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

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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


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.


(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.


)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


)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


(c) kjia kjib 2and

)1)(1()1)(1()2)(1(. ba 0


Solutions:


)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


)3,1,2(,)0,1,1(,)3,1,2( CBA


Solution: ( any shape triangle will

do )


A

B

C

Use BUTab

ba .cos



Find the cosine of angle ABC.


)3,1,2(,)0,1,1(,)3,1,2( CBA

321

31

2

011

abAB

321

31

2

011

cbCB


,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


ab

ba .cos where

Solution:

sr23

1

tr

102

and

210

21

1

e.g. Find the acute angle, , between the lines


sr23

1

tr

102

and

Solution:

ab

ba .cos where

a and

210

210

21

1



b

210

sr23

1

tr

102

and

Solution:

ab

ba .cos where

a

21

1

and

21

1

210


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)


O

F

A

C B

D E

G

We can find the angle between 2 lines even if they are skew lines.


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.


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


(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)


(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


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 . . .


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.


0. pQM


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


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


Q (1, 2, 2)

Solution:

x

M

p

112

201

s

)( psar

r

112

201

s

qmQM


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


01

12

.221

112

201

s

01

12

.2220

121

ss

s

1 s

01

12

.222

sss

0244 sss

0. pQM


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


(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


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


01024 sss

2s

Point is

)0,2,3(

01

12

.102

2

ss

s

112

2201

m


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


04025392 sss

1s

Point is

)1,5,4(

531

423

m

05

31

.85

132

ss

s


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


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.



)3,1,2(,)0,1,1(,)3,1,2( CBA


Find the cosine of angle ABC. Solutio

n: ( any shape triangle will do )


A

B

C

Use BUTab

ba .cos




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)


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.


0. pQM


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


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


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


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(

54: Applications of the Scalar Product © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core...

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Transcript of 54: Applications of the Scalar Product © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core...