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I'm having trouble with these types of questions. I have the following vector $u = (4, 7, -9)$ and it wants me to find 2 vectors that are perpendicular to this one.

I know that $(4,7,-9)\cdot (x,y,z) = 0$.

The dot product of two vectors must equal to zero in order for them to be perpendicular.

But that doesn't tell me much at this point. Any inputs?

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    $\begingroup$ What does the equation $4x + 7y -9z = 0$ mean? It certainly means any point satisfying that equation is perpendicular to your vector. What does it mean graphically? $\endgroup$ – Jacob Mar 19 '13 at 20:54
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You are correct that your first vector needs to have dot product zero with $(4,7,-9)$. Just pick any $x,y$ you like and solve for $z$. I will pick $x=9,y=0$ and find $z=4$ works, so we have $(9,0,4)$ Now you can either do the same with dot products with both vectors, or you can take the cross product, which is guaranteed to be perpendicular to both. So take $(4,7,-9) \times (9,0,4)$ getting $(28, -97, -63)$. I admit, I used Alpha do do the work.

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  • $\begingroup$ @Dimitri Topaloglou: If you don't need the two vectors to be perpendicular to each other, you can repeat my construction of the first vector to get, for example, $(4,-7,0)$ or $(0,7,9)$ more easily. $\endgroup$ – Ross Millikan Mar 19 '13 at 23:21
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Basically you can choose any two vectors that will satisfy that equation. For example, a=<0,0,0> or b=<9,9,11>.

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  • $\begingroup$ Just remember that the two vectors you choose need to be linearly independent from each other as well! $\endgroup$ – Twiceler Mar 19 '13 at 21:18
  • $\begingroup$ And $(0,0,0)$ is linearly dependent with every vector. $\endgroup$ – Daryl Mar 19 '13 at 21:46
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Take vector $n = (x,y,z)$. As you said by yourself, dot product should vanish. So $$ (4,7,-9) \cdot (x,y,z) = 4x+7y-9z = 0 $$ As you might see, all points that lie on the plane $4x+7y-9z = 0$ satisfy the condition of perpendicularity. If you want two linear independent vectors, just pick two different points. So, pick any two different triples $n_1 = (x_1, y_1, (4x_1+7y_1)/9)$ and $n_2 = (x_2, y_2, (4x_2+7y_2)/9)$ where $(x_1, y_1) \ne (x_2, y_2)$, and you're done.

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of course it depend how you define your inner product!but with ordinary product(oghlidusi). i am agree with Kaster it will be all of points on the plain $4x+7y-9z = 0$

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  • $\begingroup$ oghlidusi ro be "Euclidean" taghir bede. ;-) $\endgroup$ – Mikasa Jun 5 '13 at 5:47

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