13.5 Equations of Lines and Planes
1.Lines in R3
Let ro =< xo , yo , zo > and v =< a, b, c > be two fixed vectors.
Then there exists a unique line through < xo , yo , zo > parallel to
v. That is, a line in R3 is uniquely determined by a point it is
through and a vector it is parallel to.
2.Equation (Equations of a line)
(a) Let l be the line through a point ro =< xo , yo , zo > parallel
to the vector v =< a, b, c >. To write an equation of the
line l, let p =< x, y, z > be a point on the line and use the
essential condition for p ; p − ro / / v, equivalently
p − ro = tv,
(−∞ < t < ∞)
Note that ro and v are fixed vectors. p is on the line l
as soon as p satisfies this equation for some t. Above
equation is called a vector equation of a line. We call the
vector v =< a, b, c > the direction vector of the line l.
(b) The vector equation above can be rewritten componentwise ; < x, y, z > − < xo , yo , zo >= t <
a, b, c >, equivalently,

 x =
y =

z =
xo + ta
yo + tb ,
zo + tc
(−∞ < t < ∞)
Because each point p on l is determined as a value is assigned to t, above equations are called parametric equations of a line.
(c) We can eliminate the parameter t in the equation above to achieve an equation relating x, y and z
directly as below :
x − xo
y − yo
z − zo
=
=
a
b
c
Above equations are of a line in R3 through a point (xo , yo , zo ) parallel to the vector < a, b, c >. We
call them symmetric equations of a line.
3.Example Find (1)parametric equations and (2) symmetric equations for the line :
(a) The line through the point (1, 3, 2) whose direction vector is (1, 2, 3)
(b) The line through the points (6, 1, −3) and (2, 1, −3).
(c) The line through (1, −1, 1) and parallel to the line x + 2 =
1
1
y = z − 3.
2
(Answers)
(a) (1)
y−3
z−2
x−1
=
=
,
1
2
3
(b) (1)y = 1, z = −3,
(c) (1)
(2)x = t + 1, y = 2t + 3, z = 3t + 2.
(2)x = t, y = 1, z = −3.
x−1
y+1
z−1
=
=
,
1
2
−1
(2) x = t + 1, y = 2t − 1, z = −t + 1.
4.Planes in R3
Let ro =< xo , yo , zo > and n =< a, b, c > be two fixed vectors.
Then there exists a unique plane through < xo , yo , zo > perpendicular to v. That is, a plane in R3 is uniquely determined by a
point it is through and a vector it is perpendicular to.
5.Equation (Equations of a plane)
(a) Let α be the plane through a point ro =< xo , yo , zo > perpendicular to the vector n =< a, b, c >. Let p =< x, y, z >
be a point on the plane. Notice that any vector whose
initial and terminal points are in α is perpendicular to n.
Therefore, p − ro ⊥ n. Equivalently,
(p − ro ) · n = 0
Be reminded that ro and n are fixed vectors. Above equation is called a vector equation of a plane. We call the vector
n =< a, b, c > the normal vector of the plane α.
(b) The vector equation above can be rewritten componentwise ; < x−xo , y−yo , z−zo > · < a, b, c >= 0.
That is,
a(x − xo ) + b(y − yo ) + c(z − zo ) = 0
Above equation is an equation of a plane in R3 through (xo , yo , zo ) perpendicular to the vector
< a, b, c >. We call it a scalar equation of a plane.
(c) The scalar equation above can be written as ax + by + cz = axo + byo + czo . Notice the right side of
the equation is a constant. Generally, in R3 , an equation of the form
ax + by + cz = d
represents a plane. The plane is perpendicular to the vector < a, b, c > and passes through a point
(xo , yo , zo ) where axo + byo + czo = d. We call above equation a linear equation in x, y, and z of a plane.
Notice that the vector < a, b, c > made from the coefficients of x, y, and z is the normal vector of the
plane.
6.Example Find an equation of the plane :
(a) The plane through the point (4, 0, −3) and with normal vector j + 2k
(b) The plane through the point (−2, 8, 10) and perpendicular to the line x = 1 + t, y = 2t, z = 4 − 3t
2
(c) The plane through the point (−1, 6, −5) and parallel to the plane x + 3y − 2z + 2 = 0
(d) The plane through the points A(3, −1, 2) B(8, 2, 4), and C(−1, −2, −3)
(e) The plane through the point P (6, 0, −2) and contains the line x = 4 − 2t, y = 3 + 5t, z = 7 + 4t.
(Answers)
(a) 0(x − 4) + 1(y − 0) + 2(z + 3) = 0
;
y + 2z + 6 = 0
(b) (x + 2) + 2(y − 8) − 3(z − 10) = 0
;
x + 2y − 3z + 16 = 0
(c) 1(x + 1) + 3(y − 6) − 2(z + 5) = 0
;
x + 3y − 2z − 27 = 0
−
−
→ −→
(d) n = AB × AC =< −13, 17, 7 >
;
−13(x + 1) + 17(y + 2) + 7(z + 3) = 0
;
−13x + 17y + 7z + 42 = 0
(e) By putting t = 0 and t = 1, you find two points in the line. Let them A(4, 3, 7) and B(2, 8, 11). Then the normal vector of the plane is
−→ −
−
→
n = P A × AB =< −2, 3, 9 > × < −2, 5, 4 >=< −33, −10, −4 >
Answer; −33(x − 6) − 10(y − 0) − 4(z + 2) = 0; 33x + 10y + 4z = 190.
7.Example
(a) Find an equation of the line of intersection of the planes x + y + z = 1 and x + z = 0
(b) Find parametric equations for the line through (2, 4, 6) that is perpendicular to the plane x−y +3z =
7
(c) Find the point at which the line intersects the given plane :
z
3 − x = y − 2 = ; x − y + 2z = 9
5
(Answers)
(a)
x
z
=
, y = 1.
1
−1
(b) x = t + 2, y = −t + 4, z = 3t + 6.
(c) Parametric equations for the line : x = 3 − t, y = t + 2, z = 5t.
line intersects the plane when t = 1. The answer is (2, 3, 5).
By putting these equations into x − y + 2z = 9, we find that the
8.Formula (Distance from a point to a plane) The distance D from a point (x1 , y1 , z1 ) to a plane ax + by + cz + d = 0
is
D=
|ax1 + by1 + cz1 + d|
√
a2 + b2 + c2
(Proof)
Let (xo , yo , zo ) be a point in α and therefore axo + byo + czo + d = 0. Let
b =< xo − x1 , yo − y1 , zo − z1 > and n =< a, b, c >, the normal vector of α.
Then the distance between the plane and (x1 , y1 , z1 ) is the length of the projection vector
projn b, as seen in the figure. Hence
D
=
b · n kprojn bk = n · n n
=
|a(x0 − x1 ) + b(yo − y1 ) + c(zo − z1 )|
√
a2 + b2 + c 2
=
|ax1 + by1 + cz1 − (axo + byo + czo )|
√
a2 + b2 + c2
=
|ax1 + by1 + cz1 + d|
√
a2 + b2 + c2
9.Example
(a) Find the distance from the point (−6, 3, 5) to the plane x − 2y − 4z = 8.
(b) Find the distance between the two parallel planes 6z = 4y − 2x, 9z = 1 − 3x + 6y
(Answers)
40
(a) √ .
21
(b) From the plane 6z = 4y − 2x, choose any point such as P (3, 0, −1). Find the distance from P to −3x + 6y − 9z + 1 = 0. Answer is
1
√ .
3 14
3