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mags093
Find an equation of a plane containing the line r=(-1,5,-4) + t(9,6,-3) which is parallel to the plane -3x + 2y - 5y +28 in which the coefficient of x is -3.
if you cross the direction vector of the line with the normal of the plane you will get a new vector that you can anchor to the lines point to define another point in the parallel plane
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lol, thats a bad drawing
my idea might have a flaw in it .....
i got the normal vector which is (-3,2,-5)? and then Po=(-1,5,-4)? then used the dot product so that -3(1-x)-2(5-y)-5(-4-z)=o but it keeps saying its wrong? how do i get a new vector from the normal vector?
direction vector of the line is: <9,6,-3>, or reduced to <3,2,-1> the normal vector is as you stated: <-3,2,-5> my idea is to cross these 2 vectors to find a new vector but I cant really see it being able to define a parallel plane perse. I can develop a perpendicular place tho
the dot product of the normal and direction vectors is: 3, 2,-1 -3,2,-5 -------- -9+4+5 = 0 so the vetcors themselves are already perped to each other
so basically its asking to use the normal given and the point given to define a parallel plane
ok so how would i get the equation of the plane?
-3(x-px) + 2(y-py) - 5(z-pz) + d = 0
the +d was spurious on my part, im abit sick today :) given point is: (-1,5,-4) -3(x+1) + 2(y-5) - 5(z+4) = 0 -3x-3 +2y-10 - 5z-20 -3x + 2y -5z -33 = 0 your setup was fine except it starts with a +3x not a -3x
thank you so much :) it was right