## mazehaq 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>

1. 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.

2. mazehaq

how would you go about doing it without doing cross product

3. IrishBoy123

solve ax + by + cz = d

4. 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

5. mazehaq

What do I do next?

6. 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

7. 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!

8. 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!

9. 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?

10. 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

11. amistre64

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

12. amistre64

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

13. amistre64

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

14. 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 :)

15. IrishBoy123

cool!

16. mazehaq

Thanks so much y'all!