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Mathematics 13 Online
OpenStudy (anonymous):

Find an equation of the plane. The plane through the origin and the points (2,-4,6) and (5,1,3). does that mean i use the point (0,0,0) because i tried that and the dot product is perpendicular to everything at 0 i think

OpenStudy (mathmate):

Yes you can use (0,0,0) as a point on the plane. Here, you could also take the cross-product of the two vectors to get the normal vector.

OpenStudy (anonymous):

Thats what i did so i used the point (0,0,0) and vector A(2,-4,6) an B(5,1,3) and did AxB to get the normal vector

OpenStudy (dumbcow):

equation of plane is: \[ax +by+cz = d\] plug in the 3 points to find the coefficients since it goes through origin, this means that constant d = 0

OpenStudy (mathmate):

What did you get for the cross product?

OpenStudy (dumbcow):

those are points...not vectors

OpenStudy (anonymous):

-18i+24j-22k

OpenStudy (anonymous):

yea but i used (0,0,0) as the origin so once you subtract the other two points they are the same, is that allowed

OpenStudy (mathmate):

@dumbcow OA=<2,-4,6>, OB=<5,1,3> OAxOB gives you the normal to the plane. I get <-18.,24.22>

OpenStudy (mathmate):

You could also reduce it to <-9,12,11> and substitute in the plane equation with d=0 as @dumbcow mentioned.

OpenStudy (dumbcow):

ahh i see ...yes i get <-9,12,11> for a,b,c as well

OpenStudy (mathmate):

@carr099 : everything's cool?

OpenStudy (anonymous):

yea

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