# Thread: "Any linear combination of these vectors will be perpendicular to (a,b,c)."

1. ## "Any linear combination of these vectors will be perpendicular to (a,b,c)."

For Question 4, here ( https://www.docdroid.net/LubIxXM/e1.pdf - scroll down for the answer for Question 4, to which I am referring ), the answer says that "any linear combination of these vectors will be perpendicular to (a,b,c)."

I confirmed one such case, here ( http://www.wolframalpha.com/input/?i=(-b... ), however, I don't understand why this is the case.

Could someone please explain to me why that is always the case?

Any input would be GREATLY appreciated!

2. ## Re: "Any linear combination of these vectors will be perpendicular to (a,b,c)."

Originally Posted by s3a
For Question 4, here ( https://www.docdroid.net/LubIxXM/e1.pdf - scroll down for the answer for Question 4, to which I am referring ), the answer says that "any linear combination of these vectors will be perpendicular to (a,b,c)."
Here are three: $<0,-c ,b >,~<c, 0, -a>,~\&~<-b,a , 0>$
To see that, what is the dot product of each of those with $<a,b,c>~?$

3. ## Re: "Any linear combination of these vectors will be perpendicular to (a,b,c)."

Having said that, I see that their dot products are 0, and I know that that means that they are perpendicular. My problem is with the "linear combination" part. Could you please elaborate on that part? More specifically, I know how to find linear combinations of vectors, but I don't understand why the linear combination of a set of vectors that are each perpendicular to a certain vector (vector (a,b,c), in this case) yields another vector that is also perpendicular to that certain vector (vector (a,b,c), in this case).

4. ## Re: "Any linear combination of these vectors will be perpendicular to (a,b,c)."

Originally Posted by s3a
Suppose that each of $\vec{a}~\&~\vec{b}$ is perpendicular to $\vec{v}$.
Consider the lin. com, $\alpha\vec{a}+\beta\vec{b}$ then
$(\alpha \vec{a}+\beta\vec{b})\cdot\vec{v}\\=(\alpha \vec{a}\cdot\vec{v}+\beta\vec{b}\cdot\vec{v})\\= ( \alpha \cdot 0 +\beta\cdot 0)\\=0$