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Do Now: p.528, #27Find the measures of the angles of the triangle whose vertices areA = (–1, 0), B = (2, 1), and C = (1, –2).
3,1AB ��������������
1, 3BC ��������������
2, 2AC ��������������
3, 1BA ��������������
1,3CB ��������������
2,2CA ��������������
10AB BA ����������������������������
10BC CB ����������������������������
2 2AC CA ����������������������������
Component forms:
Magnitudes:
Do Now: p.528, #27Find the measures of the angles of the triangle whose vertices areA = (–1, 0), B = (2, 1), and C = (1, –2).
1cosAB AC
AAB AC
����������������������������
����������������������������
Angle at A:
1 3 2 1 2cos
10 2 2
1 1cos
5
63.435
Do Now: p.528, #27Find the measures of the angles of the triangle whose vertices areA = (–1, 0), B = (2, 1), and C = (1, –2).
1cosBC BA
BBC BA
����������������������������
����������������������������
Angle at B:
1 1 3 3 1cos
10 10
1 3cos5
53.130
Do Now: p.528, #27Find the measures of the angles of the triangle whose vertices areA = (–1, 0), B = (2, 1), and C = (1, –2).
1cosCB CA
CCB CA
����������������������������
����������������������������
Angle at C:
1 1 2 3 2cos
10 2 2
1 1cos
5
63.435
Vector ApplicationsSection 10.2b
Suppose the motion of a particle in a plane is represented byparametric equations. The tangent line, suitably directed,models the direction of the motion at the point of tangency.
x
yA vector is tangent or normal to a curve at a point P if it is parallel or normal, respectively, to the line that is tangent to the curve at P.
Finding Vectors Tangent and Normal to a Curve
Find unit vectors tangent and normal to the parametrized curve
12
tx 1y t 0t at the point where .4t
Coordinates of the point: 4, 1, 4 1
2x y
3,3
A graph of what we seek:
x
y
u–u
n
–n
Finding Vectors Tangent and Normal to a Curve
Find unit vectors tangent and normal to the parametrized curve
12
tx 1y t 0t at the point where .4t
Tangent slope:dy dy dt
dx dx dt
1 2
4
1 2
1 2t
t
1
2
A basic vector with a slope of 1/2:v 2,1To find the unit vector, divide v by its magnitude:
2 2
2,1u
2 1
2,1
5
2 1,5 5
Finding Vectors Tangent and Normal to a Curve
Find unit vectors tangent and normal to the parametrized curve
12
tx 1y t 0t at the point where .4t
The other unit vector:2 1
u ,5 5
To find the normal vectors (with opposite reciprocal slopes),interchange components and change one of the signs:
1 2n ,
5 5 1 2
n ,5 5
Finding Ground Speed and DirectionAn airplane, flying due east at 500 mph in still air, encounters a70-mph tail wind acting in the direction north of east. Theairplane holds its compass heading due east but, because of thewind, acquires a new ground speed and direction. What are they?
a 500Let a = airplane velocity and w = wind velocity.
60
w 70
E
N
a
wa + w70
50060
We need the magnitude of the resultant a + w and the measure of angle theta.
Finding Ground Speed and DirectionAn airplane, flying due east at 500 mph in still air, encounters a70-mph tail wind acting in the direction north of east. Theairplane holds its compass heading due east but, because of thewind, acquires a new ground speed and direction. What are they?
u 500,0Component forms of the vectors:
60
w 70cos60 ,70sin 60 35,35 3
The resultant:a w 535,35 3
Magnitude: 22a w 535 35 3 538.424
Finding Ground Speed and DirectionAn airplane, flying due east at 500 mph in still air, encounters a70-mph tail wind acting in the direction north of east. Theairplane holds its compass heading due east but, because of thewind, acquires a new ground speed and direction. What are they?
1 35 3tan535
Direction angle:
60
6.465 E
N
a
wa + w70
50060
The new ground speed of the airplane is approximately538.424 mph, and its new direction is about 6.465degrees north of east.
