Instructor: Longfei Li
Math 243 Lecture Notes
12.5 Equations of Lines and Planes
What do we need to determine a line?
2D:
• a point on the line: P0 (x0 , y0 )
• direction (slope): k
3D:
• a point on the line: P0 (x0 , y0 , z0 )
• direction (a vector parallel the line): v
Vector Equation of L:
r = r0 + tv
Proof:
P0 (x0 , y0 , z0 ) is a given (known) point on L, P (x, y, z) is an arbitrary point on L. r0 and r are the
−−→
position vectors of P0 and P . a = P0 P , then by the Triangle Law:
r = r0 + a
Since a is parallel to v, v is a scalar multiple of a:
a = tv, t ∈ R
So
r = r0 + tv, t ∈ R
Remark: r traces out the line: r0 shifts us onto the line, tv moves us along the line(t > 0 to the same
direction as v; t < 0 to the opposite direction to v)
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Scalar Equations (parametric equations)
x = x0 + at; y = y0 + bt; z = z0 + ct, t ∈ R
Proof: If we write the vectors into their components form and let v =< a, b, c >, then
< x, y, z > =< x0 , y0 , z0 > +t < a, b, c >
=< x0 + at, y0 + bt, z0 + ct >
Thus, we have the scalar equations: x = x0 + at; y = y0 + bt; z = z0 + ct.
Example: Find a vector and parametric equations for the line that passes through the points A(1, 2, 3)
and B(4, 6, 8).
Solution:
−−→
The vector of the line: v = AB =< 3, 4, 5 >
If we pick A(1, 2, 3) as the given point on the line, then
vector equation: r =< 1, 2, 3 > +t < 3, 4, 5 >=< 1 + 3t, 2 + 4t, 3 + 5t >
parametric equations: x = 1 + 3t, y = 2 + 4t, z = 3 + 5t
If we pick B(4, 6, 8) as the given point on the line, then
vector equation: r =< 4, 6, 8 > +t < 3, 4, 5 >=< 4 + 3t, 6 + 4t, 8 + 5t >
parametric equations: x = 4 + 3t, y = 6 + 4t, z = 8 + 5t
We could also a scalar multiple of the vector 2v =< 6, 8, 10 >:
vector equation: r =< 1, 2, 3 > +t < 6, 8, 10 >=< 1 + 6t, 2 + 8t, 3 + 10t >
Remark: Equations are not unique! We can change the given point, choose a different parallel vector.
Direction Numbers: If vector v =< a, b, c > is used as the direction of L, then a, b and c are called
the direction numbers.
Symmetric Equations:
x − x0
y − y0
z − z0
=
=
a
b
c
If one of a, b, c equals 0, we write the symmetric equations as:
x − x0
z − z0
=
, y = y0 , for instance, if b = 0
a
c
Proof: The symmetric equations are obtained by eliminating t from the parametric equations:
x − x0
a
y − y0
y = y0 + bt ⇒ t =
b
z − z0
z = z0 + ct ⇒ t =
c
x = x0 + at ⇒ t =
Hence,
x − x0
y − y0
z − z0
=
=
a
b
c
2
Example: Find parametric and symmetric equations for the line passes through the point (5, 1, 3) and
is parallel to i + 4j − 2k.
Solution:
parametric equations: x = 5 + t, y = 1 + 4t, z = 3 − 2t
symmetric equations:
x−5
y−1
z−3
=
=
1
4
−2
Line Segment: The line segment from r0 to r1 is given by the vector equations:
r(t) = (1 − t)r0 + tr1 , 0 ≤ t ≤ 1
Proof:
If we choose v = r1 − r0 , then the vector equation
r(t) = r0 + t(r1 − r0 ) ⇒ r(t) = (1 − t)r0 + tr1 , 0 ≤ t ≤ 1
Remark: r(t) starts at r0 , ends at r1 and traces point between.
Relations of 2 lines in space:
Intersect: pass through the same point.
Parallel: have the same direction
Skew: not parallel, don’t intersect
Example: Determine whether the lines L1 and L2 are parallel, intersecting, or skew:
(a) L1 : x = 1 + t, y = 3 + 2t, z = 5 + 3t; L2 : x = 2 − 3s, y = −6s, z = −5 − 9s
Solution:
Directions: v1 =< 1, 2, 3 >, v2 =< −3, −6, −9 >. We have v2 = −3v1 , scalar multiple ⇒ parallel
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(b) L1 : x = 1 + t, y = 1 + t, z = 2 + t; L2 : x = 3s, y = −2s, z = 0
Solution:
Directions: v1 =< 1, 1, 1 >, v2 =< 3, −2, 0 >. Not scalar multiple ⇒ Not parallel.
Intersect? If so, for some t and s:
1 + t = 3s
1 + t = −2s
2+t=0
Solve the first 2 equations ⇒ t = −1, s = 0.
However, the third equation is not satisfied! ⇒ Not intersecting. So L1 and L2 are skew lines.
x−2
y−3
z−1
x−3
y+4
z−2
=
=
, L2 :
=
=
1
−2
−3
1
3
−7
Solution::
Directions: v1 =< 1, −2, −3 >, v2 =< 1, 3, −7 >. Not scalar multiple ⇒ Not parallel.
Intersect?
The parametric equations are
(c) L1 :
L1 : x = 2 + t, y = 3 − 2t, z = 1 − 3t
L2 : x = 3 + s, y = −4 + 3s, z = 2 − 7s
If intersect, for some t and s, we have
2+t=3+s
3 − 2t = −4 + 3s
1 − 3t = 2 − 7s
Solve the first two equations ⇒ t = 2, s = 1. Check the third equation, it’s satisfied!
So the lines intersect when t = 2, s = 1, i.e., at the point (4, −1, −5).
Planes
What do we need to determine a plane?
• a point in the plane: P0 (x0 , y0 )
• direction (a vector perpendicular to the plane): n
Remark: The orthogonal vector n is called a normal vector.
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Let P (x, y, z) be an arbitrary point in the plane, P0 (x0 , y0 , z0 ) is a given point in the plane, and n be
a normal vector of plane, then n is orthogonal to r − r0 , i.e.,
n · (r − r0 ) = 0
Vector Equation of the Plane:
n · (r − r0 ) = 0 or n · r = n · r0
Scalar Equation of the Plane: If n =< a, b, c >, then
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0
Linear Equation of the Plane:
ax + by + cz + d = 0
where d = −(ax0 + by0 + czc )
Recall: the intercepts are the intersections of the graph with the coordinate axes.
Example: Find an equation of the plane through the point (2, 4, −1) with normal vector n =< 2, 3, 4 >.
Find the intersepts and sketch the plane.
Solution:
2(x − 2) + 3(y − 4) + 4(z + 1) = 0
⇒ 2x + 3y + 4z = 12
The x−intercepts is x = 6, the y−intercepts is y = 4, and the z−intercepts is z = 3.
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Example: Find an equation of the plane that passes through the points P (3, 1, 1), Q(5, −2, 2) and
R(2, 4, 7).
Solution: The vectors in the plane are
−−→
−→
a = P Q =< 2, −3, 1 > and b = P R =< −1, 3, 6 >
a × b is orthogonal to both a and b,
i
j
n = a × b = 2 −3
−1 3
thus is orthogonal to the plane and can be taken as normal.
k
1 =< −18 − 3, −1 − 12, 6 − 3 >=< −21, −13, 3 >
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So the equation of the plane is
−21(x − 3) − 13(y − 1) + 3(z − 1) = 0
Parallel of 2 Planes: Two planes are parallel if their normal vectors are parallel.
The Angle between the planes: defined as the acute angle between the normal vectors.
Example: Find the angle between the planes x + y + z = 10 and x − 2y + 3z = 2.
Solution: The normal vectors of the two planes are
n1 =< 1, 1, 1 > and n2 =< 1, −2, 3 >
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If θ is the angle between the two planes, then
n1 · n2
2
1−2+3
=√
= √ √
|n1 ||n2 |
3 14
42
2
θ = arccos √
≈ 72◦
42
cos θ =
Example: Find the parametric equations for the line of intersection L of the planes x + y + z = 1 and
x − y − z = 2.
Solution: The normal vectors of the planes are
n1 =< 1, 1, 1 > and n2 =< 1, −1, −1 >
Since the line L lies in both of the planes, it is
to L is
i
n1 × n2 = 1
1
orthogonal to both n1 and n2 . Thus, the vector parallel
j
k 1
1 =< 0, 2, −2 >
−1 −1
Pick a point on the line of the intersect:
(
x+y+z =1
x−y−z =2
set z = 0, we have x + y = 1, x − y = 2 ⇒ x = 23 , y = − 12 . So the point ( 32 , − 12 , 0) lies on L. So
far we find the direction of L: < 0, 2, −2 > and a given point ( 32 , − 12 , 0) on the line, so the parametric
equations are
3
1
x = , y = − + 2t, z = −2t
2
2
Distance: The distance D from a point P1 (x1 , y1 , z1 ) to the plane ax + by + cz + d = 0 is
D=
|ax1 + by1 + cz1 + d|
√
a2 + b2 + c2
Proof:
Let P0 (x0 , y0 , z0 ) be a point in the plane. Then the coordinates of P0 satisfies the plane equation:
ax0 + by0 + cz0 + d = 0 ⇒ d = −(ax0 + by0 + cz0 )
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−−−→
b = P0 P1 =< x1 − x0 , y1 − y0 , z1 − z0 >
From the equation of the plane, the normal vector is n =< a, b, c >. Thus,
|n · b|
|a(x1 − x0 ) + b(y1 − y0 ) + c(z1 − z0 )|
√
=
|n|
a2 + b2 + c2
|ax1 + by1 + cz1 − (ax0 + by0 + cz0 )|
|ax1 + by1 + cz1 + d|
√
√
=
=
2
2
2
a +b +c
a2 + b2 + c2
D = |compn b| =
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