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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 $(1,-2,1)$

surfaces $F$ and $G$ are tangent if and only if $nablaFtimesnablaG=mathbf0$Explain the equation of the tangent plane?Shortest distance from point and lineNormal vectors and tangent planesHow to attack this problem with a volume that is hard to interpret / sketch,Vector analysis: understanding formulas for normal and tangentFind a plane perpendicular to $yz$, passing by a point and making an angle with another planeFind all Points on the Surface at which the Tangent is Parallel to the PlaneFind all $P_0= (x_0, y_0, z_0)$ on $z = x + y^2$ so that the angle between normal vectors at $P_0$ and $(0,1,0)$ is $pi/4$.Angle between two intersecting lines that appear on a cut face

1

$begingroup$

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.
$$f(x,y,z):= 3x +z^2 -xy^2z$$

$nabla f(1,-2,1) = langle 3-y^2z,-2xyz,2z-xy^2rangle_(1,-2,1) = langle -1,4,-2 rangle$

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:

$$axcdot x_0 + bycdot y_0 + czcdot z_0 + u(x+x_0) + v(y+y_0) + w(z+z_0) +d=0$$

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

$a=langle 1,-4,2rangle$ and $b=langle 3,2,1rangle $

$Rightarrow theta= arccos(fracacdot b) = arccos(frac-37sqrt6)$

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 ?

calculus proof-verification partial-derivative vector-analysis

share|cite|improve this question

edited Jun 20 '18 at 16:19

saulspatz

17.1k31435

asked Jun 20 '18 at 15:06

So LoSo Lo

64929

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $a=langle -1,4,-2rangle$ 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 $arccosleft(frac37sqrt6right)$ 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
  
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

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.
$$f(x,y,z):= 3x +z^2 -xy^2z$$

$nabla f(1,-2,1) = langle 3-y^2z,-2xyz,2z-xy^2rangle_(1,-2,1) = langle -1,4,-2 rangle$

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:

$$axcdot x_0 + bycdot y_0 + czcdot z_0 + u(x+x_0) + v(y+y_0) + w(z+z_0) +d=0$$

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

$a=langle 1,-4,2rangle$ and $b=langle 3,2,1rangle $

$Rightarrow theta= arccos(fracacdot b) = arccos(frac-37sqrt6)$

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 ?

calculus proof-verification partial-derivative vector-analysis

share|cite|improve this question

edited Jun 20 '18 at 16:19

saulspatz

17.1k31435

asked Jun 20 '18 at 15:06

So LoSo Lo

64929

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $a=langle -1,4,-2rangle$ 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 $arccosleft(frac37sqrt6right)$ 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
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

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.
$$f(x,y,z):= 3x +z^2 -xy^2z$$

$nabla f(1,-2,1) = langle 3-y^2z,-2xyz,2z-xy^2rangle_(1,-2,1) = langle -1,4,-2 rangle$

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:

$$axcdot x_0 + bycdot y_0 + czcdot z_0 + u(x+x_0) + v(y+y_0) + w(z+z_0) +d=0$$

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

$a=langle 1,-4,2rangle$ and $b=langle 3,2,1rangle $

$Rightarrow theta= arccos(fracacdot b) = arccos(frac-37sqrt6)$

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 ?

calculus proof-verification partial-derivative vector-analysis

share|cite|improve this question

edited Jun 20 '18 at 16:19

saulspatz

17.1k31435

asked Jun 20 '18 at 15:06

So LoSo Lo

64929

$endgroup$

share|cite|improve this question

edited Jun 20 '18 at 16:19

saulspatz

17.1k31435

asked Jun 20 '18 at 15:06

So LoSo Lo

64929

share|cite|improve this question

edited Jun 20 '18 at 16:19

saulspatz

17.1k31435

asked Jun 20 '18 at 15:06

So LoSo Lo

64929

edited Jun 20 '18 at 16:19

saulspatz

17.1k31435

edited Jun 20 '18 at 16:19

saulspatz

17.1k31435

asked Jun 20 '18 at 15:06

So LoSo Lo

64929

asked Jun 20 '18 at 15:06

So LoSo Lo

64929

3 Answers
3

active

oldest

votes

0

$begingroup$

Did you go ahead and take arccos$left(frac-37sqrt6right)$? I get about 100$^o$. So your answer is no. The acute angle is 180 - 100 or about 80$^o$.

share|cite|improve this answer

answered Jun 20 '18 at 15:27

Dan Sp.Dan Sp.

30429

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  if I put $a=langle -1,4,-2rangle$ 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
  

  
  
  
  

add a comment | 

0

$begingroup$

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, ($brightarrow -b$), or equivalent
$$
theta= arccosleft(fracright)
$$
to get the acute angle

share|cite|improve this answer

answered Jun 20 '18 at 15:27

AndreiAndrei

13.2k21230

$endgroup$

add a comment | 

0

$begingroup$

We have

$$
vec n_1 = fracnabla (x y^2 z - 3 x + z^2) \
vec n_2 = fracnabla (3 x^2 - y^2 + 2 z - 1)
$$

and the sought angle is

$$
varphi = min(|arccos(pm<vec n_1,vec n_2>)|) = arccos(frac1sqrt742)approx 88^circ
$$

share|cite|improve this answer

edited Jun 20 '18 at 18:01

answered Jun 20 '18 at 16:06

CesareoCesareo

9,4713517

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $min(|arccos(pm<vec n_1,vec n_2>)|) = arccos(frac1sqrt742)approx 88^circ$
  $endgroup$
  – Cesareo
  Jun 20 '18 at 18:02
  

  
  
  
  

add a comment | 

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0

$begingroup$

Did you go ahead and take arccos$left(frac-37sqrt6right)$? I get about 100$^o$. So your answer is no. The acute angle is 180 - 100 or about 80$^o$.

share|cite|improve this answer

answered Jun 20 '18 at 15:27

Dan Sp.Dan Sp.

30429

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  if I put $a=langle -1,4,-2rangle$ 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
  

  
  
  
  

add a comment | 

0

$begingroup$

Did you go ahead and take arccos$left(frac-37sqrt6right)$? I get about 100$^o$. So your answer is no. The acute angle is 180 - 100 or about 80$^o$.

share|cite|improve this answer

answered Jun 20 '18 at 15:27

Dan Sp.Dan Sp.

30429

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  if I put $a=langle -1,4,-2rangle$ 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
  

  
  
  
  

add a comment | 

$begingroup$

Did you go ahead and take arccos$left(frac-37sqrt6right)$? I get about 100$^o$. So your answer is no. The acute angle is 180 - 100 or about 80$^o$.

share|cite|improve this answer

answered Jun 20 '18 at 15:27

Dan Sp.Dan Sp.

30429

$endgroup$

share|cite|improve this answer

answered Jun 20 '18 at 15:27

Dan Sp.Dan Sp.

30429

answered Jun 20 '18 at 15:27

Dan Sp.Dan Sp.

30429

answered Jun 20 '18 at 15:27

Dan Sp.Dan Sp.

30429

0

$begingroup$

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, ($brightarrow -b$), or equivalent
$$
theta= arccosleft(fracright)
$$
to get the acute angle

share|cite|improve this answer

answered Jun 20 '18 at 15:27

AndreiAndrei

13.2k21230

$endgroup$

add a comment | 

0

$begingroup$

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, ($brightarrow -b$), or equivalent
$$
theta= arccosleft(fracright)
$$
to get the acute angle

share|cite|improve this answer

answered Jun 20 '18 at 15:27

AndreiAndrei

13.2k21230

$endgroup$

add a comment | 

$begingroup$

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, ($brightarrow -b$), or equivalent
$$
theta= arccosleft(fracright)
$$
to get the acute angle

share|cite|improve this answer

answered Jun 20 '18 at 15:27

AndreiAndrei

13.2k21230

$endgroup$

share|cite|improve this answer

answered Jun 20 '18 at 15:27

AndreiAndrei

13.2k21230

answered Jun 20 '18 at 15:27

AndreiAndrei

13.2k21230

answered Jun 20 '18 at 15:27

AndreiAndrei

13.2k21230

0

$begingroup$

We have

$$
vec n_1 = fracnabla (x y^2 z - 3 x + z^2) \
vec n_2 = fracnabla (3 x^2 - y^2 + 2 z - 1)
$$

and the sought angle is

$$
varphi = min(|arccos(pm<vec n_1,vec n_2>)|) = arccos(frac1sqrt742)approx 88^circ
$$

share|cite|improve this answer

edited Jun 20 '18 at 18:01

answered Jun 20 '18 at 16:06

CesareoCesareo

9,4713517

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $min(|arccos(pm<vec n_1,vec n_2>)|) = arccos(frac1sqrt742)approx 88^circ$
  $endgroup$
  – Cesareo
  Jun 20 '18 at 18:02
  

  
  
  
  

add a comment | 

0

$begingroup$

We have

$$
vec n_1 = fracnabla (x y^2 z - 3 x + z^2) \
vec n_2 = fracnabla (3 x^2 - y^2 + 2 z - 1)
$$

and the sought angle is

$$
varphi = min(|arccos(pm<vec n_1,vec n_2>)|) = arccos(frac1sqrt742)approx 88^circ
$$

share|cite|improve this answer

edited Jun 20 '18 at 18:01

answered Jun 20 '18 at 16:06

CesareoCesareo

9,4713517

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $min(|arccos(pm<vec n_1,vec n_2>)|) = arccos(frac1sqrt742)approx 88^circ$
  $endgroup$
  – Cesareo
  Jun 20 '18 at 18:02
  

  
  
  
  

add a comment | 

$begingroup$

We have

$$
vec n_1 = fracnabla (x y^2 z - 3 x + z^2) \
vec n_2 = fracnabla (3 x^2 - y^2 + 2 z - 1)
$$

and the sought angle is

$$
varphi = min(|arccos(pm<vec n_1,vec n_2>)|) = arccos(frac1sqrt742)approx 88^circ
$$

share|cite|improve this answer

edited Jun 20 '18 at 18:01

answered Jun 20 '18 at 16:06

CesareoCesareo

9,4713517

$endgroup$

share|cite|improve this answer

edited Jun 20 '18 at 18:01

answered Jun 20 '18 at 16:06

CesareoCesareo

9,4713517

answered Jun 20 '18 at 16:06

CesareoCesareo

9,4713517

answered Jun 20 '18 at 16:06

CesareoCesareo

9,4713517