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anonymous
 3 years 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.
anonymous
 3 years 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.

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amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1if 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1359729417959:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1lol, thats a bad drawing

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1my idea might have a flaw in it .....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1direction 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1the 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1so basically its asking to use the normal given and the point given to define a parallel plane

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok so how would i get the equation of the plane?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.13(xpx) + 2(ypy)  5(zpz) + d = 0

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1the +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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thank you so much :) it was right
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