How do you find the equations for the tangent plane to the surface #x=y(2z-3)# through #(4, 4, 2)#?

1 Answer
Jan 21, 2017

# x-y-8z =-16 #

Explanation:

First we rearrange the equation of the surface into the form # f(x,y,z)=0#

# x=y(2z-3) #
# :. x=2yz-3y #
# :. x+3y-2yz = 0 #

And so we have our function:

# f(x,y,z) = x+3y-2yz #

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+3y-2yz))hat(i) + #
# " " ((partial)/(partial y) (x+3y-2yz))hat(j) + #
# " " ((partial)/(partial z) (x+3y-2yz))hat(k) #
# " "= (1)hat(i) + (3-2z)hat(j) + (-2y)hat(k) #
# " "= hat(i) + (3-2z)hat(j) -(2y)hat(k) #

So for the particular point #(4,4,2)# the normal vector to the surface is given by:

# grad f(4,4,2) = hat(i) + (3-2*2)hat(j) -(2*4)hat(k) #
# " " = hat(i) -hat(j) -8hat(k) #
# " " = ( (1), (-1),(-8) ) #

So the tangent plane to the surface # x=y(2z-3) # has this normal vector and it also passes though the point #(4,4,2)#. It will therefore have a vector equation of the form:

# vec r * vec n = vec a * vec n #

Where #vec r=((x),(y),(z))#; #vec n# is the normal, and #a# is any point in the plane

Hence, the tangent plane equation is:

# ((x),(y),(z)) * ( (1), (-1),(-8) ) = ((4),(4),(2)) * ( (1), (-1),(-8) ) #
# :. (x)(1) + (y)(-1) + (z)(-8) = (4)(1) + (4)(-1) + (2)(-8) #
# :. x-y-8z = 4-4-16 #
# :. x-y-8z =-16 #

We can confirm this graphically: Here is the surface with the normal vector:
enter image source here

and here is the surface with the tangent plane and the normal vector:
enter image source here