# Orthogonal matrices, scalar products, projection of a line on a plane

1. There is an infinity of orthogonal matrices $Q=[q_1,q_2,q_3]$ that have as the first two columns the vectors $q_1=\frac{1}{\sqrt{6}}(-1,2,-1)^T,q_2=\frac{1}{\sqrt{3}}(1,1,1)^T$.
I would say this is not true since there is only one third vector orthogonal on other two.
2. If $\langle u,v\rangle \ge 0$ then the measure of the angle between $u$ and $v$ is less than $\frac{\pi}{2}$
To me it looks true, since the measure can take values only in interval (0,1)
3. The projection of a line on a plane is always a line.
True?

Thank you for taking your time.

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(1) What happens if you scalar multiply one of the columns of an orthogonal matrix? (2) What is the angle between the vectors $(0,1), (1,0)$ in $\mathbb{R}^2$? (3) What happens when you project the normal of a plane onto that plane? – arjafi Jan 18 '12 at 17:16
@ArthurFischer For (1), Orthogonal means Orthonormal columns...So no scalar multiples (excepting by -1) are allowed.... – N. S. Jan 18 '12 at 17:33

1. I agree with you, false. But there is not only one. Keep in mind that the determinant has to be $\pm1$ and not only $1$.
2. What happens when $\langle u,v\rangle$ is exactly $0$? Keep in mind that $\langle u,v\rangle=\|u\|\|v\|\cos\angle(u,v)$. And the measure is not in the interval $(0,1)$, since it's an angle and not the cosine.