In order to solve problems involving angles of
elevation and depression, it is necessary to
A typical problem involving vectors and right triangles gives us information about a
vector and/or its components. We are asked to find information about unmeasured components and angles. Note that we will represent vectors either by using bold type
v or by using the
vector notation .
Suppose a
plane is taking off at a
rate of 200 mph with an
angle of 35° to the ground. What is the
rate at which it is receding from the ground? What is its ground speed?
First we make a diagram. In this case, the plane’s path makes a 35°
angle with the ground. We call this
vector p.
The
hypotenuse of the
triangle is given a
length of 200 mph which is the magnitude of the
vector p describing the plane’s path. The
vector representing climbing vertically away from the ground is
c and the
vector representing the ground
speed is
g. The
right angle is formed between vectors
c and
g which are components of
p.
An easy way to identify which vectors are the components and which vector is the resultant (or sum of those two components) is to trace their "head-to-tail" relationship.
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When adding vectors, we start by placing the head of the first vector (g) at the original of a co-ordinate system. |
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At its tail we place the origin of a second co-ordinate system and the head of the second vector (c). |
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The resultant (p) is the vector that begins at the head of vector 1 (g) and ends at the tail of vector 2 (c) as shown in the diagram to the right. |
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In the diagram shown below the known magnitude (
||p|| = 200 mph) and
angle (
q = 35º) are labeled as well as the unknown vectors corresponding to the situation: the magnitude of the climbing
speed of the
plane by
c, and the magnitude of the ground
speed of the
plane by
g.
Thus the climbing and ground speeds are approximately 114.7 mph and 163.8 mph respectively and are the vertical and horizontal components of the resultant
vector p.