Thursday, May 8, 2014

Week 16, Day 3

Today we reviewed the topic of vectors, as explained in Darron's presentation.

Physical forces and velocities are not confined to the plane, therefore it is natural to extend the concept of vectors from two-dimensional space to three-dimensional space. In space, vectors are denoted by ordered triples v = <v1, v2, v3>, otherwise known as component form. The zero vector is denoted by = < 0, 0, 0 >. Using the unit vectors = < 1, 0, 0 > in the direction of the positive z-axis, the standard unit vector notation for v is = v1+ v2+ v3k, otherwise known as unit vector form. If is represented by the directed line segment from P(p1, p2, p3) to Q(q1, q2, q3), the component form of is produced by subtracting the coordinates of the initial point from the coordinates of the terminal point, ='<v1, v2, v3> = <q1-p1, q2-p2, q3-p3>. 

Vectors in Space:
1. Two vectors are equal if and only if their corresponding components are equal. 
2. The length of = <u1, u2, u3> is ||u|| = (u1^2 + u2^2 + u3^2)^1/2 = vector length. 
3. A unit vector in the direction of is given by = v / ||v|| = unit vector.
4. The sum of = <u1, u2, u3> and = <v1, v2, v3> is = <u1 + v1, u2 + v2, u3 + ||v3>.
5. The scalar multiple of the real number c and = <u1, u2, u3> is c= <cu1, cu2, cu3>.
6. The dot product of = <u1, u2, u3> and = <v1, v2, v3> is u x v = u1v1 + u2v2 + u3v3.

Example 1: Finding the Component Form of a Vector
Find the component form and length of the vector having initial point (3,4,2) amd terminal point (3,6,4). Then find a unit vector in the direction of v
The component form of is:
= <3-3, 6-4, 4-2> = <0,2,2>
This implies that the vector's length is 
||v|| = (0^2 + 2^2 + 2^2)^1/2 = (8)^1/2 = 2(2)^1/2.
The unit vector in the direction of is:
/ ||v|| = 1 / 2(2)^1/2 <0,2,2> = <0, 1/(2)^1/2, 1/(2)^1/2>.

Example 2: Finding the Dot Product of Two Vectors
Find the dot product of <0,3,-2> and <4,-2,3>. 
<0,3,-2> x <4,-2,3> = 0(4) + 3(-2) + (-2)(3) = 0-6-6 = -12.

The angle between two nonzero vectors is the angle theta, 0 <(=) theta <(=) pi, between its respective standard position vectors. Thus, if theta is the angle between two nonzero vectors and v, then cos(theta) =u x v / ||u|| ||v||. If the dot product of two nonzero vectors is zero, the angle between the vectors is 90 degrees. Such vectors are called orthogonal. For instance, the standard unit vectors i,j, and k are orthogonal to each other. 

Example 3: Finding the Angle Between Two Vectors
Find the angle between = <1,0,2> and = <3,1,0>.
cos(theta) = u x v / ||u|| ||v|| = <1,0,2> x <3,1,0> / ||<1,0,2>|| ||<3,1,0>|| = 3 / (50)^1/2
This implies that the angle between the two vectors is 
theta = arccos 3 / (50)^1/2 = 64.9 degrees. 


Parallel Vectors:
In general two nonzero vectors and are parallel if there is some scalar c such that = cv
For example, the vectors shown below u, v, and w are parallel because = 2and = -v


Example 4: Parallel Vectors
Vector has initial point (1,-2,0) and terminal point (3,2,1). 
Which of the following vectors is parallel to w? 
a) u = <4,8,2> 
b) v = <4,8,4> 
Begin by writing in component form.
w = <3,-1, 2 -(-2), 1-0> = <2,4,1>
a) The vector is parallel to w because
= <4,8,2> = 2<2,4,1> = 2w
b) In this case, you need to find a scalar c such that <4,8,4> = c<2,4,1>.
However, equating corresponding components produces c = 2 for the first two components and c = 4 for the third. Hence, the equation has no solution, and the vectors are not parallel. 

1 comment: