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Section 9.5: Equations of Lines and Planes
Practice HW from Stewart Textbook (not to hand in)
p. 673 # 3-15 odd, 21-37 odd, 41, 47
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Lines in 3D Space
Consider the line L through the point
that is parallel to the vector v = < a, b, c >.
),,( 000 zyxP
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x
y
z
v = < a, b, c >
L
0000 ,, zyxr
),,( zyxQ
),,( 000 zyxP
zyxr ,,
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The line L consists of all points Q = (x, y, z) for which the vector is parallel to v.
Now,
Since is parallel to v = < a, b, c > ,
= t v
where t is a scalar. Thus
PQ
000 ,, zzyyxxPQ
PQ
PQ
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= t v = < t a, t b, t c >
Rewriting this equation gives
Solving for the vector gives
Setting r = , , and
v = < a, b, c >, we get the following vector
equation of a line.
PQzzyyxx 000 ,,
cbatzyxzyx ,,,,,, 000
zyx ,,
cbatzyxzyx ,, ,,,, 000
zyx ,, 0r 000 ,, zyx
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Vector Equation of a Line in 3D Space
The vector equation of a line in 3D space is given
by the equation
t v
where = is a vector whose
components are made of the point
on the line L and v = < a, b, c > are components
of a vector that is parallel to the line L.
0rr
0r 000 ,, zyx
),,( 000 zyx
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If we take the vector equation
and rewrite the right hand side of this equation as
one vector, we obtain
Equating components of this vector gives the
parametric equations of a line.
cbatzyxzyx ,, ,,,, 000
tcztbytaxzyx 000 ,,,,
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Parametric Equations of a Line in 3D Space
The parametric equations of a line L in 3D space are given by
where is a point passing through the line and v = < a, b, c > is a vector that the line is parallel to. The vector v = < a, b, c > is called the direction vector for the line L and its components a, b, and c are called the direction numbers.
tczztbyytaxx 000 , ,
),,( 000 zyx
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Assuming , if we take each
parametric equation and solve for the variable t,
we obtain the equations
Equating each of these equations gives the
symmetric equations of a line.
0 ,0 ,0 cba
c
zzt
b
yyt
a
xxt 000 , ,
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Symmetric Equations of a Line in 3D Space
The symmetric equations of a line L in 3D space are given by
where is a point passing through the line and v = < a, b, c > is a vector that the line is parallel to. The vector v = < a, b, c > is called the direction vector for the line L and its components a, b, and c are called the direction numbers.
),,( 000 zyx
c
zz
b
yy
a
xx 000
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Note!! To write the equation of a line in 3D
space, we need a point on the line and a parallel
vector to the line.
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Example 1: Find the vector, parametric, and
symmetric equations for the line through the point
(1, 0, -3) and parallel to the vector 2 i - 4 j + 5 k.
Solution:
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Example 2: Find the parametric and symmetric
equations of the line through the points (1, 2, 0)
and (-5, 4, 2)
Solution: In typewritten notes
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It is important to note that the equations of lines in
3D space are not unique. In Example 2, for
instance, had we used the point Q = (-5, 4, 2) to
represent the equation of the line with the parallel
vector v = < -6, 2, 2 >, the parametric equations
becomes
tt, zytx 2224 ,65
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Example 3: Find the parametric and symmetric
equations of the line passing through the point
(-3, 5, 4) and parallel to the line
x = 1 + 3t, y = -1 – 2t, z = 3 + t .
Solution:
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Planes in 3D Space
Consider the plane containing the point
and normal vector n = < a, b, c >
perpendicular to the plane.
),,( 000 zyxP
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x
y
z
),,( 000 zyxP
),,( zyxQ
n = < a, b, c >
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The plane consists of all points Q = (x, y, z) for
which the vector is orthogonal to the normal
vector n = < a, b, c >. Since and n are
orthogonal, the following equations hold:
This gives the standard equation of a plane.
PQ
PQ
0
PQn
0,,,, 000 zzyyxxcba
0)()()( 000 zzcyybxxa
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If we expand this equation we obtain the following
equation:
Setting gives the general form
of the equation of a plane in 3D space
We summarize these results as follows.
0
Constant
000 d
czbyaxczbyax
000 czbyaxd
0 dczbyax
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Standard and General Equations of a Plane in 3D
space
The standard equation of a plane in 3D space has the
form
where is a point on the plane and n = < a, b, c >
is a vector normal (orthogonal to the plane). If this
equation is expanded, we obtain the general equation of
a plane of the form
0)()()( 000 zzcyybxxa
),,( 000 zyx
0 dczbyax
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Note!! To write the equation of a plane in 3D
space, we need a point on the plane and a vector
normal (orthogonal) to the plane.
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Example 4: Find the equation of the plane
through the point (-4, 3, 1) that is perpendicular to
the vector a = -4 i + 7 j – 2 k.
Solution:
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Example 5: Find the equation of the plane
passing through the points (1, 2, -3), (2, 3, 1), and
(0, -2, -1).
Solution:
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Intersecting Planes
Suppose we are given two intersecting planes with angle between them.
Plane 2
Plane 1
2n
1n
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Let and be normal vectors to these
planes. Then
Thus, two planes are
1. Perpendicular if , which implies .
2. Parallel if , where c is a scalar.
1n 2n
|| ||cos
21
21
nn
nn
021 nn2
12 cnn
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Notes
1. Given the general equation of a plane
the normal vector is n = < a, b, c >.
2. The intersection of two planes is a line.
0 dczbyax
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Example 6: Determine whether the planes
and are
orthogonal, parallel, or neither. Find the angle
of intersection and the set of parametric
equations for the line of intersection of the plane.
Solution:
343 zyx 41239 zyx
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Example 7: Determine whether the planes
and are
orthogonal, parallel, or neither. Find the angle
of intersection and the set of parametric
equations for the line of intersection of the plane.
Solution: (In typewritten notes)
463 zyx 45 zyx
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Example 8: Find the point where the line
x = 1 + t, y = 2t, and z = -3t intersects the plane
Solution:
2424 zyx
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Distance Between Points and a Plane
Suppose we are given a point Q not in a plane
and a point P on the plane and our goal is to find
the shortest distance between the point Q and the
plane.
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P
Q
n
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By projecting the vector onto the normal
vector n (calculating the scalar projection
), we can find the distance D.
PQ
PQcompn
||
|||| D
plane theand
Between Distance
n
nPQPQcomp
Qn
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Example 9: Find the distance between the point
(1, 2, 3) and line .
Solution: (In typewritten notes)
42 zyx