Question

The distance from R(3,-3,1) to the plane with the equation X-2y+6z=0 is 3. Determine all possible value(s) of A for which this is true.

Answer 1

i can never recall the formulas for this so i tend to have to build something workable

Answer 2

oh ok.. please don't use equation because it's not working for me:(

Answer 3

pull the normal vector from the plane and define the line using it and the given point in parametric form

use the parametric forms to define the point on the plane where it intersects (solving for t), then plugging in t into the parametric equations defines the second point ... the distance between them is therefore apparent

Answer 4

all possible values of "A"?

Answer 5

yes

Answer 6

The distance from R(3,-3,1) to the plane with the equation X-2y+6z=0 is 3

there is no A in the post ....

Answer 7

oh!!!! my bad.. Ax-2y+6z..

Answer 8

It's assumed as a part of the equation for a plane usually :P

Answer 9

normal vector is therefore: (a,-2,6), anchored to (3,-3,1)

im thinking at first glance that this is a sphere centered at the given point :)

Answer 10

give me a a sec. i'll try

Answer 11

*watches and learns*

Answer 12

x = 3 + at
y = -3 - 2t
z = 1 + 6t

ax-2y+6z = 0

a(3+at)-2(-3-2t)+6(1+6t) = 0

3a+a^2 t + 6 + 4t + 6 +36t = 0

(40+a^2)t + 12+3a = 0

t = -(12+3a)/(40+a^2)

Answer 13

the point on the plane is therefore:

x = 3 - a(12+3a)/(40+a^2)
y = -3 + 2(12+3a)/(40+a^2)
z = 1 - 6(12+3a)/(40+a^2)

and using the distance formula between (3,-3,1) and this point in "a"; equaled to 3 should solve for all a

Answer 14

WOW.. thanks a lot.. (^_^)

Answer 15

@Psymon you understand??

Answer 16

formulas would most likely make the computations plug and chug .. but the process should still amount to the same resutls :)

Answer 17

Yeah, looks like a similar procedure to what we were doing. Im at work, sorry, haha. I work as a math tutor, Im just not as high of a level as Id like xD

Answer 18

moving the setup to the origin should subtract out the point parts leaving us with the "a" values ...

x = - a(12+3a)/(40+a^2)
y = 2(12+3a)/(40+a^2)
z = - 6(12+3a)/(40+a^2)

9 = x^2 + y^2 + z^2

Answer 19

And if youre like me, I actually save what is said on these problems xD

Answer 20

i never save anything, which means i have to reconstruct it on the fly ;)

Answer 21

Well if I remember I had a problem doneand worked out, I can refer to the steps and what the person who helped me did.