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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
