b&a
Applications
A vector perpendicular to a and b is
Corresponding unit vector
ba
|ba|
ba
kjiba ˆ17ˆ9ˆ5
Cross product
Magnitude 3952898125|| ba
Unit vector is )ˆ17ˆ9ˆ5(395
1kji
ExampleDetermine a unit vector perpendicular
to the plane of
and
)ˆ2ˆˆ5( kjia
)ˆˆ3ˆ2( kjib
Find the area of the triangle with vertices A(1,1,2) B(-1,3,2) and C(4,1,5).Find two sides of the triangle, the vectors AB and AC.
and
Now find the cross product of the two vectors:
The area is given as follows:
Find the area of the triangle with vertices A(1,1,1) B(2,3,4) and C(3,0,-1).
sq unitsArea
kjiPPPP
102
3
2
90
583221
Scalar Triple Product: Example
not.or plane same in the lie
(0,-9,18)c and (2,1,-4)b (1,4,7);a vectors threeif Determine
In here, scalar triple product can be utilized to determine whether the given points lie in same plane or not
plane. same in the lie they implying ,
0
81 9 0
4 1 2
7 4 1
)(
321
321
321
Hence
ccc
bbb
aaa
cba
The 3-dimensional Co-ordinate System
The x-y plane is horizontal in diagram and shaded green. It can also be described using the equation z = 0, since all points on that plane will have 0 for their z-value.
Normally the 'right-hand orientation’ is used for the 3 axes, with the positive x-axis pointing in the direction of the first finger of our right hand, the positive y-axis pointing in the direction of our second finger and the positive z-axis pointing up in the direction of our thumb.
Example The vector OP has initial point at the origin O (0, 0, 0) and terminal point at P (2, 3, 5). We can draw the vector OP as follows:
For the vector OP above, the magnitude of the vector is given by:
| OP | = √(22 + 32 + 52) = 6.16 units
Distance Between 2 Points in 3 Dimensions
If we have point A (x1, y1, z1) and another point B (x2, y2, z2) then the distance AB between them is given by the formula:
Distance AB =sqrt[(x2-x1)2+(y2-y1)2+(z2-z1)2]
Find the distance between the points P (2, 3, 5) and Q (4, -2, 3).
Answer: 5.74 Units
ExampleFind the lengths of the sides of triangle with vertices (0, 0, 0), (5, 4, 1) and (4, -2, 3). Then determine if the triangle is a right triangle, an isosceles triangle or neither.
Solution: First find the length of each side of the triangle by finding thedistance between each pair of vertices.
(0, 0, 0) and (5, 4, 1)
42
11625
010405 222
d
d
d
(0, 0, 0) and (4, -2, 3)
29
9416
030204 222
d
d
d
(5, 4, 1) and (4, -2, 3)
41
4361
134254 222
d
d
d
These are the lengths of the sides of the triangle. Since none of them are equal we know that it is not an isosceles triangle and since Implying it is not a right triangle. Thus it is neither.
222412942
Equation of a Sphere
If we square both sides of this equation we get:
The standard equation of a sphere is
where r is the radius and is the center.
2222r ooo zzyyxx
ooo zyx ,,
A sphere is the collection of all points equal distance from a center point.
To come up with the equation of a sphere, keep in mind that the distance
from any point (x, y, z) on the sphere to the center of the sphere,
is the constant r which is the radius of the sphere.
Using the two points (x, y, z), and r, the radius in the distance
formula, we get:
ooo zyx ,,
222r ooo zzyyxx
ooo zyx ,,
ExercisesFind the equation of the sphere with radius, r = 5 and center, (2, -3, 1).
By Just plugging above values into the standard equation of a sphere we will get:
25)1()3()2( 222 zyx
Find the equation of the sphere with endpoints of a diameter (4, 3, 1) and (-2, 5, 7).
Using the midpoint formula we can find the center and using the distance formula we can find the radius.
4,4,1
271
,2
53,
224
Center
19
919
414314Radius 222
Thus the equation is: 19)4()4()1( 222 zyx
Exercises (Contd)
Find the center and radius of the sphere, . 07864222 zyxzyx
To find the center and the radius we simply need to write the equation of the sphere in standard form, . Then we can easily identify the center, and the radius, r. To do this we will need to complete the square on each variable.
ooo zyx ,,
2222r ooo zzyyxx
36432
169471689644
7864
07864
222
222
222
222
zyx
zzyyxx
zzyyxx
zyxzyx
Thus the center is (2, -3, -4) and the radius is 6.
Adding 3-dimensional Vectors Earlier we saw how to add 2-dimensional vectors. We
now extend the idea for 3-dimensional vectors.
We simply add the i components together, then the j components and finally, the k components.
If A = 2i + 5j − 4k and B = −2i − 3j − 5k, then
determine A+B.
Answer: 2 j − 9 k
Angle Between 3-Dimensional Vectors
Earlier, we saw how to find the angle between 2-dimensional vectors. We use the same formula for 3-dimensional vectors, i.e.
ba
ba
1cos
Find the angle between the vectors a = 4i + 0j + 7k and b = -2i + j + 3k.
a• b = (4 i + 0 j + 7 j) • (-2 i + j + 3 k )
a• b = (4 × -2) + (0 × 1) + (7 × 3)
a• b = 13
Angle Between 3-Dimensional Vectors (contd)
166.3031(-2) 704|| 222222 ba
And now for denominator i.e. magnitudes
So, θ = arccos(13 ÷ 30.166)
Therefore the angle between the vectors a and b is θ = 64.47°
Find the angle between the vectors a = 3i + 4j − 7k and b = -2i + j + 3k.
Answer: θ = 135.6°
ExampleWe have a cube ABCO PQRS which has a string along the cube's diagonal B to S and another along the other diagonal C to P. What is the angle between the 2 strings?
θ
Example (contd) We will assume that we have a unit cube i.e. each side has length 1 unit, with 0 being the origin.
The unit vectors i, j, and k act in the x-, y-, and z-directions respectively. So in our diagram, since we have a unit cube
OA = I, OC = j and OS = k
From the diagram, we see that to move from B to S, we need to go -1 unit in the x direction, -1 unit in the y-direction and up 1 unit in the z-direction. Since we have a unit cube, we can write:
BS = −i − j + k
Similarly, to move from C to PCP = i − j + k
Example (contd)The dot (scalar) product for the vectors BS and CP is:BS • CP = |BS| |CP| cos θ
BS • CP = (−i − j + k) • (i − j + k) = 1
|BS| |CP| = 3
So , θ = arccos (1/3) θ = 70.5°
Therefore, the angle between the strings is 70.5°