Let \(a\) and \(b\) be real numbers. Prove that there are two orthogonal unit vectors \(u\) and \(v\) in \(\mathbb R^3\) such that \(u=(u_1,u_2,a)\) and \(v=(v_1,v_2,b)\) if and only if \(a^2+b^2\le1\).
Solution
1. If such orthogonal unit vectors \(u\) and \(v\) exist, then\[ a^2+b^2\leq1. \]
Write \[ u'=(u_1,u_2), \qquad v'=(v_1,v_2). \] Since \(u\) and \(v\) are unit vectors, \[ \norm{u'}^2=1-a^2, \qquad \norm{v'}^2=1-b^2. \] Their orthogonality gives \[ \inner{u'}{v'}=-ab. \] By the Cauchy–Schwarz inequality, \[ a^2b^2 = \abs{\inner{u'}{v'}}^2 \leq \norm{u'}^2\norm{v'}^2 = (1-a^2)(1-b^2). \] Expanding the last inequality and cancelling \(a^2b^2\) gives \[ a^2+b^2\leq1. \]
2. Assume \(a^2+b^2\leq1\) and \(a^2<1\). Then there are orthogonal unit vectors with third coordinates \(a\) and \(b\).
Set \[ r=\sqrt{1-a^2}>0 \] and define \[ u=(r,0,a), \] \[ v= \left( -\frac{ab}{r}, \sqrt{\frac{1-a^2-b^2}{1-a^2}}, b \right). \] The square root is real because \(a^2+b^2\leq1\). Clearly \[ \norm{u}^2=r^2+a^2=1. \] Also \[ \begin{aligned} \norm{v}^2 &= \frac{a^2b^2}{1-a^2} + \frac{1-a^2-b^2}{1-a^2} +b^2\\ &= 1. \end{aligned} \] Finally, \[ \inner{u}{v} = r\left(-\frac{ab}{r}\right)+ab = 0. \] Thus \(u\) and \(v\) have all the required properties.
3. Assume \(a^2+b^2\leq1\) and \(a^2=1\). Then there are orthogonal unit vectors with third coordinates \(a\) and \(b\).
The inequality \(a^2+b^2\leq1\) and \(a^2=1\) force \(b=0\). Take \[ u=(0,0,a), \qquad v=(1,0,0). \] Then both vectors are unit vectors, their inner product is \(0\), and their third coordinates are \(a\) and \(b\), respectively.
4. If \(a^2+b^2\leq1\), then the required orthogonal unit vectors exist.
The inequality implies \(a^2\leq1\). If \(a^2<1\), apply step 2; if \(a^2=1\), apply step 3.
5. The required vectors exist if and only if\[ \boxed{a^2+b^2\leq1}. \]
Step 1 proves necessity, and step 4 proves sufficiency.
6. Q.E.D.
Step 5 proves the stated equivalence.