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Mathematics 18 Online
OpenStudy (anonymous):

Given the vector p=(-1, 3, 0) and q=(1, -5, 2), determine the components of a vector perpendicular to each of these vectors.

OpenStudy (anonymous):

So you want to find a vector perpendicular to both \(p\) and \(q\). Call this vector \(v=(v_1,v_2,v_3)\). Perpendicular vectors have a zero dot product, so \[p\cdot v=0\\ -v_1+3v_2+0v_3=0\] and \[q\cdot v=0\\ v_1-5v_2+2v_3=0\]

OpenStudy (anonymous):

oh ok, thanks! I think I can handle it from here:)

OpenStudy (anonymous):

You're welcome. If I'm not mistaken, I think you'll have many solutions (or possibly none) to this, since you have two equations with three unknowns.

OpenStudy (anonymous):

yes, i agree, I'll let you know if I already have an answer...

OpenStudy (anonymous):

I got x=3y, z=y, so I the one of the vectors that is perpendicular could be (3, 1 ,1)...right?

OpenStudy (anonymous):

Yes, that's one solution. It's got me thinking though. The zero vector may have been a quicker choice to make, but I'm not sure if the zero vector is considered perpendicular to other vectors (though it does satisfy the zero dot product property).

OpenStudy (anonymous):

really? you can use the zero vector? how?

OpenStudy (anonymous):

Well the dot product of any vector with the zero vector is 0, so I would think that's enough to say it's perpendicular to any vector. It doesn't seem intuitive, is all...

OpenStudy (anonymous):

I guess intuition shouldn't be used here: http://mathworld.wolfram.com/Perpendicular.html

OpenStudy (anonymous):

oh ok,... one more question, why does zero vector does satisfy the zero dot product property?

OpenStudy (anonymous):

Take any vector \(a\in\mathbb{R}^2\), for example. Then \(a=(a_1,a_2)\) and \(\vec{0}=(0,0)\). You have \(a\cdot\vec{0}=0a_1+0a_2=0\).

OpenStudy (anonymous):

Every component of the zero vector is zero. Multiply every component in another vector by zero and add them together, you get zero.

OpenStudy (anonymous):

yeah, that makes sense......:)lol i did not think about that. thank you very much! if i can give one more medal and a fan i would give it to you..

OpenStudy (anonymous):

You're welcome!

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