If A and B are the vectors, then AXB will be a vector orthogonal to A and B.
| i j k |
|11 0|
|10 1|
Solve this determinant to get AXB.
Question 1.
Find a unit vector that is orthogonal to both and
My attempt:
Now I have 3 variables and 2 equations. What do I do?
Question 2.
If , find a vector such that
My attempt:
Once again, I havmore variables than the # of equations. What should I doo?
Thanks in advance!
Yes, because the "orthogonal to" determines only about the direction, not the length. You have the third equation . Alternatively, take any one of the variables, say x, equal to 1 and solve for the other two. Then normalize to get a unit vector.
Question 2.
If , find a vector such that
My attempt:
Once again, I havmore variables than the # of equations. What should I doo?
Thanks in advance!
Um let's step back in your example.
a + b = 0
a + c = 0
How do you know which variable to substitute with any number that doesn't equal to 0? If I let b = 1 in the 2 equations above, I get
<-1, 1, 1>
and if I let a = 1 I get
<1, -1, -1>
and if I let c = 1 I get
<-1, 1, 1>
Thus, they are not the same. They have the same magnitude, but not the same direction.
Am I having a brain fart or am I missing something?
Thanks a ton btw.
how are you getting <c, -c, -c> ? If you let c = 1 in what equations?
a + b = 0
a + c = 0
If c = 1,
a + 1 = 0
a = -1
(-1) + b = 0
b = 1
which equals:
<-1, 1, 1>
and you have <c, -c, -c> :\
Thanks!
EDIT: Just read matheagle's post
Okay that makes sense. So it doesn't matter if you let a = 1, b = 1, or c = 1. Right?
but how is Plato getting <c, -c, -c>
So I was looking through the back of the book and the answer I saw was: <1, -1, -1> * 3^(-1/2).
I am still confused why we can just plug in any # we want for one of the variables. I mean it's a vector so it doesn't have infinite # of solutions like a line has.