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					                  Topic 4: Vectors: Summary Information
LINES IN 3 DIMENSIONS
                                                                                           l 
Consider a line in 3 dimensions passing through the point (a, b, c) parallel to the vector  m 
                                                                                            
                                                                                           n 
                                                                                            
Cartesian equation of a line:                       Vector equation of a line
               xa y b z c                                               a l 
                             
                 l       m      n                                     r  b   t  m
                                                                             
                                                                          c  n 
                                                                             
                                                          or r = (ai + bj + ck) + t(li + mj + nk)

Example:
Find the equation of the line passing through the points A(3, 5, 2) and B(2, -4, 5).

Solution:
                                                                         2   3   1 
Find the direction of the line: One possible direction vector is AB   4    5    9 
                                                                             
                                                                        5   2 3 
                                                                             
                                        x3 y 5 z 2
The Cartesian equation of this line is                       (using the coordinates of point A).
                                         1      9          3
                                     x   3   1 
The equivalent vector equation is  y    5   t  9  .
                                         
                                     z   2 3 
                                         

PLANES IN 3 DIMENSIONS
There are various alternative ways to write the equation of a plane:
The equation of a plane passing through the point                  Vector equation of a plane
( a1 , a2 , a3 ) , in the direction of the two vectors                  a1   b1   c1 
 b1        c1                                                  r   a2   t  b2   s  c2 
                                                                            
 b  and  c  is …                                                   a  b  c 
 2         2                                                        3  3  3
b          c3                                                      or r = a + tb + sc
 3         
The equation of a plane perpendicular to the vector              Cartesian equation of a plane
     e 
n   f  , is …                                                         ex + fy + gz = d
      
     g 
                                                                        or     r .n  d
(n is called the normal vector)
The equation of a plane perpendicular to the vector
     e                                                                   (r  a). n  0
n   f  and passing through the point (a1 , a2 , a3 ) is
      
     g                                                                  or   r.n  a.n
      
THE SCALAR PRODUCT (or DOT PRODUCT)
The scalar product is a way of combining two vectors together to get a scalar quantity.

Key formula: a . b  a b cos        where a is the magnitude of vector a
                                           b is the magnitude of vector b
                                           θ is the angle between vectors a and b.

                                       a1   b1 
The scalar product is found using      a  . b   a b  a b  a b
                                       2  2       1 1   2 2   3 3
                                      a  b 
                                       3  3

Properties of the scalar product:
   1)      a.b=b.a
   2)      If a and b are perpendicular then a . b = 0 (and vice versa)


THE VECTOR PRODUCT (or CROSS PRODUCT)
The vector product is a way of combining two vectors together to get another vector quantity.

Key formula: a  b  (a b sin  )n
                                 ˆ                   ˆ
                                             where n is a vector perpendicular to both a and b
                                             a is the magnitude of vector a
                                             b is the magnitude of vector b
                                             θ is the angle between vectors a and b.

                                       a1   b1         i a1   b1 
The vector product is found using      a    b   det  j a    b2 
                                       2  2            k a2
                                      a  b 
                                       3  3               3   b3 
                                                                      

Properties of the scalar product:
   1) a × b = - b × a
   2) a × b gives a vector perpendicular to both of the original vectors.

Example:
                                    1        2
Find a vector perpendicular to both  4  and  0  .
                                     
                                     2      
                                              3
                                             
Solution:
We find the vector product of these two vectors:
        1   2            i 1 2                                  12 
         4    0   det  j 4 0   i 4 0  j 1 2  k 1 2   7  .
           
         2   3          k 2 3      2 3       2 3     4 0  
                                                                8 
We can check our answer by finding the scalar product of the answer with each of the original
vectors. We should get 0.
                                  COMMON QUESTIONS

FINDING THE CARTESIAN EQUATION OF A PLANE GIVEN 3 POINTS
Example: Find the Cartesian equation of the plane passing through the points (2, 0, -1), (3, 1, 2)
and (1, 2, 5).

Solution: There are 2 methods we can use:

Method 1: Finding the vector equation and then eliminating the parameters
Two possible direction vectors are:
         3   2  1          1   2   1
        1    0   1  and  2    0    2 
             
         2   1  3              
                                  5   1  6 
                                
Therefore a possible vector equation is:
                                       x  2         1       1
                                       y    0    1     2 
                                         
                                       z   1        
                                                         3      
                                                                 6 
                                                            
To find this equation in Cartesian form, we eliminate the parameters from the equations.
We first form 3 equations:
        x  2              i.e.      x2                  (1)
        y  0    2         i.e.          y    2           (2)
        z  1  3  6       i.e.      z  1  3  6          (3)

We can eliminate λ by finding (2) – (1):     y – x + 2 = 3μ
                                          1               1  1    2
                       i.e.              ( y  x  2)  y  x 
                                          3               3  3    3

                                                           1      1     2   1     2      4
From equation (2), we have:              y  2  y  2  y  x    y  x 
                                                           
                                                           3   3   3   3  3    3

                                                1    2  4     1   1   2
Substituting into equation (3) gives: z  1  3  y  x    6  y  x    3 y
                                                3    3  3     3   3   3
This gives the equation of the planes as      3y – z = 1

Method 2: Finding a normal vector using the vector product
We start by finding the vector product of the two direction vectors:
        1   1          i 1 1                                                  0 
        1    2  = det  j 1 2   i 1 2  j 1 1  k 1 1  0i  9 j  3k   9 
          
        3  6           k 3 6       3 6       3 6        1 2                      
                                                                                      3 
                                                                                 
This normal vector gives the coefficients of x, y and z in the equation of the plane. Therefore we
know that the plane has equation
                                      0x – 9y + 3z = d.

Substitute in the coordinates of one of the points, e.g. (1, 2, 5), to get -18 + 15 = d
                                                                         i.e.     d = -3
So the planes has equation -9y + 3z = -3
                       OR      3y – z = 1
FINDING THE LINE OF INTERSECTION OF TWO PLANES
Two non-parallel planes intersect to form a line. We can find the equation of that line as follows:

Example: Find the line of intersection of the planes            2x –y + 2z = -6
                                                                x + 2y + z = 2
Solution:
Because there are 3 unknowns but only two equations, we introduce a parameter t. The equations
then are
                                   2x – y + 2z = -6
                                    x + 2y + z = 2
                                             z=t

Substituting the 3rd equation into the top two equations and rearranging gives:
                                        2x – y = -6 – 2t     (1)
                                        x + 2y = 2 – t       (2)
Multiply the top equation by 2:
                                        4x – 2y = -12 – 4t
                                         x + 2y = 2 – t
Add to eliminate y:
                                        5x = -10 – 5t        i.e. x = -2 – t.

Substitute this into equation (1):
                                        2(-2 – t) – y = -6 – 2t
                                              -4 –2t = -6 – 2t + y
                                                y = 2.

The solutions therefore are x = -2 - t, y = 2 and z = t.

Writing this in vector form, we get:

         x   2  t   2   1
         y  2        2   t 0 
          
         z  t           
                         0  1 
                        

             2   1
i.e.    r  2   t 0 
               
             0  1 
               

                                                                                             1
This is the equation of a line through the point (-2, 2, 0), in the direction of the vector  0  .
                                                                                             
                                                                                            1 
                                                                                             
FINDING THE INTERSECTION OF A PLANE AND A LINE
There are several possibilities:
    the line could lie within the plane;
    the line could intersect the plane at a single point;
    the line could be parallel to the plane.

Example: Investigate which of the three situations above applies with the line
                                      x  5 y  9 z 1
                                                    
                                       1        6      2
and the plane 2x – y + 4z = 5.

Solution: To investigate this situation, it is simplest if we have the equation of the line in vector
form and the equation of the plane in Cartesian form (which it is already).

                                        x   5   1
The vector equation of the line is:     y   9  t  6 
                                            
                                        z  1   2 
                                            
i.e.           x=5–t
               y = 9 + 6t
               z = 1 + 2t

Substituting these into the equation of the plane gives:

               2x – y + 4z = 5
               2(5 – t) – (9 + 6t) + 4(1 + 2t) = 5
               10 – 2t – 9 – 6t + 4 + 8t = 5
               5=5

This equation is true for all values of t. Therefore the line must lie within the plane.

Note 1: If when we attempt to solve the equations, we get an equation which is true for no values of
t (such as 4 = -2), we know that the plane and the line do not intersect, i.e. are parallel.

Note 2: If when we attempt to solve the equations, we get an equation which is true for a single
value of t, we know that the plane and the line intersect at a single point. We can find the point of
intersection by substituting the value of t into the vector equation of the line.
FINDING THE ANGLE BETWEEN TWO PLANES
The angle between two planes is the same as the angle between the two normal vectors.

                                                                    3              1 
Example: Find the acute angle between the planes with equations: r.  0   2 and r.  2   1
                                                                     
                                                                     1             
                                                                                     5
                                                                                    
Solution: The equations of the planes are 3x + 0y – z = 2 and x + 2y + 5z = -1.
                       3        1 
The normal vectors are  0  and  2  .
                        
                        1       
                                  5
                                 
                         3  1 
The scalar product is:  0  .  2   (3  1)  (0  2)  (1  5)  2
                           
                         1  5 
                           
                        3                                  1 
The magnitudes are:      0   9  0  1  10 and  2   1  4  25  30
                         
                         1                                 
                                                             5
                                                            
Substituting into the formula a.b=abcosθ gives:
                                                                       2
                2  10  30 cos                         cos                96.6
                                                                     10  30
The acute angle between these planes therefore is:
                180 – 96.6 = 83.4°.


FINDING THE ANGLE BETWEEN A PLANE AND A LINE
To find the angle between a plane and a line, there are 2 steps:
Step 1: Find the angle between the normal vector and the direction vector of the line;
Step 2: Subtract the angle from step 1 from 90° in order to get the angle required.

                                                                              4 
                                            x2 y z 3
Example: Find the angle between the line                   and the plane r.  3   5 .
                                              2    1    1                     
                                                                               2 
                                                                               
Solution:
                                            2      4 
First we find the angle between the vectors 1  and  3  :
                                             
                                             1     
                                                      2 
                                                    
       2  4 
       1  .  3   8  3  2  13
          
        1  2 
          
       2                               4
       1  4 11  6                  3  16  9  4  29
       1                              2
Therefore:     13  6  29 cos               9.76

So the angle between the plane and the line is 90 – 9.76 = 80.2°.
FINDING THE PERPENDICULAR DISTANCE FROM A POINT TO A LINE
Example: Find the perpendicular distance from the point P(3, 5, 2) to the line with equation
                                             2  1 
                                       r   3   t  1 .
                                               
                                             1  4 
                                               

Solution:
The line passes through the point A(2, 3, -1) and has direction i – j + 4k.

Let Q be the closest point on the line to P.                          P(3, 5, 2)
Then PQ is perpendicular to the direction of the line.
                                                                                             1   
                                                                                            –1   
As Q lies on the line, the position vector of Q will take the form:                          4   
               2  1   2  t                                                  Q
        OQ   3   t  1   3  t 
                  
               1  4   1  4t    
                                   
                                                                         A(2, 3, -1)
for some t.

                        2  t   3   1  t 
Therefore PQ  q  p   3  t    5    2  t  .
                       
                        1  4t   2   3  4t 
                                                
                                               

Since PQ is perpendicular to the direction of the line, we have:

                        1  t   1 
                         2  t  .  1   0
                       
                        3  4t   4 
                                  
                                 
                      1  t  2  t  12  16t  0
                      18t  11
                            11
                      t
                            18

So, by substituting this value for t into the expression for PQ , we get
                      1  18 
                            11
                                       7 
                           11     1
                PQ   2  18        47 
                      3  44        10 
                                   18      
                           18            

The shortest distance from point P to the line is the magnitude of this vector:

                        7
                      1       1
                PQ     47   49  2209  100  2.70
                     18      18
                        10
FINDING THE PERPENDICULAR DISTANCE FROM A POINT TO A PLANE
Example: Find the perpendicular distance from the point P(3, 5, 2) to the plane with equation
                           3x – 2y + z = 4
                                                                          P(3, 5, 2)
Solution:
Let Q be the foot of the perpendicular from point P to the plane.
Then PQ will be perpendicular to the plane.

                              3 
However, the normal vector  2  is a vector
                               
                              1 
                                                                                       Q (x, y, z)

                               
perpendicular to the plane. So
                                                                3x - 2y + z = 4
                3 
       PQ  t  2  .
                 
                1 
                 
Therefore the equation of the line PQ is
           3 3 
       r   5   t  2  .
              
            2  1 
              

We can find the coordinates of Q by seeing where this line intersects the plane 3x – 2y + z = 4.
Substituting x = 3 + 3t, y = 5 – 2t, z = 2 + t into the equation of the plane, gives:
       3(3 + 3t) – 2(5 – 2t) + (2 + t) = 4
       9 + 9t – 10 + 4t + 2 + t = 4
       1 + 14t = 4
            3
So      t .
           14

                                3 
So the vector PQ is PQ     3    2  .
                           14    
                                1 
                                 

Therefore the perpendicular distance from P to the plane is
                  3
        PQ  14 2  14  9  4  1  0.802.
                3      3


                  1


Note: There is a formula (quoted in the formula book) which can also be used to find the
perpendicular distance from a point to a line. See the next page.
FINDING THE PERPENDICULAR DISTANCE FROM A POINT TO A PLANE
(FORMULA BOOK METHOD)

Formula quoted in the OCR formula book:
The perpendicular distance of ( ,  ,  ) from 1 x  2 y  3 z  d  0 is
                                             1  2   3  d
                                                  12  2  32
                                                         2




Example (continued): Find the perpendicular distance from the point P(3, 5, 2) to the plane with
equation
                           3x – 2y + z = 4

Solution: ( ,  ,  ) = (3, 5, 2).
The equation of the plane is 3x – 2y + z – 4 = 0.

Therefore the perpendicular distance is:
       1  2   3  d     (3  3)  ( 2  5)  (1  2)  4   3
                             =                                          0.802
            12  2  32
                   2
                                         32  (2)2  12             14




FINDING THE SHORTEST DISTANCE BETWEEN TWO SKEW LINES

You need to learn the following formula for finding the shortest distance between two skew lines:

If two lines l1 and l2 are given by r  a1  tb1 and r  a 2  sb 2 respectively, the shortest distance d
between l1 and l2 is given by:
              (a2  a1 ).n
        d                        where n  b1  b 2
                   n

                                                           2  1               3   2 
Example: Find the shortest distance between the lines r   3   t  1 and r   1  s  2 
                                                             
                                                          5  2                   
                                                                                   1  3 
                                                                                 
Solution:
            1   2  i 1 2                                                          1
       n    1   2   j 1 2  i 1 2  j 1 2  k 1 2   7 
               
             2   3  k 2 3                    2 3          2 3       1 2   4 
                                                                                  
                        3   2    1 1   1
       (a 2  a1 ).n   1   3   .  7    4  .  7   1  28  24  53
                                 
                        1  5    4   6   4 
                                 
Therefore the shortest distance is
             53           53
                               6.52
         1  49  16        66

				
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posted:8/26/2011
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