Steven A. Jones
March 8, 2006
BIEN 501
Geometric Interpretations of Vector Relationships
Projection into a Plane
The equation for a plane is
x  y  z  C0 ,
Eq. 1
where  ,  ,  and C 0 are constants. Such
a plane is shown in Figure 1. The plane
intersects the x ¸ y , and z axes at specific
0, C0  , 0
points (marked with ) which can be found by
setting two coordinates to zero. For example,
the intersection with the x axis is obtained by
setting y and z to zero. Then from Equation 1,
(C0  , 0, 0)

x
0, 0, C0  
x  C0  x  C0  .
Figure 1: A plane in Cartesian
If a normal vector to a plane is given, then the
space
equation for the plane can be readily obtained.




Let x  xi  yj  zk be any vector in the plane, then for that vector to be in the plane, its dot
product with the normal to the plane must be zero (because (1) the dot product of two vectors
that are normal to one another is zero and (2) any vector in the plane must be normal to the
plane’s normal vector). We can therefore write the equation for the plane in vector form as:
 
nx 0













Specifically, if n  i  j  k then i  j  k  xi  yj  zk  0 or x  y  k  0 , as in
Eq. 1.
Example: Find the equation for the x-y plane.
Solution: The normal vector to the x-y plane is n  1k . Therefore, the equation for the plane is


 


 
n  x  0  1k  xi  yj  zk  0  z  0 . This result is intuitive since it simply says that the
plane in question is the z  0 plane.
Example: You are standing up facing north. Your upper arm is extended such that it points
northeast and downward by 10 degrees. Set up a coordinate system where your elbow is the
origin, the x-axis points to the North, the y-axis points to the East, and the z-axis points upward.
Find a direction vector that describes your upper arm.
Solution: Let v  M a, b, c  represent the correct vector, where a, b, c  is a unit vector along
the correct direction.
We can find the equation for a plane that passes through two vectors that intersect at the origin
by first finding a normal vector to that plane. Since the cross product of two vectors is a vector
Steven A. Jones
March 8, 2006
BIEN 501




normal to both vectors, then if p and q are known vectors in the plane, p  q is normal to both


p and q . Therefore, an equation for the plane is:
 p  q   x  0
Example: Say you are standing up facing north. Your upper arm is extended such that it points
northeast and downward by 10 degrees, and your forearm is pointed north northwest and
upward by 20 degrees. Find an equation for the plane that includes your upper arm and
forearm.
Solution: Your upper arm and forearm are attached at the elbow. Let the elbow be the origin of
a Cartesian coordinate system. The vector describing your upper arm then goes upward by 10
degrees and southeast. It can be described as Vu  Lu sin  4 , cos 4 , sin 10

 
If the plane passes through the origin, then C0  0 . A vector within that plane, say




v  xi  yj  zk must be such that its dot product with the plane’s normal vector

 

 
n  ai  bj  ck is 0. Thus, if we know the normal vector, we can write n  v  0 , which means
that ax  by  cz  0 .




a 2  b2  c 2
1. What is the magnitude of the vector v  ai  bj  ck ? Answer:

2. If a  1 , b  2 , and c  2 , what is v ? Answer:



12  22  22  9  3
3. If p , q , and r are vectors (you may express answers in dot and cross products):

  


 
a. What is the projection of r on q ? Answer: r  q q or r  u , where u is the unit

vector in the direction of q .
b. What is the equation for the plane passing through the origin that is





 
perpendicular to r ? Answer: r  x  0 , where x  xi  yj  zk . This answer

arises because any vector in the plane must be perpendicular to r , so its dot

product with r must be zero.




  


r ? Answer: r  x  x0   0 , where x  xi  yj  zk .
c. What is the equation for the plane passing through x 0 that is perpendicular to


d. What vector is normal to the plane that passes through p and q ? Answer: A


vector that is normal to the plane is normal to both p and q , and thus can be
  
represented by n  p  q .
e. What is the equation for the plane that passes through the origin and is parallel to


both p and q ? Answer: The solution comes from a combination of b and d.
I.e., if we know a vector that is normal to the plane (question b), then we can find
 
 
the equation for the plane from n  x  0 . But the normal to the plane is p  q ,

 
from d. Therefore the equation is  p  q   x  0 .
Steven A. Jones
March 8, 2006
f.
BIEN 501
What is the equation for the line passing through the origin and directed along

the vector p ? Answer: A line in space is represented by a pair of simultaneous
equations. This representation may not be obvious when one examines the well
known line y  mx  b . However, even in that case one should note that two
equations are required, the relationship between x and y , and the statement

that the line is in the x  y plane, i.e. z  0 . The line along p must be parallel





to p . Thus, if x  xi  yj  zk is a vector along that line, its cross product with
 

p must be zero. Hence the equation for the line is p  x  0 . If this is written
out in detail, then

i
px
x

j
py
y

k



p z  0   p y z  p z y i   p x z  p z x  j   p x y  p y x  k  0 .
z
which yields three simultaneous equations:
p y z  pz y  0
px z  pz x  0
px y  p y x  0
While it may appear at first that the problem is overspecified (there are three
equations and we require only two), close inspection shows that one of these
equations is automatically satisfied if the other two are satisfied. In mathematical
terms, the third equation is a linear combination of the first two; if the first is
multiplied by p x , the second is multiplied by p y , and the two are subtracted, the
terms containing z cancel and the result is:
 px pz y  p y pz x  0  px y  p y x  0 ,
so the third equation is satisfied, as promised.
g. What is the equation for the line defined by the intersection of (1) the plane




passing through p and q and (2) the plane passing through r and s . Answer:
The line must slmultaneously obey the equations for the two planes. Thus, we
  
  
must have  p  q   x  0 and r  s   x  0 . More succinctly, a vector in the
 
direction of the line is perpendicular to both the normal to the first plane, p  q
 
and the normal to the second plane, r  s . Once we have this vector, the result
of f can be applied. Therefore, the equation for the line is:
 p  q   r  s   x  0 .


h. What is the angle between r and q . Answer: If the angle is  , then from the
 


 

dot product, r  q  r q cos  , so   cos 1 r  q r q . In terms of the cross
 
product,   sin  r  q r q  .
1
Steven A. Jones
March 8, 2006
i.
BIEN 501


What is the projection of a vector q on a plane normal to n (assume that all
three vectors begin at the origin)? Answer: The projection vector is in the plane.

It is also in a plane that passes through the vector q and is normal to the plane,


i.e. a plane that passes through both q and n . The vector therefore must be



perpendicular to both n and n  q . The only vector direction perpendicular to




both of these vectors is b  n  n  q  , and the corresponding unit vector is
  
  
n  n  q  m , where m  n  n  q  Therefore, the projection of q along this
   
  
direction is q  n  n  q  m n  n  q  m .


j. What is the angle between r and the projection of q on the plane passing


  
through r and p . Answer: n  r  p  is the normal to the plane. Let

   

s  q  r  p  , which is normal to the plane containing both q and n . The


angle between r and the projection is the complement of the angle between r
 
 

and s . Since r  s  r s cos90    ,   90  cos 1 r  s rs  , or
   
  
  90  cos 1 r  q  r  p  r q  r  p   .




4. Find a unit vector in the same direction as v  1i  1 j  2k . Answer:

 
   i  j  2k
.
uv v 
6

5. Sketch both v and the unit vector in the x, y, z coordinate system.
y
z


u

v
x
6. Let r be a vector. What geometric object is described by the equation r  2 . Answer:
A sphere.
7. Sketch the object in 6 above.
Steven A. Jones
March 8, 2006
BIEN 501
8. If f x, y, z   x sin  y e xz , find the gradient of f . Answer:



f  z sin  y  e xz i  x cos y  e xz j  x 2 sin  y  e xz k .
9. Find an equation that describes all of the planes that are perpendicular to the gradient of
f at any location in space. Answer: The gradient of f evaluated at some location
 df  df  
. A plane perpendicular to
i 
j
k
dy
dz  x  x ; y  y z  z
 dx
0
0
0

this vector is then f x0 , y0 , z0   x  0 . Note that it is necessary to first take the
x0 , y0 , z0  is f x0 , y0 , z0    df
gradient, then find its value at the given point, and then find the equation for the plane.