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I am looking for a number $\alpha\in\mathbb{R}$ such that the vectors: $(\overline a + \overline b +\alpha\overline b)\;\;{\text{ and }}\;(\overline a + \overline b -\alpha\overline b)$ are orthogonal, where $\left| {\overline a } \right| = 8$ and $\left| {\overline b } \right| = 2.$ Applying the formula: $\left( {\overline a + \overline b } \right) \cdot \left( {\overline a - \overline b } \right) = {\left| {\overline a } \right|^2} - {\left| {\overline b } \right|^2}$ I was stuck when I only find that: $\overline a \cdot \overline b + 34 = 2{\alpha ^2}$ So, I can not isolate the number $\alpha$.

What did I do wrong?

In either case, the steps in my development are here: $\begin{gathered} \left( {\overline a + \overline b + \alpha \overline b } \right) \cdot \left( {\overline a + \overline b - \alpha \overline b } \right) = 0 \\ {\left( {\overline a + \overline b } \right)^2} - {\left( {\alpha \overline b } \right)^2} = 0 \\ {\left| {\overline a } \right|^2} + 2\overline a \overline { \cdot b} + {\left| {\overline b } \right|^2} - {\alpha ^2}{\left| {\overline b } \right|^2} = 0 \\ {8^2} + 2\overline a \overline { \cdot b} + {2^2} - {\alpha ^2}{2^2} = 0 \\ 68 + 2\overline a \overline { \cdot b} = 4{\alpha ^2} \\ 34 + \overline a \cdot \overline b = 2{\alpha ^2} \\ \end{gathered} $

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    @mathsalomon I think Robert has a point. When you say "Find the number of $\alpha$", then it seems like you are asking for the number of solutions for $\alpha$ such that the given condition is met. That will be $2$ or $1$ or $0$, depending on $\mathbf a \cdot \mathbf b$. (Actually, you can show that it will always be $2$..) If possible, can you quote the exact text of the question?2011-09-03

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Hint: given a real number $c$, how many $\alpha$ are there such that $\alpha^2 = c$? How does the answer depend on $c$?

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    @Gerry Myerson , Ok thanks2011-09-04