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Gradient vector, tangent
planes and normal lines

In this section we want to revisit tangent planes only this time we’ll look at them in light of the
gradient vector
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Let’s first recall the equation of a plane that contains the point ( x0 , y0 , z0 ) with normal vector

n = a,b, c is given by,
a ( x − x0 ) + b ( y − y0 ) + c ( z − z0 ) = 0
When we introduced the gradient vector in the section on directional derivatives we gave the
following fact
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Likewise, the gradient vector ∇f ( x0 , y0 , z0 ) is orthogonal to the level
surface f ( x, y, z ) = k at the point ( x0 , y0 , z0 )
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This says that the gradient vector is always
orthogonal, or normal, to the surface at a point
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Note however, that we can also get the equation from the previous section using this more general
formula
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In
order to use the formula above we need to have all the variables on one side
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All we need to do is subtract a z from both sides to get,

f ( x, y ) − z = 0

Now, if we define a new function

F ( x, y, z ) = f ( x, y ) − z

we can see that the surface given by z = f ( x, y ) is identical to the surface given by

F ( x, y, z ) = 0 and this new equivalent equation is in the correct form for the equation of the
tangent plane that we derived in this section
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∇F = Fx , Fy , Fz = f x , f y , −1
Notice that

∂
Fx = ( f ( x, y ) − z ) = f x
∂x

∂
Fy = ( f ( x, y ) − z ) = f y
∂y

∂
Fz = ( f ( x, y ) − z ) = −1
∂z
The equation of the tangent plane is then,

f x ( x0 , y0 )( x − x0 ) + f y ( x0 , y0 )( y − y0 ) − ( z − z0 ) = 0

Or, upon solving for z, we get,

z = f ( x0 , y0 ) + f x ( x0 , y0 )( x − x0 ) + f y ( x0 , y0 )( y − y0 )

which is identical to the equation that we derived in the previous section
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We might on
occasion want a line that is orthogonal to a surface at a point, sometimes called the normal line
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Since we want a line that is at the point ( x0 , y0 , z0 ) we know that
this point must also be on the line and we know that ∇f ( x0 , y0 , z0 ) is a vector that is normal to
the surface and hence will be parallel to the line
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2

2

2

Solution
For this case the function that we’re going to be working with is,

F ( x, y, z ) = x + y + z
2

2

2

and note that we don’t have to have a zero on one side of the equal sign
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To finish this problem out we simply need the gradient evaluated at the point

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