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  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.

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
up vote 5 down vote accepted

Ok, let's see:

  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.

  3. What if the line is orthogonal to the plane.

I guess the main intention of these questions is to make you read them carefully and precisely and not judge by just pure intuition. Since they prove false when looking at the special cases after seeming obviously true at first.

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