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?
$\endgroup$ 14 Answers
$\begingroup$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.
$\endgroup$ 1 $\begingroup$Basically you can choose any two vectors that will satisfy that equation. For example, a=<0,0,0> or b=<9,9,11>.
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ $\begingroup$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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