Transcript IB HL Maths - ISA-International School of Athens, Kifissia

IB HL Maths
Summary of key points for vectors
of planes and lines in space
Key points:
•
Vector equation of a plane.
•
Write the cartesian equation of a plane.
•
Normal vector to a plane.
•
Shortest distance from a point to a plane.
•
Cross product of 2 vectors.
•
Angle made between 2 intersecting planes.
•
Cartesian equation of a line in 3d.
•
Equation of a line made between 2 intersecting
planes.
Vector (or parametric) equation of
a plane:
r  a  b  c
a is a position vector in the plane.
 and  are parameters and b and c are nonparallel vectors that lie within the plane.
Cartesian equation of a plane:
ax  by  cz  d
a, b, c, and d are constant real numbers
Normal vector to a plane:
The normal is a line that meets a plane or another line at 90
If ax+by+cz=d (the equation of the plane) is known then the
normal to the plane will be
a
 .
b
c 
 

The dot product of any vector in the plane and the normal to
the plane will be 0.
The shortest distance from a point to a plane:
Find the shortest distance
from the plane x+2y-3z = 10
to the point A(2,-1,6).
The normal to the plane will
have the vector:
1 
 
 2 
3 
 
.

The equation
of the normal
will be:
x 
 
y  
z 
 

 2  1 
   
1 t 2 
 6  3
   
Now form parametric
equations and find t.
x=2+t, y=-1+2t, z=6-3t
2+t+2(-1+2t)-3(6-3t) = 10
so t=2.
If t=2, then the coordinate is
(4,3,0).
Now find the distance between 2
points (2,-1,6) and (4,3,0).
(4  2)2  (3  (1))2  (0  6)2
 56
The cross product of 2 vectors (in 3d) - axb
The cross product is used to find a perpendicular vector:
axb gives a vector that is perpendicular to a and b. This is
useful for vectors in a plane.
1
2 
 
 
a  2, b  3 
3
1
 
 
i j k 


a  b  1 2 3 
2 3 1


(2  9)i  (6  1) j  (3  4)k
11i  7 j  k
Vector is


11


7


 1 


Angle between two planes
The angle made between two planes is the same as the
angle made by the planes respective normals.
P1 has the equation 2x+y+3z=4 and P2 has the equation
3x-4y+z=5.
Find the angle made between the two planes.
Normal to P1 is 2i+j+3k, and normal to P2 is 3i-4j+k.
Angle between them is:
cos  
6  4  3

 14 

 26




  74.8

Equation of a line in 3d
Cartesian equation can be written in the form:
xa y b zc


l
m
n
This can be confusing, it is always much easier to write in
vector form:
a
 
r  b
c 
 
 l 
 
m
n 
 
Can you see the connection?
Line of intersection made between 2 planes
Find the line formed at the intersection of the planes:
P1: x-3y+2z=3 and P2: 2x+2y+z=5
Let x=t, P1: t-3y+2z=3 and P2: 2t+2y+z=5
Eliminate z by mulltiplying P2 by 2 and P2-P1:
3t+7y=7
t 
7  7y
3
7  3t 
21  7z
z

5

2t

2

t



Now find z from P2:
8
 7 

x 7  7y 21  7z


t
Cartesian equation is:
1
3
8


 