Find the acute angle between the surfaces $xy^2z = 3x + z^2 $ and $3x^2-y^2+2z=1$ at the point $(1,-2,1)$

Question

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 ?

Answer 1

Did you go ahead and take arccos$\left(\frac{-3}{7\sqrt{6}}\right)$? I get about 100$^o$. So your answer is no. The acute angle is 180 - 100 or about 80$^o$.

Answer 2

The angle between the tangent planes is the angle between normals. Note that if the scalar product between the normals is positive, the angle is acute. If the scalar product is negative, the angle is obtuse. In this case, just take the opposite direction for one of the normals, ($b\rightarrow -b$), or equivalent $$ \theta= \arccos\left(\frac{|a\cdot b|}{|a||b|}\right) $$ to get the acute angle

Answer 3

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

$$ \varphi = \min(|\arccos(\pm<\vec n_1,\vec n_2>)|) = \arccos(\frac{1}{\sqrt{742}})\approx 88^{\circ} $$