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Finding Angles between Lines

With lines instead of vectors, we have 2 possible angles. We usually give the acute angle.

a

We use the 2 direction vectors only since these define the angle.

( If the obtuse angle is found, subtract from . ) 180

e.g. Find the acute angle, a, 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.

e.g. The line through C and A and the line through O and F

To 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.

SUMMARY 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

Exercise

1. 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

Solutions:

(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)

Vectorsconsolidation

• Find the coordinates of the foot of the perpendicular from the point given to the line given

• Recap and secure all work on vectors

Another 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

Another 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 . . .

Another 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

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)

Solution:

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(

SUMMARYTo 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

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