Alg1131W4T2 - Projections
Friday, 23 March 2018
1:29 PM
Created with Microsoft OneNote 2016.
![Untitled picture.png Machine generated alternative text:
Consider w = proju(v), the projection of the vector v onto the vector u. An intuitive explanation is that w is the multiple of u which is the closest to v _ This means that (v
Play witn the GeoGebra applet to give approximate answers to the following questions, to the nearest 0.1.
0.0
i) It u
ii) It u
iii) It u
iv) It u
and v =
3
and v =
4
-4
and v =
3
—3
and v =
4
1.2
4
then
1
2
then
4
-3
then
—4
3
then
3
0.5
proju(v)
0.4
0.3
0.2
-0.4-0.35-0.3-0.25-0.2 -0.15 1-0.05
Note: You might find the zooming in teature useful.
proju(v) z
proju(v) z
proju(v) z
proju(v) z
Note: the syntax for the vector
3.4
is < 1.2, 3.4>
Untitled picture.png Machine generated alternative text:
The formula for the projection of the vector v onto the (non-zero) vector u is given by
proj u (v)
Note that the projection of the vector v onto u is: [select all that apply]
parallel to u
C' parallel to v
a multiple of u
C' a multiple of v.
Also note that vector v only occurs once in the formula, in the dot product in the numerator. Finally note that the denominator only depends on u, and is in tact the quadrance Q (u), also known as the square of
the length ot u.
So tor example in two dimensions, it u =
and v =
4
4
1
4
then
and v =
proju (v)
In three dimensions, the same formula works. So it u =
Note: enter your vectors using Maple syntax, e.g. < 1, 2
—1
3
—5
then
proju(v)
<-2418. 24/18, -96/18>
Untitled picture.png Machine generated alternative text:
Projections are useful to find the closet point on a line to a given point
4
3
2
a) It I is the line tnrougn the points [0,0] and [4, 2] , then the closest point on I to the point [2, 3] is
Note: the Maple syntax tor a point is of the form [1, 2]
7
6
5
4
3
2
1
2
3
4
5
6
0123456
b) The line m shown passes tnrougn the point [3, 1] in the direction of the vector
_ What is the closest point on m to the point [4, 7]?
1
The closest point is [08, 351
Hint: try to rephrase these questions in terms ot projecting one vector onto another.
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![Untitled picture.png Machine generated alternative text:
Projections are useful to find the closet point on a line to a given point
4
3
2
a) It I is the line tnrougn the points [0,0] and [4, 2] , then the closest point on I to the point [2, 3] is
Note: the Maple syntax tor a point is of the form [1, 2]
7
6
5
4
3
2
1
2
3
4
5
6
0123456
b) The line m shown passes tnrougn the point [3, 1] in the direction of the vector
_ What is the closest point on m to the point [4, 7]?
1
The closest point is [08, 351
Hint: try to rephrase these questions in terms ot projecting one vector onto another.
Untitled picture.png Machine generated alternative text:
6
5
4
3
2
2
2
Consider a block of ice of mass m = 0.5 kg sitting on an inclined plane with a slope of
The force due to gravity (measured in Newtons) is given by the well known formula
6
7
mg
8
-9.8
9
m/sec2
F gr = mg
where the vector g is acceleration due to gravity (measured in meters per second per second), namely g
Let's use our projection formula to determine F, the force acting parallel to the plane.
i) From the picture we see AO =
Note: make sure to enter your vectors in Maple notation (e.g. < 1, 2
ii) The force parallel to the plane is (to the nearest 0.1)
F = (Fgr)
So we see that the magnitude of the force F is (to the nearest 0.1)
O Newtons (kg m/sec2)_
It is the force F which determines the actual motion of the block down the inclined plane.
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![Untitled picture.png Machine generated alternative text:
6
5
4
3
2
2
2
Consider a block of ice of mass m = 0.5 kg sitting on an inclined plane with a slope of
The force due to gravity (measured in Newtons) is given by the well known formula
6
7
mg
8
-9.8
9
m/sec2
F gr = mg
where the vector g is acceleration due to gravity (measured in meters per second per second), namely g
Let's use our projection formula to determine F, the force acting parallel to the plane.
i) From the picture we see AO =
Note: make sure to enter your vectors in Maple notation (e.g. < 1, 2
ii) The force parallel to the plane is (to the nearest 0.1)
F = (Fgr)
So we see that the magnitude of the force F is (to the nearest 0.1)
O Newtons (kg m/sec2)_
It is the force F which determines the actual motion of the block down the inclined plane.
Untitled picture.png Machine generated alternative text:
You may find this question challenging.
Given the three points A = [6, —2], B = [—3, 0] and C = [3, —5], consider the triangle A ABC, and let M be the point on AC so that BM is the altitude through B perpendicular to
AC. Likewise let N be the point on AB so that CN is the altitude through C perpendicular to AB
(diagram not to scale)
i) Find M
ii) Find N
1
Note: in Maple notation the point [
Challenge (worth 2 points):
M
N
3
—l is entered as [1/2, 3/4]
[25, -55]
[321/85: -128/85]
iii) It is known that the three altitudes of any triangle meet at a common point called the orthocenter For A ABC, this orthocenter is
H = [31/11,-64/111
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![Untitled picture.png Machine generated alternative text:
There is 4 steps to calculate the orthocenter of any triangle as described below.
Step 1: Calculate the slope of the sides of the triangle using following formula:
Y2—YI
Slope of a line —
X2—X1
Step 2: Calculate the perpendicular slope of the sides using following formula:
Perpendicular slope of the line
Slope of a line
Step 3: Calculate the equation for any two altitudes with their respective coordinates using following formula:
Step 4: Solving altitude equations (any two altitude equation of Step 3)
Untitled picture.png Machine generated alternative text:
You may find this question challenging.
Given the three points A = [6, —2], B = [—3, 0] and C = [3, —5], consider the triangle A ABC, and let M be the point on AC so that BM is the altitude through B perpendicular to
AC. Likewise let N be the point on AB so that CN is the altitude through C perpendicular to AB
(diagram not to scale)
i) Find M
ii) Find N
1
Note: in Maple notation the point [
Challenge (worth 2 points):
M
N
3
—l is entered as [1/2, 3/4]
[25, -55]
[321/85: -128/85]
iii) It is known that the three altitudes of any triangle meet at a common point called the orthocenter For A ABC, this orthocenter is
H = [31/11,-64/111
http://eguruchela.com/math/Calculator/triangle-orthocenter-calculator
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