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anonymous
 one year ago
Find a Cartesian equation of the the plane that contains the point (3,7,2) and the line <x,y,z> = <3,5,4> + t<2,1,3>
anonymous
 one year ago
Find a Cartesian equation of the the plane that contains the point (3,7,2) and the line <x,y,z> = <3,5,4> + t<2,1,3>

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4from the equation of the line, you can generate two more points on the plane , say let t= 0, t= 1 then solve ax + by + cz = d with your three points: or calculate the crossproduct aka normal for the plane, depending upon what you are currently learning.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how would you go about doing it without doing cross product

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4solve ax + by + cz = d

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I think that's what I don't get... so I let t=1, I got another point (1,6,1) My parametric equations are: x=32t, y=5+t, z=4+3t

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4make 3 points, then solve for the plane equation you have 4 unknowns  a,b,c,d  but if you plough through with the solution  you will see that you can still get something

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm still very confused, I'm so sorry. I'm attempting at the cross product right now. Let me see if I get somewhere. I suppose I'm not understanding the concept too well. Thank you for your help though!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4ok if the plane is ax + by + cz = d it is also a/d x + b/d y + c/d z = 1 you do the cross product, i'll try the other way, see where we get!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4i've done it the other way you are solving \[\left[\begin{matrix}3 & 7 & 2\\ 3 & 5 & 4\\ 1 & 6 & 1\end{matrix}\right]\left(\begin{matrix}\frac{a}{d} \\ \frac{b}{d} \\ \frac{c}{d}\end{matrix}\right) = \left(\begin{matrix}1 \\ 1 \\ 1\end{matrix}\right)\] thats the given point and 2 generated off the line at t= 0 and t= 1 i get 7x + 2y + 4z = 15 do we agree?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1cross product using 2 noncolinear vectors in the plane, one vector is the direction vector of the line (2,1,3) the other is a vector from a point on the line, and the stated point ... the anchor point is convenient: (3,7,2) (2,1,3)  (5,8,1) seems useful cross the vectors (5z24x+2y) x y z x y 2 1 3 2 1 5 8 1 5 8 +(x+15y+16z) 23x, 13y,11z should be the cross product for our normal not all vectors (x^,y^,z^) that perp to the normal (a,b,c) form the inner product (a,b,c).(x^,y^,z^) = 0 ax^ + by^ + cz^ = 0

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1bu is early and i mighta messed up the cross of course

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i see my error, i used the vector as the anchor point ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1(3,7,2) (3,5,4)  (0,12,6) ... or (0,2,1) as the other vector .. maybe

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1(4z1x+0y) x y z x y 0 2 1 0 2 2 1 3 2 1 +(6x+2y+0z) 7x, 2y, 4z ... now we agree :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks so much y'all!
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