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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 ?

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  • $\begingroup$ If the cosine is negative, the angle can't be acute, right? Recall that $\arccos$ takes values in $[0,\pi]$ $\endgroup$ – saulspatz Jun 20 '18 at 15:20
  • $\begingroup$ @saulspatz So I need to subtract it from 180 degrees? what if the angle lies between 180 to 270 degrees? $\endgroup$ – So Lo Jun 20 '18 at 15:31
  • $\begingroup$ It won't. $\arccos$ will always give you an angle between $0$ and $180$ degrees. $\endgroup$ – saulspatz Jun 20 '18 at 15:36
  • $\begingroup$ @saulspatz if I put $a=\langle -1,4,-2\rangle$ then $cos(\theta ) $ is positive. But that will not match the answer I got previously. I am getting confused here $\endgroup$ – So Lo Jun 20 '18 at 16:08
  • $\begingroup$ You'll get $\arccos\left(\frac{3}{7\sqrt{6}}\right)$ which is the supplement of what you had before. What is confusing you? (I haven't checked your arithmetic.) $\endgroup$ – saulspatz Jun 20 '18 at 16:22
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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

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

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  • $\begingroup$ This is not an acute angle. $\endgroup$ – saulspatz Jun 20 '18 at 16:25
  • $\begingroup$ Also, I think you must have made an arithmetic error. I get the same numerical answer as the OP. $\endgroup$ – saulspatz Jun 20 '18 at 16:33
  • $\begingroup$ @saulspatz Really the answer is $\min(|\arccos(\pm<\vec n_1,\vec n_2>)|) = \arccos(\frac{1}{\sqrt{742}})\approx 88^{\circ}$ $\endgroup$ – Cesareo Jun 20 '18 at 18:02
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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$^o$.

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  • $\begingroup$ if I put $a=\langle -1,4,-2\rangle$ then $cos(\theta ) $ is positive. But that will not match the answer I got previously. I am getting confused here $\endgroup$ – So Lo Jun 20 '18 at 16:08
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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

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