Question

Find the normal vector of the tangent plane to `x^2yz+3y^2-2xz^2+8z=0` at `(1,2,-1)` ?

Answer 1

`-6hat(i) +12hat(j)+14hat(k)`

Explanation:

First we rearrange the equation of the surface into the form `f(x,y,z)=0`, this is already done for us:

`x^2yz+3y^2-2xz^2+8z=0`

And so we define our surface function, `f`, by:

`f(x,y,z) = x^2yz+3y^2-2xz^2+8z`

In order to find the normal at any particular point in vector space we use the Del, or gradient operator:

`grad f(x,y,z) = (partial f)/(partial x) hat(i) + (partial f)/(partial y) hat(j) + (partial f)/(partial z) hat(k)`

remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:

`grad f = ((partial)/(partial x) (x^2yz+3y^2-2xz^2+8z))hat(i) +`
`" " ((partial)/(partial y) (x^2yz+3y^2-2xz^2+8z))hat(j) +`
`" " ((partial)/(partial z) (x^2yz+3y^2-2xz^2+8z))hat(k)`
`" "= (2xyz-2z^2)hat(i) + (x^2z+6y)hat(j) + (x^2y-4x+8)hat(k)`

So for the particular point `(1,2,-1)` the normal vector to the surface is given by:

`grad f(1,2,-11) = (-4-2)hat(i) +(-1+12)hat(j) +(2-4+8)hat(k)`
`" " = -6hat(i) +12hat(j)+14hat(k)`

We can confirm this graphically: Here is the surface with the normal vector: