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I have a question in my work where I have to find the values of a and b such that (a,2,3)^T and (1,2,b-2)^T are not perpendicular. (I have chosen random values as an example, as I'm not asking anyone to do my work for me)

The way I was going to work it out if they didn't need to be transposed was to find the values a and b to be perpendicular and then state that they cannot equal those values.

Doing this using the dot product of any two vectors (a1,a2,a3) and (b1,b2,b3) =a1b1+a2b2+a3b3. Then simplifying the expression to solve for a and b.

Where I need help however (assuming that the above is appropriate) is transposing a vector and then finding the two vectors to NOT be perpendicular.

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You can exploit the fact that the dot product between two non-zero vectors is 0 if and only if the two vectors are perpendicular in order to set up an inequality.

In this way, you obtain ${a}_{1}{b}_{1}+{a}_{2}{b}_{2}+{a}_{3}{b}_{3} \neq 0$, which can be rewritten as a 2x2 system if there are two unknowns ( ${a}_{1}$ and ${b}_{3}$, as in your example).

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  • $\begingroup$ So does the fact that the vectors are transposed make a difference? $\endgroup$ – Lucy May 18 '16 at 10:10
  • $\begingroup$ Not at all. A transposed vector is still a vector, and the property I mentioned holds for any choice of vectors. $\endgroup$ – A. Darwin May 18 '16 at 10:12

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