anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
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
amistre64
  • amistre64
|dw:1359729417959:dw|
amistre64
  • amistre64
lol, thats a bad drawing

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
my idea might have a flaw in it .....
anonymous
  • anonymous
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?
amistre64
  • amistre64
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
amistre64
  • amistre64
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
amistre64
  • amistre64
so basically its asking to use the normal given and the point given to define a parallel plane
anonymous
  • anonymous
ok so how would i get the equation of the plane?
amistre64
  • amistre64
-3(x-px) + 2(y-py) - 5(z-pz) + d = 0
amistre64
  • amistre64
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
anonymous
  • anonymous
thank you so much :) it was right

Looking for something else?

Not the answer you are looking for? Search for more explanations.