Suppose that F(x,y) = x[tex]^{2}[/tex]+y[tex]^{2}[/tex]. By using vector geometry, find the Cartesian equation of the tangent plan to the surface z = F(x,y) at the point where (x,y,z) = (1,2,5). Find also a vector n that is normal to the surface at this point.(adsbygoogle = window.adsbygoogle || []).push({});

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Now the step by step working is given for this question, I however get confused at one part.

First they intersect the surface z = x[tex]^{2}[/tex]+y[tex]^{2}[/tex] with the plane x = 1

Therefore z = 1 + y[tex]^{2}[/tex]

The gradient F[tex]_{y}[/tex](1,2) = 4

By using the point gradient formula for a straight line, the Cartesian eq for the tangent is:

z -5 = 4(y - 2), x =1

If [tex]\lambda[/tex] = y - 2, then the equation of the tangent line in parametric vector form is

(x y z) = (1 2 5) + [tex]\lambda[/tex](0 1 4)

Now my question is, where did the (0 1 4) come from?

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# Tangent planes to surfaces

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