Show that the two surfaces $z=7x^2−12x−5y^2$ and $xyz^2=2$ intersect orthogonally at the point (2,1,−1). [This means that their tangent planes at (2,1,−1) are orthogonal]
[Math] Show that two surfaces intersect orthogonally at a point
calculusmultivariable-calculusorthogonalitypartial derivativetangent line
Related Question
- [Math] Finding a parameter such that two curves intersect orthogonally
- [Math] Prove that the two lines intersect orthogonally
- [Math] Proving that for any point on an intersection of two surfaces, their respective tangent planes at the point are orthogonal.
- [Math] Find the acute angle between the surfaces $xy^2z = 3x + z^2 $ and $3x^2-y^2+2z=1$ at the point $(1,-2,1)$
Best Answer
If you calculate the partial derivatives for the first surface $z=f(x,y)$, you get $$ f_x=14x-12,\ \ f_y=-10y. $$ So $f_x(2,1)=16$, $f_y(2,1)=-10$, and the tangent plane is $$ 16(x-2)-10(y-1)-(z+1)=0, $$ with normal vector $$\tag1 v=(16,-10,-1).$$
For the second surface $z=g(x,y)$, since we need the point with $z<0$ we may write it as $$ z=-\sqrt{\frac2{xy}}. $$ Then $$ g_x=\frac1{\sqrt 2}\,x^{-3/2}\,y^{-1/2},\ \ \ g_y=\frac1{\sqrt 2}\,x^{-1/2}\,y^{-3/2}, $$ so $$ g_x(2,1)=\frac1{\sqrt2}\,\frac1{2^{3/2}}=\frac14,\ \ \ g_y(2,1)=\frac1{\sqrt2}\,\frac1{\sqrt2}=\frac12. $$ So the second tangent plane is $$ \frac14\,(x-2)+\frac12\,(y-1)-(z+1)=0. $$ The normal vector for the plane is then $$\tag2 w=(1/4,1/2,-1). $$ Now, looking at $(1)$ and $(2)$, $$ v\cdot w=\frac{16}4-\frac{10}2+1=0. $$ So $v$ and $w$ are orthogonal, and the two surfaces are orthogonal at $(2,1,-1)$.