© Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules 6: Differentiating.
54: Applications of the Scalar Product © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core...
-
Upload
basil-mathews -
Category
Documents
-
view
224 -
download
2
Transcript of 54: Applications of the Scalar Product © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core...
54: Applications of the 54: Applications of the Scalar ProductScalar Product
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 2: A2 Core Vol. 2: A2 Core ModulesModules
Applications of the Scalar Product
Module C4
"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"
Applications of the Scalar Product
the product of 2 vectors divided by
Finding Angles between Vectors
cos. abba ab
ba .cos
The scalar product can be rearranged to find the angle between the vectors.
Notice how careful we must be with the lines under the vectors.
The r.h.s. is
the product of the 2 magnitudes of the vectors
Applications of the Scalar Product
Solution:
)1)(1()1)(1()2)(1(. ba 2
,3111 222 a 6112 222 b
ab
ba .cos
e.g. Find the angle between
kjia kjib 2
and
63
2cos 961
( 3s.f. )
Tip: If at this stage you get zero, STOP.
The vectors are perpendicular.
Applications of the Scalar Product
(a)
122
a
086
band
(b)
231
a
21
2band
(c) kjia kjib 22 and
Exercise1. Find the angle between the following
pairs of vectors.
Applications of the Scalar Product
)0)(1()8)(2()6)(2(. ba 4
39122 222 a
1010086 22 b
)10)(3(
4cos 382
( 3s.f. )
Solutions:
(a)
086
band
122
a
Applications of the Scalar Product
)2)(2()1)(3()2)(1(. ba 9
14231 222 a
39212 222 b
143
9cos
143
( 3s.f. )
Solutions:
(b)
231
a
21
2band
Applications of the Scalar Product
(c) kjia kjib 2and
)1)(1()1)(1()2)(1(. ba 0
The vectors are perpendicular.
Solutions:
Applications of the Scalar Product
)3,1,2(,)0,1,1(,)3,1,2( CBA
When solving problems, we have to be careful to use the correct vectors.e.g. The triangle ABC is given by
Find the cosine of angle ABC. Solutio
n: ( any shape triangle will do )
We always sketch and label a triangle
A
B
C
Use BUTab
ba .cos
the a and b of the formula are not the a and b of the question.
We need the vectors and AB CB
Applications of the Scalar Product
)3,1,2(,)0,1,1(,)3,1,2( CBA
When solving problems, we have to be careful to use the correct vectors.e.g. The triangle ABC is given by
Solution: ( any shape triangle will
do )
We always sketch and label a triangle
A
B
C
Use BUTab
ba .cos
the a and b of the formula are not the a and b of the question.
We need the vectors and AB CB
Find the cosine of angle ABC.
Applications of the Scalar Product
)3,1,2(,)0,1,1(,)3,1,2( CBA
321
31
2
011
abAB
321
31
2
011
cbCB
Applications of the Scalar Product
,14321 222 AB
321
AB
321
CB
14321 222 CB
1414
4cos
321
321
. .CBAB
4941
CBAB
CBAB
.cos
7
2cos
14
4cos
2
7
Applications of the Scalar ProductFinding Angles between
Lines
e.g.
With lines instead of vectors, we have 2 possible angles. We usually give the acute angle.
We use the 2 direction vectors only since these define the angle.
( If the obtuse angle is found, subtract from . )
180
Applications of the Scalar Product
ab
ba .cos where
Solution:
sr23
1
tr
102
and
210
21
1
e.g. Find the acute angle, , between the lines
Applications of the Scalar Product
sr23
1
tr
102
and
Solution:
ab
ba .cos where
a and
210
210
21
1
e.g. Find the acute angle, , between the lines
Applications of the Scalar Product
b
210
sr23
1
tr
102
and
Solution:
ab
ba .cos where
a
21
1
and
21
1
210
e.g. Find the acute angle, , between the lines
Applications of the Scalar Producte.g. Find the acute angle, , between the lines
sr23
1
tr
102
and
Solution:
ab
ba .cos where
a
21
1
and
b
21
1
)2)(2()1)(1()1)(0(. ba 5
,521 22 a 6211 222 b
65
5cos
156 24
210
210
(nearest degree)
Applications of the Scalar Product
O
F
A
C B
D E
G
We can find the angle between 2 lines even if they are skew lines.
Applications of the Scalar Product
O
F
A
C B
D E
G
We can find the angle between 2 lines even if they are skew lines. e.g. The line
through C and A and the line through O and FTo define the angle we just draw a line parallel to one line meeting the other.
Applications of the Scalar Product
O
F
A
C B
D E
G
We can find the angle between 2 lines even if they are skew lines. e.g. The line
through C and A and the line through O and FTo define the angle we just draw a line parallel to one line meeting the other.The direction vector of the new line is the same
as the direction vector of one of the original lines so we don’t need to know whether or not the lines intersect.
Applications of the Scalar ProductSUMMARY To find the angle between 2 vectors
• Use the direction vectors only and apply the method above.
• Form the scalar product.
• Find the magnitude of both vectors.
• Rearrange toand substitute.
cos. abba ab
ba .cos
To find the angle between 2 lines
• If the angle found is obtuse, subtract from .
180
Applications of the Scalar Product
(a)
201
321
sr
213
01
2trand
(b) and
Exercise1. Find the acute angle between the following
pairs of lines. Give your answers to the nearest degree.
111
1
2
1
tr
2
1
1
2
2
1
sr
Applications of the Scalar ProductSolutions
:(a)
201
a
213
bab
ba .cos where and
)2)(2()1)(0()3)(1(. ba 7
,521 22 a 14213 222 b
145
7cos 33
(nearest whole degree)
Applications of the Scalar Product
(b)
211
a
111
bab
ba .cos where and
)1)(2()1)(1()1)(1(. ba 2
,6211 222 a 3111 222 b
36
2cos
118
62 (nearest whole degree)
Applications of the Scalar ProductAnother Application of the Scalar Product
x Q
M
and a point not on the line.If we draw a perpendicular from the point to the line . . .we can find the coordinates of M, the foot of the perpendicular.The scalar product of QM . . .
Suppose we have a line, . . .psar
Applications of the Scalar ProductAnother Application of the Scalar Product
x Q
MIf we draw a perpendicular from the point to the line . . . p
Suppose we have a line, . . .psar and a point not on the line.
we can find the coordinates of M, the foot of the perpendicular.The scalar product of QM
and the direction
vector of the line . . .
Applications of the Scalar ProductAnother Application of the Scalar Product
x Q
MIf we draw a perpendicular from the point to the line . . .
and the direction vector of the line
p
equals zero ( since the vectors are perpendicular )
Suppose we have a line, . . .psar and a point not on the line.
we can find the coordinates of M, the foot of the perpendicular.The scalar product of QM
0. pQM
Applications of the Scalar Product
x Q
M
p
M is a point on the line so its position vector is given by one particular value of the parameter s.
0. pQM
So, qmQM wher
e psam
psar
We can therefore substitute into and solve for s.
0. pQM
Applications of the Scalar Product
112
201
sr
e.g. Find the coordinates of the foot of the perpendicular from the point to the line
Q (1, 2, 2)
Applications of the Scalar ProductSolution:
Q (1, 2, 2)x
M
p
)( psar
r
112
201
s
qmQM
Applications of the Scalar Product
Q (1, 2, 2)
Solution:
x
M
p
112
201
s
)( psar
r
112
201
s
qmQM
Applications of the Scalar Product
Q (1, 2, 2)
Solution:
x
M
p
221
112
201
s
)( psar
r
112
201
s
0
112
.221
112
201
s
qmQM
0. pQM
Applications of the Scalar Product
01
12
.221
112
201
s
01
12
.2220
121
ss
s
1 s
01
12
.222
sss
0244 sss
0. pQM
Applications of the Scalar Product
Finally we can find m by substituting for s in the equation of the line.
r
112
201
s
1s
111
112
201
m
The coordinates of M are . )1,1,1(
Applications of the Scalar ProductSUMMARYTo find the coordinates of the foot of the perpendicular from a point to a line:
0. pQM Substitute into , where
• is the direction vector of the linep
This is because it is so easy to substitute the wrong vectors into the equation.
Solve for the parameter, s Substitute for s into the equation of the
line Change the vector m into coordinates.
Sketch and label the line and point Q psar
• M is the foot of the perpendicular and is
a value of r so . psam qmQM
Applications of the Scalar Product
(a)
821
q
112
201
srand
(b) )4,1,1( Q and
Exercise1. Find the coordinates of the foot of the
perpendicular from the points given to the lines given:
531
423
szyx
Applications of the Scalar Product
112
201
sr
01
12
.8
21
112
201
s
01
12
.102
2
ss
s
0. pQM
(a)
Q (1, 2, 8)
Solution:
x
M
p
Applications of the Scalar Product
01024 sss
2s
Point is
)0,2,3(
01
12
.102
2
ss
s
112
2201
m
Applications of the Scalar Product
531
423
szyx
05
31
.4
11
531
423
s
05
31
.85
132
ss
s
0. pQM
(b)
Q (1, 1, 4)
Solution:
x
M
p
Applications of the Scalar Product
04025392 sss
1s
Point is
)1,5,4(
531
423
m
05
31
.85
132
ss
s
Applications of the Scalar Product
Applications of the Scalar Product
The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.
Applications of the Scalar ProductSUMMARY To find the angle between 2 vectors
• Use the direction vectors only and apply the method above.
• Form the scalar product.
• Find the magnitude of both vectors.
• Rearrange toand substitute.
cos. abba ab
ba .cos
To find the angle between 2 lines
• If the angle found is obtuse, subtract from .
180
Applications of the Scalar Product
Solution:
)1)(1()1)(1()2)(1(. ba 2
,3111 222 a 6112 222 b
ab
ba .cos
e.g. Find the angle between
kjia kjib 2
and
63
2cos
961( 3s.f. )
Tip: If at this stage you get zero, STOP.
The vectors are perpendicular.
Applications of the Scalar Product
)3,1,2(,)0,1,1(,)3,1,2( CBA
When solving problems, we have to be careful to use the correct vectors.e.g. The triangle ABC is given by
Find the cosine of angle ABC. Solutio
n: ( any shape triangle will do )
We always sketch and label a triangle
A
B
C
Use BUTab
ba .cos
the a and b of the formula are not the a and b of the question.
We need the vectors and AB CB
Applications of the Scalar Product
321
31
2
011
abAB
321
31
2
011
cbCB
,14321 222 AB 14321 222 CB
1414
4cos
CBAB .
4941
CBAB
CBAB
.cos
107 ABC angle ( 3 s.f. )
Applications of the Scalar ProductFinding Angles between
Lines
e.g.
With lines instead of vectors, we have 2 possible angles. We usually give the acute angle.
We use the 2 direction vectors only since these define the angle.
( If the obtuse angle is found, subtract from . )
180
Applications of the Scalar Producte.g. Find the acute angle, , between the lines
sr23
1
tr
102
and
Solution:
ab
ba .cos where
a
21
1
and
b
21
1
)2)(2()1)(1()1)(0(. ba 5
,521 22 a 6211 222 b
65
5cos
156 24
210
210
(nearest degree)
Applications of the Scalar ProductAnother Application of the Scalar Product
x Q
MIf we draw a perpendicular from the point to the line
and the direction vector of the line
p
equals zero ( since the vectors are perpendicular )
Suppose we have a line, psar and a point not on the line.
we can find the coordinates of M, the foot of the perpendicular.The scalar product of QM
0. pQM
Applications of the Scalar Product
x Q
M
p
M is a point on the line so its position vector is given by one particular value of the parameter s.
0. pQM
So, qmQM wher
e psam
psar
We can therefore substitute into and solve for s.
0. pQM
Applications of the Scalar ProductSUMMARYTo find the coordinates of the foot of the perpendicular from a point to a line:
0. pQM Substitute into , where
• is the direction vector of the linep
Solve for the parameter, s Substitute for s into the equation of the
line Change the vector m into coordinates.
Sketch and label the line and point Q psar
• M is the foot of the perpendicular and is
a value of r so . psam qmQM
Applications of the Scalar Product
112
201
sr
e.g. Find the coordinates of the foot of the perpendicular from the point to the line
Q (1, 2, 2)
Q (1, 2, 2)
Solution:
x
M
p
)( psar
r
112
201
s
Applications of the Scalar Product
221
112
201
s
01
12
.221
112
201
s
qmQM
0. pQM
01
12
.2220
121
ss
s
1 s
01
12
.222
sss
0244 sss
Applications of the Scalar Product
Finally we can find m by substituting for s in the equation of the line.
r
112
201
s
1s
111
112
201
m
The coordinates of M are . )1,1,1(