Question

Please help anyone :0

Answer 1

http://screencast.com/t/r20DYWNnToCf

@AccessDenied @theEric

Answer 2

cross OA and OB to find the normal vector.

Answer 3

okay is it -8 , 6, -2

Answer 4

yes

Answer 5

and your initial point is (0,0,0).

Answer 6

ok?

Answer 7

a(x - xo) + b(y - yo) + c(z - zo) = 0

where (a,b,c) is the normal vector and (xo, yo, zo) = (0,0,0). Plug them in and simplify

Answer 8

-8x +6y -2z

Answer 9

-8x +6y -2z = 0

Answer 10

you might want to divide both sides by 2 to simplify further

Answer 11

but tht isnt the answer
http://screencast.com/t/jNDfcaSm

Answer 12

What I did, following the second method they listed, was take the cross product of OA and OB like you did above to find the normal vector to the plane OAB
and then take the cross product with that vector and the vector in the direction of the line, which is OA - OB or OB - OA. to get the vector normal to the plane we want.
That would get you something like 4(2i + 5j + 7k), which is equivalent to their answer provided...

Answer 13

oh then whats the point on the plane needed to be substituted

Answer 14

You could choose either A or B, which was given in vector form but is also a point on the line AB, but it was contained in our plane just by our requirements (find the plane which contains AB)

Answer 15

oh okay

Answer 16

thanks

Answer 17

could u help me with another one

Answer 18

Sure, I will try. :)