Non-collinear vectors

[elleocin]

Non-collinear vectors (2 q's)

Find three non-collinear vectors perpendicular to (2, -3, 1)

So far I've found the dot product of this vector with vector #2 which we'll call (x, y, z) and got the equation 2x - 3y + z

But how exactly are you supposed to find 3 other vectors using this equation Are there infinite possibilities?


This is also a similar question I am trying to figure out, but can't because I don't understand the first one:sad:

Find a unit vector that is parallel to xy-plane and perpendicular to the vector 4i - 3j + k

Step by step explanations please. Thanks in advance!

 

[pka]

Find three non-collinear vectors perpendicular to (2, -3, 1)
So far I've found the dot product of this vector with vector #2 which we'll call (x, y, z) and got the equation 2x - 3y + z

Can you find three solutions to \(\displaystyle 2x-3y+z=0~?\)

I can give you three: \(\displaystyle <3,0,-2>,~<3,2,0>,~<0,1,3>\)

Are those three non-collinear vectors?

 

[elleocin]

Yes! (3,2,0) and (0,1,3) are two of the answers. The third answer listed is (3,4,6).

What method did you use to come up with those solutions?

 

[pka]

Yes! (3,2,0) and (0,1,3) are two of the answers. The third answer listed is (3,4,6).

What method did you use to come up with those solutions?

Again the question implies that you are to find any three non-collinear vectors.
There are infinitely many possibles.

I use no method. I just randomly picked three that worked.

 

[HallsofIvy]

A vector, <x, y, z>, is perpendicular to <2, -1, 3> if and only if the dot product is 0, 2x- y+ 3z= 0.

One obvious choice is to take x= 3, y= 0, z= -2 so that 2(3)- (0)+ 3(-2)= 0.

Another is to take x= 1. ,y= 2, z= 0 so that 2(1)- (-2)+ 3(0)= 0.

A third is to take x= 0, y= 3, z= 1 so that 2(0)- (3)+ 3(1)= 0.

Do you see how I chose those? In the first, I took y= 0 and chose x and z so that 2x+ 3z= 0 or 2x= -3z. In the second, I took z= 0 and chose x and y so that 2x- y= 0 or y= 2x. For the last, I took x= 0 and chose y and z so that -y+ 3z= 0 or y= 3z. Taking one of the three numbers equal to 0 lets us change from trying to choose three numbers to just choosing two numbers.