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- mazehaq

Find a Cartesian equation of the the plane that contains the point (3,-7,2) and the line = <3,5,-4> + t<-2,1,3>

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- mazehaq

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- IrishBoy123

from 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.

- mazehaq

how would you go about doing it without doing cross product

- IrishBoy123

solve ax + by + cz = d

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- mazehaq

Yeah 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=3-2t, y=5+t, z=-4+3t

- mazehaq

What do I do next?

- IrishBoy123

make 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

- mazehaq

I'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

ok
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

i'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

cross 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
-(5z-24x+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

bu is early and i mighta messed up the cross of course

- amistre64

i see my error, i used the vector as the anchor point ...

- amistre64

(3,-7,2)
-(3,5,-4)
---------
(0,-12,6) ... or (0,2,-1) as the other vector .. maybe

- amistre64

-(-4z-1x+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 :)

- IrishBoy123

cool!

- mazehaq

Thanks so much y'all!

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