What is the relationship between vectors perpendicular to (1,1,1) and (1,2,3)?

AI Thread Summary
Vectors that are perpendicular to (1,1,1) and (1,2,3) lie on a line, as established by the relationship derived from their dot products. The condition A dot (1,1,1) = 0 and A dot (1,2,3) = 0 leads to the equation 2c = -b, allowing for the representation of vector A as <c, -2c, c>. This implies that all such vectors can be expressed in terms of a single variable, indicating they lie along a line. Although there are two unit vectors perpendicular to both, they share the same direction, reinforcing the concept of a unique line of perpendicular vectors. Thus, the relationship between the vectors is fundamentally linear.
Yosty22
Messages
182
Reaction score
4

Homework Statement



The vectors that are perpendicular to (1,1,1) and (1,2,3) lie on a ____.

Homework Equations


The Attempt at a Solution


This is really straight forward, but I cannot validate the answer to myself. The textbook says that they should lie on a line, but why is this? Obviously if a vector, say A = <a,b,c> is perpendicular to (1,1,1) and (1,2,3), A dot (1,1,1) = A dot (1,2,3) = 0. This means a+b+c = a + 2b + 3c, or 2c = -b.

How does this result let you know that any vector A such that A is perpendicular to both (1,1,1) and (1,2,3) has components <a,-2c,c>? Is it because A can be anything and it only depends on the other two components?

Thanks in advance.
 
Physics news on Phys.org
Try visualizing it, or holding up two pencils, it really doesn't matter which direction you point them in, there is a unique unit vector that is perpendicular to both. That is there is a unique "direction" in which a vector can point such that, that vector is perpendicular to your two vector. i.e. a line.
 
Yosty22 said:

Homework Statement



The vectors that are perpendicular to (1,1,1) and (1,2,3) lie on a ____.

Homework Equations


The Attempt at a Solution


This is really straight forward, but I cannot validate the answer to myself. The textbook says that they should lie on a line, but why is this? Obviously if a vector, say A = <a,b,c> is perpendicular to (1,1,1) and (1,2,3), A dot (1,1,1) = A dot (1,2,3) = 0. This means a+b+c = a + 2b + 3c, or 2c = -b.

How does this result let you know that any vector A such that A is perpendicular to both (1,1,1) and (1,2,3) has components <a,-2c,c>? Is it because A can be anything and it only depends on the other two components?

Thanks in advance.
You haven't quite finished. Now that you know that 2c = -b, which means b = -2c, you can replace every occurrence of the variable b with its equivalent value of -2c. This means, for the first equation a + b + c = 0, that a - 2c + c = 0, which allows you to solve for the value of a in terms of the single variable c. What does this imply about the character of A?
 
So if you know that b = -2c, and you know that a+b+c = 0, then a = c. This means that the vector A is made up of components <c,-2c,c>. So is this saying that since you can describe vector A with just a single variable (c) that is is described by a line?
 
Kind of, because a vector of that form will be a scalar multiple of the vector <1,-2,1>, but the generalization you made doesn't quite hold true. Could you think pf any examples where a vector is in terms of only one variable but does not describe a line?
 
MostlyHarmless said:
Try visualizing it, or holding up two pencils, it really doesn't matter which direction you point them in, there is a unique unit vector that is perpendicular to both..

Actually there are two unit vectors perpendicular to both but they line on the same line.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

Similar threads

Replies
3
Views
1K
Replies
69
Views
8K
Replies
32
Views
2K
Replies
12
Views
2K
Replies
4
Views
2K
Replies
3
Views
2K
Replies
3
Views
2K
Back
Top