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I am trying to find values for a and b for the following set of vectors:

$$\left(\left[\begin{matrix} 1\\2\\3\end{matrix}\right],\left[\begin{matrix} 4\\1\\-2\end{matrix}\right], \left[\begin{matrix} a\\b\\3\end{matrix}\right]\right) = 0$$

I know the idea is that, if I take the dot product of all the column vectors in the above vector space, I should end up with zero. But I am not sure how to get started solving for a and b, such that the below set of vectors is orthogonal. A hint as to how to get started?

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The dot product is an operation that is applied to two vectors, not three. You mean, every pairwise dot product equal to $0$? – Arturo Magidin Mar 19 '12 at 1:50
Sorry, you are right. I should have said take the dot product of all pairs of vectors. – Dylan Mar 19 '12 at 2:13
up vote 2 down vote accepted

In order for $(1,2,3)^t$ to be orthogonal to $(a,b,3)^t$, we must have $$a + 2b + 9 = 0.$$ In order for $(4,1,-2)^t$ to be orthogonal to $(a,b,3)^t$, we must have $$4a + b - 6 = 0.$$ $(1,2,3)^t$ and $(4,1,-2)^t$ are orthogonal, regardless of the values of $a$ and $b$.

So what you need to do is solve the $2\times 2$ system of linear equations $$\begin{array}{rccccl} a & + & 2b & = & -9\\ 4a & + & b & = & 6. \end{array}$$

Do you know how to solve systems of linear equations? (Gaussian-Jordan elimination, for instance?)

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Yes, I know how to solve with Gauss-Jordan. I think I can figure it out now. Thanks. – Dylan Mar 19 '12 at 2:12
Solving the above system I found the following: $\left[\begin{matrix}1&0&3\\0&1&-6\end{matrix}\right]$ . I made sure to check each pairewise dot product, where $a=3$ and $b=-6$, and it does check out. Thanks for the tip. – Dylan Mar 19 '12 at 19:50

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