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Angle Between Two Planes

jk8985

New member
Dec 16, 2013
12
The graph of z=x^2+y^2+1 and the graph of x+y+z=e^(xyz) have a common point (0,0,1). Find the angle between the two corresponding tangent planes respect to these two graphs at point (0,0,1). In addition, find the tangent line of the curve intersected by these two graphs at point (0,0,1).
 

jk8985

New member
Dec 16, 2013
12
So I tried this:

x^2+y^2-z=-1
and I know the angle=arccos(dot product of normal unit vectors)

But I don't know exactly what to do :(
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
The graph of z=x^2+y^2+1 and the graph of x+y+z=e^(xyz) have a common point (0,0,1). Find the angle between the two corresponding tangent planes respect to these two graphs at point (0,0,1). In addition, find the tangent line of the curve intersected by these two graphs at point (0,0,1).
We can think of the two surfaces as "equi-potential" surfaces of [tex]f(x,y,z)= z- x^2- y^2[/tex] and [tex]g(x,y,z)= x+ y+ z- e^{xyz}[/tex]. Their gradients, [tex]\nabla f= -2x\vec{i}- 2y\vec{j}+ \vec{k}[/tex] and [tex]\nabla g= (1- yze^{xyz})\vec{i}+ (1- xze^{xyz})\vec{j}+ (1- xye^{xyz})\vec{k}[/tex] are perpendicular to the surfaces and perpendicular to the tangent planes at each point.

In particular, at (0, 0, 1), [tex]\nabla f(0, 0, 1)= \vec{k}][/tex] so the tangent plane is [tex]0(x- 0)+ 0(y- 0)+ 1(z- 1)= 0[/tex] or z= 1. And [tex]\nabla g(0, 0, 1)= \vec{i}+ \vec{j}+ \vec{k}[/tex] so the tangent plane is [tex]1(x- 0)+ 1(y- 0)+ 1(z- 1)= x+ y+ z- 1= 0[/tex] or [tex]x+ y+ z= 1[/tex]. The angle between those two planes is the same as the ange between the two perendicular vectors and you can use [tex]\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)[/tex] to find that angle.

The length of [tex]\vec{k}[/tex] is 1, the length of [tex]\vec{i}+ \vec{j}+ \vec{k}[/tex] is [tex]\sqrt{3}[/tex], and their dot product is 1. So [tex]1(\sqrt{3})cos(\theta)= 1[/tex].
 

jk8985

New member
Dec 16, 2013
12
We can think of the two surfaces as "equi-potential" surfaces of [tex]f(x,y,z)= z- x^2- y^2[/tex] and [tex]g(x,y,z)= x+ y+ z- e^{xyz}[/tex]. Their gradients, [tex]\nabla f= -2x\vec{i}- 2y\vec{j}+ \vec{k}[/tex] and [tex]\nabla g= (1- yze^{xyz})\vec{i}+ (1- xze^{xyz})\vec{j}+ (1- xye^{xyz})\vec{k}[/tex] are perpendicular to the surfaces and perpendicular to the tangent planes at each point.

In particular, at (0, 0, 1), [tex]\nabla f(0, 0, 1)= \vec{k}][/tex] so the tangent plane is [tex]0(x- 0)+ 0(y- 0)+ 1(z- 1)= 0[/tex] or z= 1. And [tex]\nabla g(0, 0, 1)= \vec{i}+ \vec{j}+ \vec{k}[/tex] so the tangent plane is [tex]1(x- 0)+ 1(y- 0)+ 1(z- 1)= x+ y+ z- 1= 0[/tex] or [tex]x+ y+ z= 1[/tex]. The angle between those two planes is the same as the ange between the two perendicular vectors and you can use [tex]\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)[/tex] to find that angle.

The length of [tex]\vec{k}[/tex] is 1, the length of [tex]\vec{i}+ \vec{j}+ \vec{k}[/tex] is [tex]\sqrt{3}[/tex], and their dot product is 1. So [tex]1(\sqrt{3})cos(\theta)= 1[/tex].
This is exactly what I ended up getting, with theta equalling ~1 radian. How would I go about finding the tangent line? I would always think it's a plane, so I didn't know how to approach it.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
How would I go about finding the tangent line? I would always think it's a plane, so I didn't know how to approach it.
The curve is the intersection of the two graphs, and the tangent line to the curve (at the point (0,0,1)) will be the intersection of the tangent planes to the graphs at that point. HallsofIvy has already found the equations of the two planes, so all you need to do now is to find the equation of their intersection.
 

jk8985

New member
Dec 16, 2013
12
The curve is the intersection of the two graphs, and the tangent line to the curve (at the point (0,0,1)) will be the intersection of the tangent planes to the graphs at that point. HallsofIvy has already found the equations of the two planes, so all you need to do now is to find the equation of their intersection.
How would I make that at the point (0,0,1) [i think it was that point]

- - - Updated - - -

How would I make that at the point (0,0,1) [i think it was that point]
My apologies for the bad grammar, by the way, I'm on my phone.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
This is exactly what I ended up getting, with theta equalling ~1 radian. How would I go about finding the tangent line? I would always think it's a plane, so I didn't know how to approach it.
What tangent line are you talking about? The tangent line to the curve of intersection? "I would always think it's a plane" confuses me. What do you always think is a plane"?

The tangent planes of the two surfaces, at (0, 0, 1), are, as I said before, z= 1 and x+ y+ z= 1. Since z= 1, x+ y+ z= x+ y+ 1= 1 so x+ y= 0 and y= -x. Taking x as parameter, we can write x= t, y= -t, z= 1 as parametric equations for the tangent line to the curve of intersection at (0, 0, 1).