Given the vector p=(-1, 3, 0) and q=(1, -5, 2), determine the components of a vector perpendicular to each of these vectors.
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 \[q\cdot v=0\\ v_1-5v_2+2v_3=0\]
oh ok, thanks! I think I can handle it from here:)
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.
yes, i agree, I'll let you know if I already have an answer...
I got x=3y, z=y, so I the one of the vectors that is perpendicular could be (3, 1 ,1)...right?
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).
really? you can use the zero vector? how?
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...
I guess intuition shouldn't be used here: http://mathworld.wolfram.com/Perpendicular.html
oh ok,... one more question, why does zero vector does satisfy the zero dot product property?
Take any vector \(a\in\mathbb{R}^2\), for example. Then \(a=(a_1,a_2)\) and \(\vec{0}=(0,0)\). You have \(a\cdot\vec{0}=0a_1+0a_2=0\).
Every component of the zero vector is zero. Multiply every component in another vector by zero and add them together, you get zero.
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..
You're welcome!
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