Direction Perpendicular
To The Plane
Take a pencil, stand it on table top.
Perpendicular.
Tilt the table and keep the pencil as it
was.
Still perpendicular.
This allows us to find a different vector
form of the equation of a plane.
You need to know the position vector of a
point on the plane.
You need to know the direction,
n=n1i+n2j+n3k, perpendicular to the
plane.
We want to find an expression for the
position vector r of a point R on the plane.
n
A
AR is on the plane. Therefore AR is
perpendicular to direction n.
R
Position vector r
Position
vector a
Vector AR is r-a
Since AR = r – a and AR.n=0
Then (r – a ).n=0
r.n-a.n=0
 x   n1 
  
 y . n2  − a.n = 0
z n 
  3 
n1 x + n2 y + n3 z + d = 0
where d = -a.n
- a.n will always be a number, the constant scalar.
Example
Write down the equation of the plane through the
point(5,2,1) given the vector 35  is perpendicular to
 −1
 
the plane.
5 
3 
 
 
a =  2  and n =  5 
1 
 − 1
 
 
n1x+n2y+n3z - a.n=0
3x+5y-z-(5x3+2x5+(-1)x1)=0
3x+5y-z-24=0
Intersection of a plane and a line
Find the point of intersection of the plane
1   2 
3x+2y-z=3 and the line
   
r =  4  + λ 1 
 x  1   2 
     
r =  y  =  4  + λ 1 
 z   3   − 1
     
3
 
 − 1
 
x = 1 + 2λ
y=
z=
Substitute into the equation of the plane.
Distance of a point from a plane
The shortest distance of a point, A, from a
plane is the distance AP, where P is the
point where the line through A
perpendicular to the plane intersects the
plane.
A
P
Example
a.
b.
c.
A is the point (5,2,1) and the plane has
equation 4x+2y-z=8. Find:
The equation of the line through A
perpendicular to the plane.
P, the point of intersection with the
plane.
The distance AP
a.
The direction, n, is taken from 4x+2y-z=8 and
the line through (5,2,1) gives us:
5   4 
   
r =  2  + λ 2 
1   − 1 
   
b. x= 5+4λ
y=
z=
Subin to the equation of the plane to find λ.
Then use this value to get the point P.
c. The vector AP is given by:


AP = p − a = 


| AP |=
 
 
 −
 
 
 
 
=
 
 





Page 322 Ex 11f 6 onwards