In order to solve problems involving vectors and their resultants and dot products, it is necessary to
A typical problem involving vectors which can be solved by adding and/or subtracting vectors, using a dot product, or applying the Law of Sines or Law of Cosines to give us information about vectors that form a nonright triangle. Usually you 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 .
Let's look at two introductory examples of this type of problem.
First we make a diagram in standard position. The attempted flight path of the
plane is
vector OX. Notice that the
angle shown is 50° because the 40° bearing is measured clockwise from North. The
vector XY represents the wind from the South. The
vector OY is the resultant path of the plane, affected by the wind. In the diagram shown below, the known magnitude and
angle are labeled. It is useful to put such a diagram in “standard position” with North being at the top of the vertical axis. We then measure a bearing clockwise from North to label the measurement of an angle.
We need to find
vector OY (ground speed) which is the
vector sum of
OX (air speed) and
XY (wind speed). One way to do this is to add the components of
OX and
XY and find the magnitude and direction of
OY, their resultant..We have the following :
The
chart given above could also have been written as:
We find the magnitude of OY, the plane’s ground speed, using the Pythagorean Theorem:
The
angle this
vector makes with the horizontal
axis in the diagram can be found as follows:
The
plane has been pushed (54.194°  50º) = 4.194º off its attempted path if no corrections are made for the wind. The plane's new bearing is
90° – 54.194° = 35.806°
The final bearing and ground speed of the plane are 35.806° at 439.479 mph.
We first make a
vector diagram in standard position.
The
angle made by
vector OY with the horizontal
axis is 50° because bearings are measured clockwise from North. The new bearing is the direction the
plane should take in order to allow the wind to push it upward to the desired 40° bearing, which is the
angle made by
vector OY with the vertical (North) axis.
We do not know the
angle formed by the
vector OX and the horizontal axis, labeled as (50º  θ) in our diagram.
In our diagram, we labeled
angle YOX as
angle θ. Using
triangle YOX, we can apply the Law of Sines as follows:
The
plane must therefore aim 4.6° East of the desired 40° bearing, or at a new bearing of
(40º + θ) = 44.6°
To find the magnitude of
vector OY, we must first know the measure of the
angle opposite
OY in the
triangle OXY. To do this, we will start with the fact that the sum of the angles in
triangle equals 180º. Note that
angle b = 40º due to
vertical angles being equal.
ÐOXY = 180°  b  θ = 180  40°  4.6° = 135.4°
We will now use the Law of Sines once again to find the magnitude of the
vector OY.
Remember that when working vector problems that involve currents the vector equation is:
air speed + wind speed = ground speed
OX + XY = OY
In order to reach its desired destination, our plane will fly with
an air speed of 400 mph on a bearing of 44.6º.
His resultant ground speed will be 436.9 mph at a bearing of 40º.
Now that we can determine the components of the vectors
OY and
OX we can use their dot
product to check our calculations and verify that the
angle YOX (θ in our diagrams) is 4.6º.
OY = (436.9cos(50°), 436.9sin(50°)) = (280.8, 334.7)
OX = (400cos(45.4°), 400sin(45.4°)) = (280.9, 284.8)
Recall that in general, if
v = (v
_{1}, v
_{2}) and
u = (u
_{1}, u
_{2}), are two vectors, their
dot product can be expressed as either


where
v and
u represent the
magnitudes of vectors
v and
u.
Applying the second formula to our situation results in the formula
Since this is a cumbersome calculation, you should use a calculator. Entering this into a calculator is shown in the screen below.
The result is cos(θ)
0.9967788785. This gives θ
4.6°.
In some situations, the Laws of Sines and Cosines work very well. In other situations, it might be necessary to first calculate the
vector dot product.