What is the equation of the tangent plane to the function: 2x^2 +y^2 +2z^2 +2xz =7. At the point (-1,1,2) ?

- amistre64

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- amistre64

it begins with the gradient....

- anonymous

yeah thatl giv u d direction ratios of the tangent plane

- anonymous

4x+2z, 2y, 4z+2x
thats 0, 2, 6

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## More answers

- amistre64

goody goody goody :)

- anonymous

but then how do u get the direction ratios of the normal?

- amistre64

the normal is <0,2,6>; so use the point to calibrate the equation

- anonymous

thats the ratios parallel to the plane

- amistre64

0(x+1)+2(y-1)+6(z-2)=0

- anonymous

thats wrong

- anonymous

those coefficients have to be in the direction of the normal, and not the tangent

- amistre64

can you explain?

- anonymous

ur using the standard eqn of a plane passing through a given pt a,b,c
A(x-a) + B(x-b) +C(x-c)=0
where A,B,C are direction ratios of the NORMAL

- amistre64

x-x0; y-y0 and z-z0 are the components of the vector on the plane

- anonymous

u get the cross product of <0,2,6> and <-1, 1,2>

- anonymous

then use the standard equation

- amistre64

-1,1,2 is a point on the plane right? not a vector parallel to it tho.

- anonymous

yeah

- amistre64

the vector that is parallel to the plane is (x-x0,y-y0,z-z0) where P(x0,y0,z0) is on the plane right?

- anonymous

-2(x+1) +6(y-1) +2 (z-2) = 0

- amistre64

-2,6,2 is not the normal to the tangent plane tho. if anything it too is parallel to the plane

- anonymous

arent you suppose to find Fx(X,Y) and Fy(X,Y) use the cross product to find the N?

- anonymous

i told u get the vector product of the direction of the tangent and the point

- anonymous

weve already found the gradient, i suggest we cross that with the point to get a vector normal to the plane

- amistre64

the normal was found by the gradient; isnt the gradient and the normal parallel vectors?

- anonymous

no
the gradient is parallel to the plane

- anonymous

If you found the N and have a point whats the problem?

- amistre64

hmmm.... whats the gradient of 0 = 2x -y +4

- amistre64

just trying to understand concepts :)

- amistre64

<2,-1,0> right?

- amistre64

is that perp or // to y=2x+4 ; z=0?

- anonymous

||

- amistre64

so the vector <2,-1>; whose slope is -1/2 is // with the plne y=2x+2; z=0?

- amistre64

(-1/2)*2=-1.....

- anonymous

u cant define a gradient vector for a line

- amistre64

i didnt; i defined a plane; hence the z=0

- anonymous

fine

- anonymous

bt m saying i think wt i did ws right

- anonymous

Do you guys know about green theorem?

- amistre64

i can tell :)

- anonymous

ive heard f it..nt studied it..

- anonymous

hey sorry..i confirmed..the gradient vector is normal to the surface..rly sorry

- amistre64

if we dot the gradient to a vector parallel to the plane y=2x+4; z=0, we will know if its perp or //

- anonymous

no i confirmed

- amistre64

g=<2,-1,0>
v=<1, 2,0>
--------
2-2+0=0

- anonymous

yes m tellin u its normal to the surface

- anonymous

wt u were doing is right...i feel u alrdy know all this nd ur just testing us out

- amistre64

;) gotta keep entertained ya know lol

- anonymous

mean guy

- amistre64

id give you another medal if i could ;)

- anonymous

wts stopping u??

- amistre64

the only option available to me is the undo lol

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