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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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Can you check the final equation? I am getting some other constant instead of $34$. Note: $|a|^2 = 8$, $|a|\neq 8$. – Srivatsan Sep 2 '11 at 18:25
@Srivatsan Narayanan I have corrected my post, it is $|a|^2 = 8$ but $|a|=8$ and is also $|b |^2=2 $ but $|b| = 2$. – mathsalomon Sep 2 '11 at 18:53
@Srivatsan Narayanan , that means there is not solution for this problem? – mathsalomon Sep 3 '11 at 15:46
@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? – Srivatsan Sep 3 '11 at 16:13

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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Sorry, but I do not understand your answer. – mathsalomon Sep 2 '11 at 18:49
The vectors are orthogonal if and only if $\alpha^2$ is a certain number $c$ (which depends on $\overline{a} \cdot \overline{b}$). So, how many real numbers have $c$ as their square? This does depend on the sign of $c$, but you can discover what that sign is. – Robert Israel Sep 2 '11 at 19:07
@mathsalomon I think Robert is asking you to solve for $\alpha$ in terms of $\bar a \cdot \bar b$. So it is not possible to get an answer that is an absolute constant not involving $\bar a$ and $\bar b$. – Srivatsan Sep 2 '11 at 19:35
mathsalomon, think about it: you've been given a formula for $\alpha$ that depends on the dot product. That means there can't be a formula that doesn't depend on the dot product, doesn't it? If the dot product is 16, $\alpha$ will be $\pm5$. If the dot product is $6.5$, then $\alpha$ will be $\pm4.5$. To find $\alpha$, you need to know something about $\bf a$ and $\bf b$; in particular, you need to know enough about them to be able to compute their dot product. Got it? – Gerry Myerson Sep 3 '11 at 7:53
@mathsalomon Yes (as long as you a specific number as the answer). – Srivatsan Sep 3 '11 at 16:09

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