Find the acute angle between the surfaces $xy^2z = 3x + z^2 $ and $3x^2-y^2+2z=1$ at the point $(1,-2,1)$
Angle between curves at a point is given by the angle between their tangent planes at the point.
$$f(x,y,z):= 3x +z^2 -xy^2z$$
$\nabla f(1,-2,1) = \langle 3-y^2z,-2xyz,2z-xy^2\rangle_{(1,-2,1)} = \langle -1,4,-2 \rangle$
Equation of tangent plane to $xy^2z = 3x + z^2 $ will be
$-x + 4y -2z + d=0$ Putting $(1,-2,1)$ and solving for $d$ we have
$$x-4y+2z=11 \; \; \; (1)$$
Also, a nice way to write equation of tangent plane to curve $ax^2 + by^2 + cz^2 + 2ux + 2vy + 2wz + d =0$ at $P(x_0,y_0,z_0)$ would be:
$$ax\cdot x_0 + by\cdot y_0 + cz\cdot z_0 + u(x+x_0) + v(y+y_0) + w(z+z_0) +d=0$$
Hence tangent plane to $3x^2-y^2+2z=1$ will be $3x(1) -y(-2)+(z+1)=1$
$$ \Rightarrow 3x +2y+z=0 \; \; \; (2)$$
One of the reason for posting this is, how'd you follow this method to write equation of tangent plane to the first curve $xy^2z = 3x + z^2 $? Basically, is it possible to extend this method to equations where degree is greater than $2$ and contains terms such as $xyz$?
Now, angle between tangent planes is angle between their normals,
Direction ratios of normal to $(1)$ and $(2)$ respectively are
$a=\langle 1,-4,2\rangle$ and $b=\langle 3,2,1\rangle $
$\Rightarrow \theta= \arccos(\frac{a\cdot b}{|a||b|}) = \arccos(\frac{-3}{7\sqrt{6}})$
Problem here is, how'd I know if this is acute or not, i.e, if this is the answer that I was looking for ?
Best Answer
We have
$$ \vec n_1 = \frac{\nabla (x y^2 z - 3 x + z^2)}{||\nabla(x y^2 z - 3 x + z^2)||} \\ \vec n_2 = \frac{\nabla (3 x^2 - y^2 + 2 z - 1)}{||\nabla(3 x^2 - y^2 + 2 z - 1)||} $$
and the sought angle is
$$ \varphi = \min(|\arccos(\pm<\vec n_1,\vec n_2>)|) = \arccos(\frac{1}{\sqrt{742}})\approx 88^{\circ} $$