find equation of plane through the origin and the points and

Question

find equation of plane through the origin and the points <4,-2,5> and <6,1,1>

calculusfunctions · Answer

Which one. Vector, parametric, symmetric, or scalar equation of the plane?

anonymous · Answer

simple equation of plane

experimentx · Answer

ax+by+cz+d=0
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for passing through origin, out x=y=z=0 and get d=0
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then put a = 6-4, b=1-(-2) ...

anonymous · Answer

thanx

anonymous · Answer

find r(t) if r'(t) = 4t i + 3tj +t k and r(1) = i+j

experimentx · Answer

integrate it ... use the initial value to get rid of constant.

anonymous · Answer

"then put a = 6-4, b=1-(-2) ... "

How does this work?

anonymous · Answer

If I cross the vectors to get a normal, I get (-7,26,16) so plane is

-7(x-4) + 26(y+2) + 16(z-5) = 0  which satisfies all three points given.

experimentx · Answer

a(x_1) + b(y_1)+ c(z_1) = 0
a(x_2) + b(y_2)+ c(z_2) = 0
-----------------------
        a                          b                      c
------------   =  -----------   =  ----------
y1z2 - z1y2             z1x2 - x1z2         x1y2 - y1x2
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