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I was watching this MIT video.

At that particular point in the video I linked, he says you can derive the formula $$z=z_0 + a(x-x_0) + b(y-y_0)$$ by finding parametric equations for the following two lines: $$z=z_0 + a(x-x_0) $$ and $$z=z_0 + b(y-y_0)$$. Once you have the parametric equations for the two lines, take vectors along them and then take the cross product, which gives you the normal vector to the plane where the two lines sit and then get the equation of the plane.

I tried to do that as an exercise but I had no luck! I don't even understand how to get the parametric equations in the first place. Can please someone help?

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    $\begingroup$ Tangent plane to...? $\endgroup$ – Chappers Sep 23 '15 at 20:39
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The parametric equation ofa line typically takes the form $\underline r=\underline pt+ \underline q$, where $\underline p$ is the "direction vector" of the line and $\underline q$ is the position vector of a known point on the line.

In the situation you describe, the line $z=z_0+a(x-x_0)$ has gradient $a$ and passes through the point $(x_0, y_0, z_)$.

Its parametric or vector equation is $\underline r=\left(\begin{array}{r} 1 \\ 0 \\ a \end{array}\right)t + \left(\begin{array}{r} x_0 \\ y_0 \\ z_0 \end{array}\right) $

Similarly, the line $z=z_0+b(y-y_0)$ has gradient $b$ and passes through the point $(x_0, y_0, z_)$.

Its parametric or vector equation is $\underline r=\left(\begin{array}{r} 0 \\ 1 \\ b \end{array}\right)t + \left(\begin{array}{r} x_0 \\ y_0 \\ z_0 \end{array}\right) $

The vector product of the two direction vectors is $\underline n=\left(\begin{array}{r} 1 \\ 0 \\ a \end{array}\right) \times \left(\begin{array}{r} 0 \\ 1 \\ b \end{array}\right)=\left(\begin{array}{r} -a \\ -b \\ 1 \end{array}\right) $

The equation of the plane is given by $\underline r .\underline n = c$

Substitute $\underline n=\left(\begin{array}{r} -a \\ -b \\ 1 \end{array}\right) $ and the known point $\underline r=\left(\begin{array}{r} x_0 \\ y_0 \\ z_0 \end{array}\right) $ to get $c=z_0-ax_0-by_0$

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