anonymous
  • anonymous
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.
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
i can never recall the formulas for this so i tend to have to build something workable
anonymous
  • anonymous
oh ok.. please don't use equation because it's not working for me:(
amistre64
  • amistre64
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

Looking for something else?

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

More answers

amistre64
  • amistre64
all possible values of "A"?
anonymous
  • anonymous
yes
amistre64
  • amistre64
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 ....
anonymous
  • anonymous
oh!!!! my bad.. Ax-2y+6z..
Psymon
  • Psymon
It's assumed as a part of the equation for a plane usually :P
amistre64
  • amistre64
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 :)
anonymous
  • anonymous
give me a a sec. i'll try
Psymon
  • Psymon
*watches and learns*
amistre64
  • amistre64
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)
amistre64
  • amistre64
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
anonymous
  • anonymous
WOW.. thanks a lot.. (^_^)
anonymous
  • anonymous
@Psymon you understand??
amistre64
  • amistre64
formulas would most likely make the computations plug and chug .. but the process should still amount to the same resutls :)
Psymon
  • Psymon
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
amistre64
  • amistre64
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
Psymon
  • Psymon
And if youre like me, I actually save what is said on these problems xD
amistre64
  • amistre64
i never save anything, which means i have to reconstruct it on the fly ;)
Psymon
  • Psymon
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.

Looking for something else?

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