[SOLVED] Tangent plane

**Spec** · September 5th 2008, 11:24 AM

At what angle does the tangent plane to the surface $S: z=2x^3-x^2y-y^2-2y+11$ in the point (2,3,0) meet the xy-plane?

$F=2x^3-x^2y-y^2-2y+11 - z=0$

$grad F(x,y,z) = (6x^2-2xy,-x^2-2y-2,-1)$
$grad F(2,3,0) = (12,-12,-1)$

The normal vector to the xy-plane, n, is (0,0,1).

$grad F \cdot n = |17||1|cos\theta$
$grad F \cdot n = -1$

So we have $cos\theta = \frac{-1}{17} \implies \theta = arccos \frac{-1}{17}$

The correct answer is $\theta = arccos \frac{1}{17}$ though, which is what you get if you select $(0,0,-1)$ as the normal vector. So are both answers correct? Why is it that you choose (0,0,-1) as the normal vector?

**NonCommAlg** · September 5th 2008, 12:38 PM

Originally Posted by Spec

At what angle does the tangent plane to the surface $S: z=2x^3-x^2y-y^2-2y+11$ in the point (2,3,0) meet the xy-plane?

$F=2x^3-x^2y-y^2-2y+11 - z=0$

$grad F(x,y,z) = (6x^2-2xy,-x^2-2y-2,-1)$
$grad F(2,3,0) = (12,-12,-1)$

The normal vector to the xy-plane, n, is (0,0,1).

$grad F \cdot n = |17||1|cos\theta$
$grad F \cdot n = -1$

So we have $cos\theta = \frac{-1}{17} \implies \theta = arccos \frac{-1}{17}$

The correct answer is $\theta = arccos \frac{1}{17}$ though, which is what you get if you select $(0,0,-1)$ as the normal vector. So are both answers correct? Why is it that you choose (0,0,-1) as the normal vector?

we usually take $0 \leq \theta \leq \frac{\pi}{2}.$ so if $n_1, n_2$ are the normal vectors of two planes, then $\theta,$ the angle between the planes, is definied to be $\theta = \arccos \left|\frac{n_1 \cdot n_2}{||n_1|| \cdot ||n_2||} \right|.$

what you've found is the supplement of $\theta.$

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