Notesale: Turn your study into money

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Finishing Cross Products
Recall we had


 
 
x2 y3 − x3 y2
y1
x1
 x2  ×  y2  =  −(x1 y3 − x3 y1 ) 
...
First x×y = −y×x
...

There are some nice, basic, properties
(ax) × y = a(x × y) ,

0×x=0 ,

x×x=0

(x + y) × w = x × w + y × w
Finally
∥x × y∥2 + (x · y)2 = (∥x∥ ∥y∥)2
...


Some Geometry
The relevant section in the text is 3
...
, xn )
if we want to be clear it’s a point, or in a more vector type form
...
xn ]
...
,

...

xn
A vector in these spaces can be described as the displacement between two points, so
[ ]
−→
−
1
x=
= OX,
using O = (0, 0), and X = (1, 2)
...
The
−→
−
−→
−
order is important: OX = −XO
...

As a result, there is very little distinction to be made between these, especially mechanically
...
You can, however,
add vectors, as we have seen
...


This misses out on some options, it’s better to forget the ‘y dependent on x’ setup and use
ay + bx = c
which allows, for instance, x = 2, a vertical line
...


...


...


...


...

Example: Here are two simple lines

   

x1
2
1
 x2  =  0  +  2  r,
x3
1
−1



 
 

x1
0
−2
 x2  =  −4  +  −4  s
...
To confirm that there are two procedures we can take, first,
we can simply equate the lines, confirm that the set of points is the same in each line
...

−1
2
−2


This is then a two by three system, just solve it
...
Here another way
...
To
find if two lines are the same, make sure they have the same direction to them, the vectors
should be simple multiples of each other
...

2
−1
They then have the same shape, are parallel, etc
...
For instance,
  
 

2
0
−2
 0  =  −4  −  −4  ,
so, s = −1
...
Just equate the
lines, so if we have
x = a + bt
,
y = c + ds
then we want to find if there are t, s such that
x=y

=⇒

a + bt = c + ds

which works out to the system
bt − ds = c − a
...

1
2
1
2
0
1
1
We get the set of equations
−1 = −t =⇒
1 = −s =⇒
0=s−t

t=1
s = −1

and if you check the last row, they do not match, so the lines do not intersect
...

-A single point of intersection, meaning they cross each other
...

Note that two lines in R2 that are not parallel will intersect
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3

Planes
We will stick to R3 for a while here, since planes have a particularly nice property in R3
...
They are FLAT
...

x = a + bt + cs
=⇒
x3
a3
b3
c3
That’s the parametric form of a plane
...
As before, we are on a plane, at the point a
...
We are in R3 , so there has to be a third
direction
...
We call this direction
the normal of the plane, usually written n
...

Expanding that out with the column vector notation leads to

 

x1
n1
 x2  ·  n2  = n1 x1 + n2 x2 + n3 x3
x3
n3

= a · n
...

This is frequently written
ax + by + cz = d
or something similar, depending on what variables we are using
...
The semi-official name for this expression:
the scalar equation of a plane
...

 This is not a big deal, usually
...
There are three values in  −2 , it’s not = 0, so there will be 2 vectors orthogonal
1
1
to it
...
The best way to make sure
they aren’t in the same direction is to make them cover different variables, i
...
, use only x
and y for one, y and z for the other
...
Those are clearly not in
0
1
2
the same direction
...
Recall that we need to match the right hand side
...
We just need a single point
that matches
...

2
0
0
2
So, any will do, here’s one final set
     
 
x
1
1
0
 y  =  1  +  2  t +  1  s
...
We only need
the orthogonal vectors, since we have a zero on the right
...
The remaining term is  0 
...

z
0
1
Note: To get the normal vector for the equation, just use the cross product on the two
vectors in the parametric form, then get a starting point
...

Example: Convert



    
−1
3
1
x1
 x2  =  0  +  1  t +  −1  s
2
2
1
x3


into the scalar equation form
...

−3 + 1
−2
2
2
This give us an equation of the form
4x1 − 8x2 − 2x3 = something
so we need to find a right hand side value
...
If we use  0 , the left hand
1
side will be
4(1) − 8(0) − 2(0) = 2
5

so our final answer is
4x1 − 8x2 − 2x3 = 2
...
If you are using the parametric equations, it is like solving
for intersections of lines, except there are more variables to worry about
...

Example: find the intersections of 2x − 3y + z = 2 and 4x − z = −1
...

First, we can re-arrange the second equation for z = 4x + 1
...

3
3
z
4x + 1
4
1
We get a line
...

-A line (they intersect, but are not parallel)
...

Compare with the available answers for two lines in R2 , where any line can be written
ax + by = c
...
The text uses a slightly different
scalar equation of a line:
  
x = 3+t
x
 y =
y = 1−t
equivalent to
z = −3 + 2t
z
4
...

7
...
(note: no answer at the back)
10
...
b),d)
17
...

6

format some times, called the
 

3
1
1  +  −1  t
−3
2



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