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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: $|a|^2 = 8$, $|a|\neq 8$.2011-09-02
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    @Srivatsan Narayanan I have corrected my post, it is $|a|^2 = 8$ but $|a|=8$ and is also $|b |^2=2 $ but $|b| = 2$.2011-09-02
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    @Srivatsan Narayanan , that means there is not solution for this problem?2011-09-03
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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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    Sorry, but I do not understand your answer.2011-09-02
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    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.2011-09-02
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    @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 $\bar b$.2011-09-02
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    @Robert, Or can you determine the real number $\alpha $ that meets that condition?2011-09-02
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    The question as I interpreted it was not "determine $\alpha$ ...", but rather "find the number of $\alpha$ ...".2011-09-02
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    the question is, 'how to find a real number $\alpha$ that complies with the conditions: $|a|=8$, $|b|=2$, y $(a+b+\alpha b)$, $(a+b-\alpha b)$ are orthogonal? ' I tried to find $\alpha$ and I could not leave do you understand?2011-09-03
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    From $34+{\bf a}\cdot{\bf b}=2\alpha^2$ you get $\alpha=\pm\sqrt{17+(1/2){\bf a}\cdot{\bf b}}$. If you know ${\bf a}\cdot{\bf b}$ you can give a numerical answer for $\alpha$, but as people are trying to tell you if you don't know $\bf a$ and $\bf b$ the best you can do is give a formula for $\alpha$ in terms of ${\bf a}\cdot{\bf b}$2011-09-03
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    @Gerry Myerson, thanks Gerry, but I have not the value of a.b 'Can not we find α without a.b?'2011-09-03
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    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?2011-09-03
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    @Gerry Myerson, Thanks. 'Is it impossible to determine $a.b$ with only the data from this exercise?2011-09-03
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    @mathsalomon Yes (as long as you a specific number as the answer).2011-09-03
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    mathsalomon, you tell me. The only data in the exercise are the lengths of the two vectors. If you know the lengths of two vectors, is that enough to determine their dot product? Can you find an example of vectors of lengths 8 and 2 with dot product 16? and another example with dot product 0? and another example with dot product $-16$?2011-09-04
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    @ Gerry Myerson . Lengths is not the only data. What about the other, ie: $(\overline a + \overline b +\alpha\overline b)\;\; \bot \;(\overline a + \overline b -\alpha\overline b)$ 'Is that enough data to calculate the dot product $a.b$?2011-09-04
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    **NO!** I'm sorry, but you just aren't trying.2011-09-04
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    @Gerry Myerson , Ok thanks2011-09-04