Calculus and Vectors – How to get an A+

9.6 Distance from a Point to a Plane ©2010 Iulia & Teodoru Gugoiu - Page 1 of 2

9.6 Distance from a Point to a Plane A Distance from a Point to a Plane Let consider a plane π with a normal vector nr and a point ),,( 0000 zyxP on this plane. The distance from a point ),,( 1111 zyxP to the plane π is given by the

scalar projection of the vector 10PP onto the normal vector n

r :

|||||| 10

nnPPd r

r⋅

= (1)

Ex 1. For each case, find the distance between the given plane and the given point. a) )1,0,2()2,1,0()2,0,1( str ++=

r , )0,3,2(B

2117

4161|4121|

||||||

)2,4,1(0210210210

)2,3,1()2,0,1()0,3,2(

0

0

=++++

=⋅

=∴

−==×=

−=−=

nnBPd

jikjivun

BP

r

r

rrrrr

rrr

b) ⎪⎩

⎪⎨

⎧

−+−=−−=+−=

stzstystx

32122

1 )2,0,1( −M

591

9149|320|

||||||

)3,1,7()12,32,43(2132111211

)3,2,1();2,1,1()1,2,0()1,2,1()2,0,1(

0

0

=++

−+=

⋅=∴

−=+−+=−−−−−−−=×=

−−=−−=−−=−−−=

nnMPd

n

jikjivun

vuMP

r

r

r

rrrrr

rrr

rr

B Distance from a Point to a Plane (II) If the plane π is given by the Cartesian equation

0: =+++ DCzByAxπ , then the distance from a point ),,( 1111 zyxP to the plane is given by:

222111 ||

CBA

DCzByAxd++

+++= (2)

Indeed,

DCzByAxCzByAxCzByAxCBAzzyyxxnPP

DCzByAxP

+++=−−−++=⋅−−−=⋅

=+++⇒∈

111000111

01010110

0000

),,(),,(

0rπ

Ex 2. Find the distance between the point )3,0,2(−R and the plane 0632: =−+− zyxπ .

214

147

194|634|

1)3(2

|63)0(3)2(2|222

==++

−+−=

+−+

−+−−=∴d

C Distance between two Parallel Planes To get the distance between two parallel planes: a) Find a specific point into one of these planes. b) Find the distance between that specific point and the other plane using one of the formulas above.

Ex 3. Find the distance between the parallel planes. 03963:1 =−−+ zyxπ , 04642:2 =−−+ zyxπ

141

562

36164|4)0(6)0(4)1(2|

)0,0,1( 11

==++

−−+=∴

∈

d

P π

Calculus and Vectors – How to get an A+

9.6 Distance from a Point to a Plane ©2010 Iulia & Teodoru Gugoiu - Page 2 of 2

Ex 4. Consider a plane π with the x-, y-, and z-intercepts equal to a , b , and c respectively. a) Find the Cartesian equation of the plane π .

01

0

int

int

int

0

=−++

=+−

+−

+−

−=⇒=−=−

−=⇒=−=−

−=⇒=−=−

=+++

cz

by

ax

DzcDy

bDx

aD

cDCc

CDz

bDBb

BDy

aDAa

ADx

DCzByAx

b) Find the distance from the origin to the plane π .

222222111

1111

|1000|

cbacba

cbad++

=++

−++=

Ex 5. Consider the plane 01032: =+−+− zyxπ and the point )0,3,2(P . a) Find the vector equation of the line ⊥L that passes through the point P and is perpendicular to the plane π . The line ⊥L is perpendicular to the plane π and therefore is parallel to the normal vector )3,2,1( −−=n

r to the plane π . The normal vector )3,2,1( −−=n

r may be used as direction vector for the line ⊥L . The vector equation of the line ⊥L is:

RttrL ∈−−+=⊥ ),3,2,1()0,3,2(:r

b) Find the point of intersection F between the perpendicular line ⊥L and the plane π .

)3,1,3(3)1(31)1(233)1(2

:11414

010)3(3)23(2)2(323

2:

Fzyx

Ftt

ttttztytx

L

∴

⎪⎩

⎪⎨

⎧

=−−==−+==−−=

⇒−=⇒−=

=+−−++−−

⇒⎪⎩

⎪⎨

⎧

−=+=−=

⊥ π

c) Find the shortest distance between the point P and the plane π .

14941||)3,2,1(|||||| =++=−== PFd

Reading: Nelson Textbook, Pages 542-549 Homework: Nelson Textbook: Page 549 # 1, 2, 3, 5, 6