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a) $u$ and $w$ are parallel

b) $u$ is parallel to $v + w$

c) $v$ and $w$ are orthogonal

d) $u$ is orthogonal to $v + w$

I chose d) since $u\cdot(v+w) = u\cdot v + u\cdot w$.

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It is correct.. –  Babak S. Nov 1 '12 at 11:46
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2 Answers 2

Sometimes things depend on what space you are working in!

a) Not true. A simple counter-example takes the unit vectors in 3-space $\hat{i},\hat{j},\hat{k}$. All of them dot to zero, but none of them are parallel.

b) Not true. Take the same three vectors as above, $\hat{i}+\hat{j}$ lies in the $xy$-plane which is orthogonal to $\hat{k}$

c) Not true. Take $\hat{i}\cdot \hat{k}=0, (\hat{i}+\hat{j})\cdot \hat{k}=0$.

d) As you have said, $\cdot $ is associative, so the sum of two vectors orthogonal to $u$ is orthogonal to $u$.

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Can you give an example in $\mathbb{R}^2$ which has $u\cdot v=0$ and $u\cdot w=0$, but $u$ parallel to $v+w$? In $\mathbb{R}^2$, you would be required to have $w=kv$, as the space orthogonal to $u$ can be at most one dimensional. –  Daryl Nov 1 '12 at 12:14
Hi @Daryl. I misread the question... $u$ looks like $v$, I'll edit! –  npfedwards Nov 1 '12 at 12:28
I think the question I originally answered for a) is actually more interesting as it's a common misconception that two vectors that dot with a third to zero must be parallel to each other. –  npfedwards Nov 1 '12 at 12:39

You're correct. Specifically, assuming $v \ne w$, $u$ is perpendicular to two independent vectors. These vectors span a plane that contains all vectors of form $<av+bw>$. $u$ is perpendicular to any vector in this plane($u$ is a normal vector to the plane). This can be shown as follows: $$<av+bw>\cdot u=av\cdot u+bw\cdot u=a(0) + b(0) = 0$$

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