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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

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amistre64Best ResponseYou've already chosen the best response.1
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
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
dw:1359729417959:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
lol, thats a bad drawing
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
my idea might have a flaw in it .....
 one year ago

mags093Best ResponseYou've already chosen the best response.0
i got the normal vector which is (3,2,5)? and then Po=(1,5,4)? then used the dot product so that 3(1x)2(5y)5(4z)=o but it keeps saying its wrong? how do i get a new vector from the normal vector?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
so basically its asking to use the normal given and the point given to define a parallel plane
 one year ago

mags093Best ResponseYou've already chosen the best response.0
ok so how would i get the equation of the plane?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
3(xpx) + 2(ypy)  5(zpz) + d = 0
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
the +d was spurious on my part, im abit sick today :) given point is: (1,5,4) 3(x+1) + 2(y5)  5(z+4) = 0 3x3 +2y10  5z20 3x + 2y 5z 33 = 0 your setup was fine except it starts with a +3x not a 3x
 one year ago

mags093Best ResponseYou've already chosen the best response.0
thank you so much :) it was right
 one year ago
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