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Problem

If we have surfaces $$ z=8-x^3-y^2 $$ $$z= \frac{23}{8}+\frac{y}{16}-\frac{\sin(\pi x)}{4\pi} $$ Show that these surfaces intersect in point $(1,2,3)$ and that they are perpendicular at this point.

Attempt to solve

I know that if two planes are perpendicular their normal vector's dot product is zero. If plane equation is form $z=ax+bx+cx+d$. Unit normal vector can be defined: $$ \hat{n}=\begin{bmatrix} \frac{a}{\sqrt{a^2+b^2+c^2}} \\ \frac{b}{\sqrt{a^2+b^2+c^2}} \\\frac{c}{\sqrt{a^2+b^2+c^2}} \end{bmatrix} $$

Now this wouldn't work since i am not dealing with planes here. These are two surfaces.

Now i could try to just compute normal vectors for both planes in $p(1,2,3)$

I'll make two new functions. I'll call them $w_1$ & $w_2$

$$ w_1(x,y,z)=8-x^3-y^2-z $$ $$ w_2(x,y,z)=\frac{23}{8}+\frac{y}{16}-\frac{\sin(\pi x)}{4 \pi}-z $$ I'll mark $\hat{n_1}$ & $\hat{n_2}$ as normal vectors corresponding to $ (\text{function},\text{vector})\rightarrow (w_1(x,y,z),\hat{n_1})\text{ & }(w_2(x,y,z),\hat{n_2}) $

$$ \hat{n_1}=\nabla w_1(x,y,z)=\begin{bmatrix} 3x^2 \\ 2y \\-1 \end{bmatrix} $$ $$ \hat{n_2}=\nabla w_2(x,y,z)=\begin{bmatrix} -\frac{1}{4} \cos(\pi x) \\ \frac{1}{16} \\ -1 \end{bmatrix} $$

and normal vectors on point $p(1,2,3)$ are

$$ \hat{n_1}=\nabla w_1(1,2,3)=\begin{bmatrix} 3 \\ 4 \\ -1 \end{bmatrix}, \quad \hat{n_2}=\nabla 2_2(1,2,3)=\begin{bmatrix} \frac{1}{4} \\ \frac{1}{6} \\ -1 \end{bmatrix} $$

now if these are indeed perpendicular to each other these vectors should satisfy condition $\hat{n_1}\cdot \hat{n_2} = 0$.

$$\hat{n_1}\cdot \hat{n_2}= \begin{bmatrix} 3 \\ 4 \\ -1 \end{bmatrix} \cdot \begin{bmatrix} \frac{1}{4} \\ \frac{1}{16} \\ -1 \end{bmatrix} = 3\cdot \frac{1}{4}+4\cdot \frac{1}{16}+(-1)\cdot(-1)=2 $$


These two vectors should have dot product of 0 since they should be perpendicular. The problem was only to show that they are indeed perpendicular so something went clearly wrong. I've tried to find the error but it seems i cannot find it (reason i am posting this here).

If i was able to show these two vectors have dot product of 0 it would also mean that they do intersect right ?

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  • $\begingroup$ The normal vectors are actually $\hat n_1=\begin{bmatrix} -3\\ -4\\ -1\end{bmatrix}$ and $\hat n_2=\begin{bmatrix} 1/4\\ 1/16\\ -1\end{bmatrix}$ with their dot product being $$\hat n_1\cdot\hat n_2=\frac{-3}4-\frac 14+1=0$$ $\endgroup$ – Prasun Biswas Jan 22 '18 at 18:16
  • $\begingroup$ You fipped some signs here $\nabla w_1 = \begin{bmatrix} -3x^2\\-2y\\-1 \end{bmatrix}$ Fix that and you should be good. $\endgroup$ – Doug M Jan 22 '18 at 18:19

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