# x y z - Penn Math ```Math 114 – Rimmer
12.5 Equations of Lines and Planes
12.5 Equations of Lines and Planes
In order to find the equation of a line, we need :
A ) a point on the line P0 ( x0 , y0 , z0 )
B ) a direction vector for the line v = a, b, c
r = r0 + tv vector equation of line L
z
x, y, z = x0 , y0 , z0 + t a, b, c
P0 P = tv
P ( x, y, z )
P0 ( x0 , y0 , z0 )
equating components:
r
L
parametric equations of the line L
r0
v
r0 = x0 , y0 , z0
r = x, y , z
x
x = x0 + at , y = y0 + bt , z = z0 + ct
P0 P = tv = t a, b, c
y
eliminating the parameter t (solve for t
in each, then equate the results)
x − x0 y − y0 z − z0
=
=
a
b
c
symmetric equations of the line L
Math 114 – Rimmer
12.5 Equations of Lines and Planes
Find parametric equations of the line containing ( 5,1,3) and ( 3, −2, 4 ) .
In order to find the equation of a line, we need :
A ) a point on the line P0 ( x0 , y0 , z0 )
Pick either point
B ) a direction vector for the line v = a, b, c
Use the vector from
one point to the other
L:
x =
y =
z =
point
direc.
x0
+ at
y0
z0
+ bt
+ ct
v = 3 − 5, −2 − 1, 4 − 3
v = −2, −3,1
L:
x = 5 − 2t
y = 1 − 3t
z = 3 + t
1
Two lines in 3 space can interact in 3 ways:
Math 114 – Rimmer
12.5 Equations of Lines and Planes
A) Parallel Lines - their direction vectors are scalar
multiples of each other
B) Intersecting Lines - there is a specific t and s, so that the
lines share the same point.
C) Skew Lines their direction vectors are not parallel and
there is no values of t and s that make the
lines share the same point.
Math 114 – Rimmer
12.5 Equations of Lines and Planes
Determine whether the lines L1 and L2 are parallel, skew
or intersecting. If they intersect, find the point of intersection.
L1
L2
x = 3 − t
y = 5 + 3t
z = −1 − 4t
x = 8 + 2s
y = −6 − 4 s
z = 5 + s
Check to make sure that the z
values are equal for this t and s.
Set the x ' s =
3 − t = 8 + 2s
−t = 5 + 2 s
t = −5 − 2 s
Set the y ' s =
−1 − 4t = 5 + s
−1 − 4 ( −1) = 5 + ( −2 )
3 = 3 check
This should happen at
the same time t ,
so plug in t = −5 − 2 s
5 + 3t = −6 − 4s
5 + 3 ( −5 − 2 s ) = −6 − 4s
5 − 15 − 6s = −6 − 4s
−10 − 6s = −6 − 4s t = −5 − 2 ( −2 )
−2 s = 4
t = −1
s = −2
Now find the pt. of intersection.
using L1
x = 3 − ( −1)
y = 5 + 3 ( −1)
z = −1 − 4 ( −1)
( 4, 2,3)
2
Math 114 – Rimmer
12.5 Equations of Lines and Planes
Determine whether the lines L1 and L2 are parallel, skew
or intersecting. If they intersect, find the point of intersection.
L1
L2
x = 4 + t
y = −8 − 2t
z =
12t
x = 3 + 2s
y = −1 + s
z = −3 − 3s
Check to make sure that the z
values are equal for this t and s.
Set the x ' s =
4 + t = 3 + 2s
t = −1 + 2 s
??
12 ( −3 ) =− 3 − 3 ( −1)
−36 ≠ 0
Set the y ' s =
−8 − 2t = −1 + s
−8 − 2 ( −1 + 2s ) = −1 + s
This should happen at
This means that the x and y
values are the same for t = −3
and s = −1, but the z values are
different.
the same time t ,
so plug in t = −1 + 2 s
−8 + 2 − 4 s = −1 + s
t = −1 + 2 s
−6 − 4s = −1 + s
−5s = 5 s = −1 t = −3
The lines are skew
Planes
Math 114 – Rimmer
12.5 Equations of Lines and Planes
In order to find the equation of a plane, we need :
A ) a point on the plane P0 ( x0 , y0 , z0 )
B ) a vector that is orthogonal to the plane n = a, b, c
n
z
P0 P = r − r0
n ⋅ ( r − r0 ) = 0
n ⋅ r = n ⋅ r0
P ( x, y, z )
vector equation of the plane
r
P0 ( x0 , y0 , z0 )
n ⋅ ( r − r0 ) = 0 ⇒
r0
x
this vector is called
the normal vector
to the plane
y
r0 = x0 , y0 , z0
r = x, y , z
P0 P = x − x0 , y − y0 , z − z0
n = a , b, c
a ( x − x0 ) + b ( y − y0 ) + c ( z − z0 ) = 0
scalar equation of the plane
ax − ax0 + by − by0 + cz − cz0 = 0
ax + by + cz − ( ax0 + by0 + cz0 ) = 0
ax + by + cz + d = 0
linear equation of the plane
3
Determine the equation of the plane that contains the lines L1 and L2 .
L1
L2
x = 3 − t
y = 5 + 3t
z = −1 − 4t
x = 8 + 2s
y = −6 − 4 s
z = 5 + s
We have two points in the plane:
Math 114 – Rimmer
12.5 Equations of Lines and Planes
In order to find the equation of a plane, we need :
A ) a point on the plane
B ) a vector that is orthogonal to the plane
n = a, b, c
from L1 : ( 3,5, −1) and from L2 : ( 8, −6,5 )
We have two vectors in the plane:
from L1 : −1,3, −4 and from L2 : 2, −4,1
Let u = −1,3, −4 and v = 2, −4,1 Find u &times; v.
i
u &times; v = −1
2
j
3
k
−4 =
−4
1
3
−4
−4
1
i + ( −1)
−1 −4
2
1
j+
−1
3
2
−4
k
= ( 3 − 16 ) i + ( −1)( −1 + 8 ) j + ( 4 − 6 ) k
u &times; v = −13i − 7 j − 2k = n
−13 x − 7 y − 2 z + 72 = 0
−13 x − 7 y − 2 z + d = 0
−13 ( 3) − 7 ( 5 ) − 2 ( −1) + d = 0
13 x + 7 y + 2 z − 72 = 0
−39 − 35 + 2 + d = 0 ⇒ d = 72
Math 114 – Rimmer
12.5 Equations of Lines and Planes
Determine the equation of the plane that passes through
(1, 2,3) , ( 3, 2,1) , and ( −1, −2, 2 ) .
R
Q
P
PQ = 3 − 1, 2 − 2,1 − 3
PQ = 2, 0, −2
In order to find the equation of a plane, we need :
A ) a point on the plane
we have 3 to choose from
PR = −1 − 1, −2 − 2, 2 − 3
B ) a vector that is orthogonal to the plane
PR = −2, −4, −1
n = a, b, c
we need 2 vectors in the plane
Let u = 2, 0, −2 and v = −2, −4, −1 Find u &times; v.
i
u&times;v = 2
−2
j
0
−4
k
0
−2 =
−4
−1
−2
−1
= ( 0 − 8) i
i + ( −1)
2
−2
−2 −1
j +
2
0
−2 −4
k
+ ( −1)( −2 − 4 ) j + ( −8 − 0 ) k
u &times; v = −8i + 6 j − 8k = n or n = 4i − 3 j + 4k ( in lowest terms )
4x − 3 y + 4z + d = 0
4 (1) − 3 ( 2 ) + 4 ( 3) + d = 0
4 − 6 + 12 + d = 0
4 x − 3 y + 4 z − 10 = 0
⇒ d = −10
4
Math 114 – Rimmer
12.5 Equations of Lines and Planes
Two distinct planes in 3-space either are
parallel or intersect in a line.
z
y
x
Find the line of intersection of the two planes
Use the elimination method
x − 2y + z = 0
2 ( x − 2 y + z = 0)
2x + 3y − 2z = 0
2x + 3y − 2z = 0
Math 114 – Rimmer
12.5 Equations of Lines and Planes
2x − 4 y + 2z = 0
2x + 3y − 2z = 0
4x − y = 0
y = 4x
Take the first equation:
x − 2 y + z = 0 and plug in y = 4 x
x − 2 ( 4x ) + z = 0 ⇒ z = 7 x
x can be anything and we have y and z as functions of x.
Let x = t
then y = 4t and z = 7t
L:
x =
t
y = 4t
z = 7t
5
Math 114 – Rimmer
12.5 Equations of Lines and Planes
If two planes intersect, then you can determine the angle between them.
between planes = between their normal vectors
n1
n2
cos θ =
θ
n1 ⋅ n 2
n1 n 2
θ
Find the angle between the planes
x − 2y + z = 0
2x + 3 y − 2z = 0
cos θ =
n1 ⋅ n 2
n1 n 2
6
102
=
θ ≈ 54
⇒ n1 = 1, −2,1 and n 2 = 2,3, −2
n1 = 6 and n 2 = 17
⇒ n1 ⋅ n 2 = 2 − 6 − 2 = −6
Distance between a point and a plane:
Math 114 – Rimmer
12.5 Equations of Lines and Planes
1. Take any point in the plane call it P0 ( x0 , y0 , z0 ) .
2. Let b = P0 P1 = x1 − x0 , y1 − y0 , z1 − z0
( x1 , y1 , z1 )
projn b
3. Take projn b. The length of this vector = D
Remember projnb = compn b =
n⋅b
n
4. Since D must be positive,
take the absolute value.
D=
n ⋅b
ax + by + cz + d = 0
n
n ⋅ b = a, b, c ⋅ x1 − x0 , y1 − y0 , z1 − z0 = a ( x1 − x0 ) + b ( y1 − y0 ) + c ( z1 − z0 )
= ax1 + by1 + cz1 + ( − ax0 − by0 − cz0 )
⇒ n ⋅ b = ax1 + by1 + cz1 + d
D=
d
ax1 + by1 + cz1 + d
a 2 + b2 + c2
6
Math 114 – Rimmer
12.5 Equations of Lines and Planes
Distance between a point and a line:
Math 114 – Rimmer
12.5 Equations of Lines and Planes
7
Math 114 – Rimmer
12.5 Equations of Lines and Planes
8
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