Question

Determine the point on the plane with equation x+2y+3z=6 that is closest to (1,2,3)

Answer 1

i know the short way of finding the distance, which is \[\large D =
\frac{\left|ax+by+cd+d\right|}{\sqrt{a^2+b^2+c^2}}\] but my professor wanted us to use the partial derivative method... and the formula for the distance \(\large D(x,y) =f_{xx}f_{yy} -(f_{xy})^2\)

Answer 2

I would think that theres so much of an easier way :/ I would have an idea how to do it if it werent partial deriv method, haha x_X

Answer 3

partial derivative in terms of x twice multiplied by partial derivative in terms of y twice

Answer 4

Made an error, supposed to say \[\large D= \frac{\left|ax+by+cz+d\right|}{\sqrt{a^2+b^+c^2}}\]

Answer 5

take the derivative in terms of x first and then do it again in terms of x for Fxx. treat y as a constant. leave it alone

Answer 6

Yeah but isnt the second partial going to = 0?

Answer 7

ya

Answer 8

Fx = 1
Fxx = 0

Answer 9

then multiplying it to fyy would make the distance 0...technically.. fxy=0 and fxxfyy = 0 as well O_o

Answer 10

yeahhhhhh... : /

Answer 11

my answer is supposed to be (something)/7 xD

Answer 12

w ell tell the prof that you tried it and got o's in retgurn

Answer 13

Haha thing is she's already solved it...hmm. but wait.-feeling lazy to type it all out-

Answer 14

\[\large d= \sqrt{(x-1)^2+(y-2)^2+(z-3)^2}\]\[\ \ \ \large -3z=x+2y-6\]\[\ \ \ \ \ \ \ \large z=-\frac{x}{3} -\frac{2}{3}y+2\]\[\large d=\sqrt{(x-1)^2+(y-2)^2+(-\frac{1}{3}x-\frac{2}{3}y-1)^2}\]\[\large d^2=(x-1)^2 +(y-2)^2 +(-\frac{1}{3}x-\frac{2}{3}y-1)^2\] and now we solve for\( \large f_x \) and \(\large f_y\)?

Answer 15

chain rule... and only take derivatives in terms of x

Answer 16

geomtrical but long method
x+2y+3z=6 1,2,3 direction cosines of line normal to this plane
which passes through 1,2,3
(x-1)/1=(y-2)/3=(z-3)/3 =>x=y/2=z/3 =k putting in eqn of plane gives point
k+4k+9k=14k=6 =>k=6/14=3/7
x=3/7, y=6/7, z=9/7
distance= sqrt((1-3/7)^2+(2-6/7)^2+(3-9/7)^2)=ans

Answer 17

point is (3/7,6/7,9/7)

Answer 18

Could you explain step 2? How did you find
"(x-1)/1=(y-2)/3=(z-3)/3 =>x=y/2=z/3 =k putting in eqn of plane gives point
k+4k+9k=14k=6 =>k=6/14=3/7"?

Answer 19

Oh, so we're trying to use parametric equations to find the point on the tangent line

Answer 20

yes