Transcript Document

Vectors in space
Two vectors in space will either be parallel, intersecting or skew.
Parallel lines
Intersecting lines
Skew lines
Lines in space can be parallel with
an angle between them of 0o.
Lines that meet in space will
intersect each other. The angle at
the point of intersection is q °.
Lines that do not meet each other
in space, but are not parallel, are
called skew.
These lines can still have an angle
between them.
Parallel 3d lines
The parametric equations of two lines in space are given as:
L1 : x = -3 + 2s, y = 2 - s, z = 1 + 3s
L2 : x = 2 - 6t, y = 4 + 3t, z = -1 - 9t
Show that the two lines are parallel.
Turn the parametric equations into vector equations:
æ-3ö
ç ÷
L1 : ç 2 ÷ +
ç1÷
è ø
æ2ö
ç ÷
sç-1÷
ç3÷
è ø
æ2ö
ç ÷
L2 : ç 4 ÷ +
ç-1÷
è ø
æ-6ö
ç ÷
tç 3 ÷
ç-9÷
è ø
Take the vector parts of these equations and note that one is a
multiple of the other:
æ2ö
ç ÷
a = ç-1÷
ç3÷
è ø
æ-6ö
ç ÷
b =ç3÷
ç-9÷
è ø
b is -3a therefore the lines are parallel.
Intersecting 3d lines
Show that the following 2
straight lines intersect and find
the angle between them.
L1 : x = 2 + s, y = 1 + 2s, z = 2 - 2s
L2 : x = 5 + t, y = 4 - t, z = 3 + 5t
If the lines intersect each other
there must be coordinate
(x,y,z) at the point of
intersection.
So the task is to find values of s
and t that satisfy all the
equations.
Set up two simultaneous
equations:
2 + s = 5 + t Solving these
equations gives:
1 + 2s = 4 - t s=2 and t=-1
This equation works for the x and y
parts of the lines. Check it also
works for the z part.
The
coordinate
at
the
2 - 2s = -2
point of intersection is
3 + 5t = -2 (4,5,-2).
To find the angle
between them use
the formula:
Write out the
vector parts
of the lines:
æ1ö
æ1ö
ç ÷
ç ÷
a = ç 2 ÷, b = ç-1÷
ç-2÷
ç5÷
è ø
è ø
a = 3, b = 3 3
a · b = (1 ´ 1) + (2 ´ -1) + (-2 ´ 5) = -11
q = 135°
Skew 3d lines
Show that the following 2
straight lines are skew and find
the angle made between them.
L1 : x = 1 - t, y = 3 + 4t, z = 4 + t
L2 : x = 3 + u, y = 2 - 3u, z = 11 - u
As before, set up two
simultaneous equations and
solve them.
1- t = 3+ u
from the x and y
3 + 4t = 2 - 3u parts of the line.
Solving these will give t=5 and
u=-7.
Now check this works for the z
part of the line:
4+5=9
11 - (-7) = 18
Not equal.
As the simultaneous equations do
not satisfy all 3 coordinates there is
no point of intersection and
therefore the lines are skew.
To find the angle
between them use
the formula:
Write out the
vector parts
of the lines:
æ-1ö
æ1ö
ç ÷
ç ÷
a = ç 4 ÷, b = ç-3÷
ç1÷
ç-1÷
è ø
è ø
a = 18, b = 11
a · b = (-1 ´ 1) + (4 ´ -3) + (1 ´ -1) = -14
q = 174°