Two statements regarding orthogonal unit vectors and orthogonal complements respectively

Question

I am new to linear algebra, and I was confused regarding the following question. I would really appreciate it, if anybody could give some feedback...

True or False?

1. $\left(\frac{1}{\sqrt14},\frac{-2}{\sqrt14},\frac{3}{\sqrt14}\right)$ is an orthogonal unit vector for $Sp\{(1,-1,-1),(2,7,4)\}$.

2. If ${U=\{(x,y,z) \in R^3|x+2y+3z=0\}}$, then $U^\bot=Sp\{(2,4,6)\}$.

For 1, I tried the Gram Schmidt process, which did not lead me to $\left(\frac{1}{\sqrt14},\frac{-2}{\sqrt14},\frac{3}{\sqrt14}\right)$, so my interim conclusion is that it is wrong.

For 2, I reasoned that $x=-2y-3z$, so we can rewrite the vector as $(-2y-3z,y,z)$, and indeed $(-2y-3z)·2+4y+6z=0$, so the answer is "true".

Thank you!

Answer 1

1. The given vector is orthogonal to $(1,-1,-1)$ as well as $(2,7,4)$ and hence it is orthogonal to the span of these two. There is no need to use Gram Schmidt.
2. Use some geometry. The equation $x+2y+3z=0$ represents a plane for which $(1,2,3)$ is a normal vector. Since the orthogonal complement of a plane is one dimensional $U^{\perp}$ must be $span \{(1,2,3)\}=span \{(2,4,6)\}$. Hence 2) is true.