# Given two vectors $a$, $b$ with only strictly positive coordinates, can those two vectors be orthogonal?

I have a couple of questions to answer, and I am unsure if i argue correctly:

Given two vectors $$a$$, $$b$$ with only strictly positive coordinates, can those two vectors be orthogonal? My answer would be no. As $$\langle a,b\rangle=0$$ this can only be the case if all $$ab$$-coordinates are $$0$$, which is not the case because the coordinates have to be strictly positive, so in ordered to get to $$0$$ some ab products have to be negative.

Can $$ca$$ and $$db$$ be orthogonal, for $$c$$,$$d$$ elements of $$\mathbb R$$?

No if $$a$$ and $$b$$ are not orthogonal? I'm not sure if this question refers to the question above....my second question would be if $$c$$ and $$d$$ are $$0$$, $$\langle ca,db\rangle$$ would also be $$0$$ would this be than count as orthogonal??

How many vectors build a orthonormal basis in $$\mathbb R^n$$ - n

How many of these vectors (of the basis above) can have strictly positive Coordinates, how many strictly negative?

I would guess only $$1$$, because if all vectors are orthogonal than there can be only one with strictly positive coordinates..

• In your second question, yes, if $c=d=0$ then $ca$ and $bd$ are orthogonal; indeed, the zero vector is always orthogonal to everything. There are other possibilities though; what if $c=-d$ for example? – Greg Martin Dec 11 '18 at 9:22
• @GregMartin They’re orthogonal if $c=0$ or $d=0$. – amd Dec 11 '18 at 18:53

$$c\vec a$$ and $$d\vec b$$ are orthogonal for non-orthogonal vectors $$\vec a,\vec b$$ iff $$c=0$$ or $$d=0$$.

This is because:

$$1|\ \ \ \ c\vec a\cdot d\vec b=0\implies cd=0\ (\vec a\cdot\vec b\ne0)$$

$$2|\ \ \ \ cd=0\implies cd\cdot(\vec a\cdot\vec b)=0\implies c\vec a\cdot d\vec b=0$$

If the coordinates are strictly positive, they cannot be $$0$$! Therefore, the dot product between the two vectors is always a sum of positive products, so it is never $$0$$.
The second question is asking whether, if you multiply each vector by a real number, it is possible that the vectors are orthogonal ($$(c\textbf{a}) \cdot (d\textbf{b}) = 0$$). Using the properties of the dot product, because $$c\textbf{a} \cdot d\textbf{b} = cd(\textbf{a} \cdot \textbf{b})$$, and because $$\textbf{a} \cdot \textbf{b} \neq 0$$, the answer is no unless one of $$c$$ and $$d$$ is $$0$$.