Hedwig and Errol
The vector addition and subtraction we have done so far has been quite easy, since we could do
them component-wise. That is, we just added or subtracted each component separately. But if
vectors are given in magnitude-direction form, this process becomes somewhat more difficult.
We must first convert into component form so we can use the easy component-wise methods.
1.
Starting from Hogwarts, Ron’s clumsy owl Errol flies 5 miles at a 45° angle (remember,
these are standard position angles, so they are measured counterclockwise from East).
Then he realizes he’s flying the wrong way, so he adjusts his path and flies 3 miles at a
330° angle (again, counterclockwise from East). On a separate piece of paper, carefully
draw Errol’s flight, letting 1 cm = 1 mile. Use a protractor to make your angles accurate.
Errol’s flight is shown in blue.
2.
Hedwig has a better sense of direction, so she flies from Hogwarts directly to Errol’s
ending point. On the same paper, draw Hedwig’s path. Use a ruler and protractor to
estimate its magnitude and direction.
Hedwig’s flight is shown in red. Estimates will vary, but should be close to 6.5 miles at
an angle of 18°.
Hedwig’s flight vector is the sum of the vectors representing the two parts of Errol’s flight. You
have just used the tail-to-end method for adding two vectors, by placing the tail of Errol’s
second part at the end of his first part. This method obviously makes sense for a two-part flight,
but we will see vectors in a variety of contexts.
3.
Let’s go back to Hedwig’s flight. We could only estimate Hedwig’s path using the
tail-to-head method graphically. Let’s use trigonometry to get more accurate results.
Complete the table below to convert both parts of Errol’s flight into component form:
Magnitude-Direction Form
(owl form)
magnitude
direction
Part 1
5 miles
45°
Part 2
3 miles
330°
Component Form
(Harry’s Form)
horizontal
vertical
5cos45°=
5sin45°=
3.54 mi E
3.54 mi N
3cos330°=
3sin330°=
2.60 mi E
1.50 mi S
Show your work here:
Students who recognize that the law of cosines works well here should be reminded to
be careful about finding the angles they use. In fact, this is a great opportunity to
remind students about the geometry in finding angles and to check their work using
the law of cosines.
4.
How far East or West must Hedwig fly to get to Errol’s location from Hogwarts? How
far North or South?
3.54 + 2.60 = 6.14 mi East
3.54 + (-1.50) = 2.04 mi North
5.
Find the magnitude and direction of Hedwig’s path. Compare to your estimates from #2.
Comment:
As before, since this vector points northeast, it is easier to determine the direction.
Some problems in later tasks require an extra step.
Solution:
magnitude: (6.142 + 2.042) = 6.47 miles
direction: arctan(2.04/6.14) = 18.38°
6.
How far did Errol fly altogether? Did Hedwig fly the same distance? Why does this
make sense?
Errol: 5 + 3 = 8 miles
Hedwig flew only 6.47 miles (see #5), so he flew less distance than Errol. This makes
sense because Errol did not take a direct path to the final destination, whereas Hedwig
did.
Geometrically, this is the triangle inequality: Errol’s two distances must add up to
more than Hedwig’s distance.
7.
There is one particular situation where the sum of the magnitudes of two (or more)
vectors equals the magnitude of their sum. Describe this situation by drawing a picture
and explaining why the sum of the magnitudes equals the magnitude of the sum.
If the vectors point in the same direction, then there is no difference between the two
paths—they are both direct paths; one just has a “break” in it.
Geometrically, this is the degenerate form of the triangle inequality. Students might
have also see this as the segment addition property in geometry.
8.
Is the direction (angle) of Hedwig’s path equal to the sum of the angles of Errol’s paths?
Comment:
The previous series of questions (#6-8) are intended to help students see the advantages
of vectors given in component form, and that there are no similar shortcuts for
problems presented in magnitude-direction form, except in very specific cases.
No. Errol’s directions add up to 45° + 330° = 375° (or 15°), which is definitely not the
same as Hedwig’s direction of 18.38°.
Let’s see how we can modify the tail-to-end method of addition to apply to vector subtraction.
9.
Use component-wise subtraction to find 3, -7 – 4, -5.
3 – 4, -7 – (-5) = -1, -2
10.
Draw the vectors 3, -7 and 4, -5 tail-to-end. Describe a modified tail-to-end method
that will make the resultant vector equal to your answer from #10.
We must “flip” the second vector (rotate it 180°) so we move in the opposite direction.
u-v
u
-v
v
11.
How is subtracting vectors similar to the common “add the opposite” method of
subtracting integers?
We keep the first vector, then we change the subtraction problem into an addition
problem by changing the direction of the second vector.
In other words, u – v = u + (-v)
12.
What is the advantage of presenting this subtraction problem in component form rather
than direction-magnitude form?
We were able to do the subtraction component-wise without needing to convert to
component form first.
Now let’s look at scalar multiplication in each form.
13.
From Hogwarts, Harry walks two blocks East and three blocks North then rests. He
repeats this five times total. Describe how Ron can walk to Harry’s ending position from
Hogwarts without resting.
10 blocks East and 15 blocks North.
14.
From Hogwarts, Hedwig flies four miles at a direction of 70° then rests. He repeats this
five times total. Describe how Errol can fly to Hedwig’s ending position from Hogwarts.
fly 20 miles at a direction of 70°
15.
We have seen that vector addition and subtraction is easier when the problem is presented
in component form. Is the same true for scalar multiplication? Explain your answer.
No, scalar multiplication is quite easy in both forms.
Summarize:
In each box, explain how to perform the indicated operation depending on the form in which the
problem is presented. Two boxes have been completed for you.
given in
component
form
given in
directionmagnitude
form
add
u+v
subtract
u–v
multiply
au
(a is a real number)
add corresponding
components
subtract corresponding
components
multiply each
component by the
scalar
convert to component
form using trig
convert to component
form using trig
multiply magnitude by
the absolute value of
the scalar
add corresponding
components
subtract corresponding
components
(convert back if
required)
(convert back if
required)
u
v
v
u
geometric
representation
keep the same
direction, or add 180°
if the scalar was
negative (flips
direction)
u
½u
-v
2u
u–v
u+v
v
-u
u
stretch if |a| > 1
shrink if |a| < 1
reverse direction if a <
0
Practice:
Let p = 3, 5; q = -1, 6; r = 4, -3; s = -2, -6. Find each of the following:
1.
p+q
3 + -1, 5 + 6 = 2, 11
2.
p + 5r + 3s
3 + 5(4) + 3(-2), 5 + 5(-3) + 3(-6) = 17, -28
3.
s – 2(p + r)
-2 – 2(3 + 4), -6 – 2(5 + -3) = -16, -10
4.
Find t so that p + t = s
3 + x, 5 + y = -2, -6  -5, -11
OR s – p = -2 – 3, -6 – 5 = -5, -11
5.
Find u so that 3u – 4(p + q) = r
3x – 4(3 + -1), 3y – 4(5 + 6) = 4, -3  4, 41/3
Let a, b, c, and d be the vectors shown below. Sketch each of the following:
b
a
c
d
Comment:
These vectors are intentionally not presented on a grid. If they were, students
could estimate their components and perform operations component-wise. Presenting
the vectors without a grid forces students to interpret the operations geometrically.
1.
b+c
b
c
2.
-¼ a
a
3.
2d – b
d
d
b
4.
Find e so c + e =
option 1: subtraction: e =
–c
c
option 2: c and resultant start at same point
5.
Find f so that 2f + a = 0.
f = ½(-a) = -½a
a