Math
Exercise 10
Choose the best answer.
Question 1:
An aviator is steering his light plane due north at an air speed of 120 miles per hour with a wind blowing towards the east at 50 mph. Find the direction and magnitude of the resultant velocity of the airplane, relative to the ground. Let north be 0o, east be 90o, south be 180o and west be 270o.
(a) |
130 mph at 23o | (b) |
130 mph at 67o |
(c) |
110 mph at 23o | (d) |
130 mph at 337o |
(a) |
130 mph at 23o |
|
Draw these situations for yourself! It makes understanding what the question is about much easier. (0,120) + (50, 0) = (50,120) magnitude = √[(50)2 + (120)2] = 130 tan x = 120/50, so a = 67o, so the plane flies at 23o. |
Question 2:
Find the vector sum: (1, 3, 4) + (2, 1, 1). The third coordinate in the result is:
(a) |
3 | (b) |
4 |
(c) |
5 | (d) |
9 |
(a) |
5 |
|
< 1+2, 3+1, 4+1> = <3,4,5> |
Question 3:
Find the difference: (2, -4, 5) – (-4, -7, 9). The second coordinate in the result is:
(a) |
2 | (b) |
3 |
(c) |
4 | (d) |
6 |
(b) |
3 |
|
<2 + 4, -4 + 7, 5 – 9> = < 6, 3, -4> |
Question 4:
Assume O is the origin of a 3 dimensional coordinate system. Let C be the point (2, -3, 1) and R be the point (4, -6, 2). Find 
.
(a) |
(2, -3, 1) | (b) |
(-2, 3, -1) |
(c) |
(6, -9, 3) |
|
(a) |
(2, -3, 1) |
|
Find the difference. |
Question 5:
If K is the point (1, 2, 3) and L is the point (-2, 3, 1):
i) Find the vector 
(a) |
(3, -1, 2) | (b) |
(-3, 1, -2) |
(c) |
(-1, 5, 4) | (d) |
(-3, -1, 2) |
(b) |
(-3, 1, -2) |
|
Find the difference. |
ii) What is the magnitude of the vector ? Choose the correct answer.
(a) |
2.3 | (b) |
3.7 |
(c) |
5.4 | (d) |
10.3 |
(b) |
3.7 |
|
 |
Question 6:
Find the difference: (2i – 3j – 4k) – (2i + 5j + k) The first coordinate in the result is:
(a) |
0 | (b) |
2 |
(c) |
4 | (d) |
8 |
(a) |
0 |
|
The result is: (0, -8j, -5k) |
Question 7:
In order for the vectors A = (-1, 3, 5) and B = (2, -1, k) to be perpendicular k must be equal to:
(a) |
-5 | (b) |
0 |
(c) |
1 | (d) |
5 |
(c) |
1 |
|
A∙B = 0, so -2 -3 + 5k = 0, so k =1 |
Question 8:
Let O be the origin of a 3 dimensional coordinate system. Let C be the point (2, -3, 1) and R be the point (-9, 4, -1):
i) Find 
. The second coordinate of the result is:
(a) |
-7 | (b) |
-1 |
(c) |
1 | (d) |
7 |
(c) |
1 |
|
OC + OR = (-7,1,0) |
ii) What is the magnitude of 
?
(a) |
0 | (b) |
14 |
(c) |
3.7 | (d) |
(4, 9, 1) |
(c) |
3.7 |
|
 |
iii) Find
The second coordinate of the result is:
(a) |
-7 | (b) |
-1 |
(c) |
1 | (d) |
7 |
(d) |
7 |
|
Take the difference: (-11, +7, -2) |
iv) Find 
. The second coordinate of the vector is:
(a) |
-7 | (b) |
-1 |
(c) |
1 | (d) |
7 |
(d) |
7 |
|
Is the same vector! (Make a sketch if you don’t see it right away.) |
Question 9:
If the points A = (3, 0, 0), B = (3, 4, 0) and P = (3, 4, 5) (O is at the origin) find the vector 
. The option closest to the second coordinate is:
(a) |
0 | (b) |
3 |
(c) |
4 | (d) |
5 |
(c) |
4 |
|
Vector OA = <3,0,0> The resultant of the three is therefore <3,4,5> |
Question 10:
A plane is flying NW at a speed of 110 miles per hour:
i) Write this vector in the form <x, y> where the positive y-axis points north and the positive x-axis points east. Round the values off to the nearest integer.
(a) |
(55, 55) | (b) |
(78, -78) |
(c) |
(-78, 78) | (d) |
(-78, -78) |
(c) |
(-78, 78) |
|
Draw the picture! Northwest = 45o, so sin 45o = x/110, so x = 78 |
ii) If the wind is blowing from the south at 65 mph, what vector describes the motion of the plane relative to the ground?
(a) |
(-13, 143) | (b) |
(-78, 143) |
(c) |
(-13, 78) | (d) |
(-78, 13) |
(b) |
(-78, 143) |
|
Since the wind vector = (0,65), the resultant = <78,78> + <0,65> = < -78, 143> |
Question 11:
A balloon is initially sighted at a position given by the vector (5, 4, 2) in a 3 dimensional space where the positive x-axis points true north and the positivey-axis points true west. After one minute the balloon had drifted to a position given by (9, 2, 7):
i) What is the magnitude of the horizontal drift? Choose the option closest to the correct answer.
(a) |
4 | (b) |
(4, -2, 5) |
(c) |
7 | (d) |
3 |
(a) |
4 |
|
Movement: <5,4,2> - <9,2,7> = < -4,2,-5>, where the first two digits refer to the horizontal (on the ground) movement, and the last digit to the vertical movement (balloon going up) = 4.5 min/unit |
ii) What is the balloon's vertical velocity (in units per minute)? Choose the option closest to the correct answer.
(a) |
-5 | (b) |
3 |
(c) |
5 | (d) |
7 |
(c) |
 |
Question 12:
If 
, find a two-dimensional vector perpendicular to 
(a) |
(2, -3) | (b) |
(-8, 12) |
(c) |
(-4, 6) | (d) |
all of the above vectors are perpendicular |
(d) |
all of the above vectors are perpendicular |
|
Draw the vectors. |
Question 13:
Find a unit vector perpendicular to the vector 
. Which option is the closest to the first coordinate?
(a) |
-0.8 | (b) |
-0.5 |
(c) |
0.5 | (d) |
0.8 |
| Both (a) and (d) are correct | |
|
Unit vector has a length of 1, so √( a2 + b2) = 1 Now we have an equation system, with two unknown variables, which we can solve
by taking either a or b out of the second equation and substitute it into the first. |
Question 14:
Find the dot or scalar product of 
and 
(a) |
-11 | (b) |
6 |
(c) |
11 | (d) |
29 |
(a) |
-11 |
|
2(-2) + 3(3) + 4(-4) = -11 |
Question 15:
Find the scalar equation of the plane through the point P = (1, 2, 3) having the normal vector 
(a) |
x + 2y + 3y = 32 | (b) |
x + 2y + 3z = 4 |
(c) |
4x + 5y – 6z = 32 | (d) |
4x + 5y – 6z = – 4 |
(d) |
4x + 5y – 6z = –4 |
|
<4,5,-6> ∙ (<x,y,z> - <1,2,3> = 0 <4,5,-6> ∙ (<x-1, y-2, z-3> = 0 4x + 5y -6z + 4 = 0 |
Question 16:
If the following planes are to be perpendicular, what must be the value of k?
2x – y + 3z + 4 = 0
3x + ky – z – 11 = 0
(a) |
3 | (b) |
1 |
(c) |
9 | (d) |
the planes are not perpendicular for any value of k |
(a) |
3 |
|
plane 1 x plane 2 = 0,
( 2,-1,3) x (3,k,-1) = 0, so 6-k-3 = 0 |
Question 17:
Two theodolites (optical instruments that give an azimuth and an elevation angle) are positioned 100 metres apart on level ground. A balloon is held at a launch point of a atmospheric experiment. The azimuth is the angle measured between the line joining the two theodolites and the line from the theodolite to the balloon:
i) The initial readings from the left hand unit gives an azimuth of 60o and an elevation of 3o. The right hand unit gives an azimuth of 30o and an elevation of 1.7o. Use the left hand unit as the origin of a three dimensional coordinate system with z = 0 at the height of the two theodolites and the positive x-axis going through the right hand unit. Determine the vector from the origin to the balloon.
(a) |
2.6 | (b) |
10 |
(c) |
25 | (d) |
43 |
(c) |
25 |
|
Draw the situation, both looking at it from the side, as well as from above. Put all the data you have in the sketch. Looking from above: the third angle is not given, but you know that all angles in the triangle together should be 1800. So we know
Given: dist. T1-T2 = 100 meter So, dist. T1-baloon =50 meter. Let's call the point where the line from the balloon makes a straight angle with the line between the theodolites x, our x-coordinate for the balloon. The line we just drew, connecting the balloon with the line between the theodolites is then our y-coordinate. The height of the balloon is our z-coordinate. sin 600 = y/50 =>y=43.5 cos 600 = x/50 =>x=25 Now let's look at it from the side. tan 30=height/25=>height=1.3 meter. Our vector is<25, 43.5, 1.3> |
ii) After 1 minute the left hand unit gives an azimuth of 65o and an elevation of 20o, while the right hand unit gives an azimuth of 70o and an elevation of 20.7o. The z coordinate of the balloons position now is.
(a) |
36 | (b) |
48 |
(c) |
80 | (d) |
171 |
(b) |
48 |
|
Use the same procedure, withy the new data:
This gives us T1 - ballon = 133 meter tan 200 = height/133, so the height is 48 meters. |
iii) What is the average vertical velocity during this one minute (in metres per minute)?
(a) |
36 | (b) |
47 |
(c) |
55 | (d) |
128 |
(b) |
47 |
|
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