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Mathematics 15 Online
OpenStudy (anonymous):

Find the plane through the origin (0,0,0) and perpendicular to the vector (v1) <1, -2, 5> . So I thought to find the normal vector by the cross products of (v1) and some other vector. But I can not seem to find the other vector. Any ideas?

OpenStudy (rational):

<1, -2, 5> itself is the normal vector

OpenStudy (rational):

Can you write the equation of plane using the given point ?

OpenStudy (anonymous):

\[n _{1}(x _{1}-x _{0})+n _{2}(y _{1}-y _{0})+n _{3}(z _{1}-z _{0})=0\] Thus: x-2y+5z=0 So they gave me the normal vector. Makes more sense. I think I was thrown off by the perpendicular part of the question for some reason. Thanks for the help rational!

OpenStudy (rational):

yw!

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