# Angle between two surfaces

I have the next question:
Find the angle between two surfaces: $$x^2+y^2+z^2=9$$ and $$z=x^2+y^2-3$$ at the point $$(2,-1,2)$$.
I have the next formula: $$\cos\theta =\frac{\nabla \Phi _{1}\cdot\nabla \Phi _{2}}{\left | \nabla \Phi _{1} \right |\cdot \left | \nabla \Phi _{2} \right |}$$
But I don't understand why we need to do gradient for the two surfaces, I would be very happy if someone would explain this to me. Thank You.

• How would you do it if you couldn't use the gradient ?
– user65203
Jun 6, 2019 at 19:04
• I do not know, but that's what we learned in class. I just do not understand how the gradient action solves the problem Jun 6, 2019 at 19:09
• It is useful to ask yourself how you would solve, starting from scratch.
– user65203
Jun 6, 2019 at 19:10

Locally, two smooth surfaces can be approximated as two planes. To determine the angle between two planes, you can draw a line in both planes, from a point on the intersection, and measure the angle they form.

But this definition is not satisfactory because the angle will depend on the particular lines. To work around this problem, we observe that the angle is minimized when the lines are perpendicular to the intersection, and this minimum angle is uniquely defined.

As the gradient of a surface is perpendicular to it, the two gradients are perpendicular to the intersection, and the angle between the planes is also the angle between the gradient vectors.

Intuition of angle between two planes

• But the gradient gives us the direction of the fastest change of function, and to find the angle between to surfaces we need to find two direction vectors per surface and then to do the dot product. Jun 6, 2019 at 19:25

Let $$P=x^2+y^2+z^2=9$$ and $$Q=x^2+y^2-z=3$$

Now take the gradient of the above surfaces

$$\vec VA=\dfrac{dA}{dx}i+\dfrac{dA}{dy}j+\dfrac{dA}{dz}k=2xi+2yj+2zk=\vec R$$

and $$\vec VB=2xi+2yj-k=\vec S$$

Given vectors at point $$(2,-1,2)$$ is

$$\vec R=4i-2j+4k$$

Now the dot product of vectors $$R$$ and $$S$$ is

$$\vec R\cdot \vec S=|R||S|\cos\theta=(4i-2j+4k)(4i-2j-k)=\sqrt{4^2+2^2+4^2}\sqrt{4^2+2^2+1^2}\cos\theta$$

$$16=6\sqrt{21}\cos\theta$$

$$\cos\theta=\dfrac{8}{3\sqrt{21}}$$

$$\theta=\arccos\left(\dfrac{8}{3\sqrt{21}}\right)$$

Therefore, the angle between the two surfaces is $$\theta=\arccos\left(\dfrac{8}{3\sqrt{21}}\right)$$

• Thank you for the full answer, so if I understood this right the gradient gives us a direction vector at a specific point, and then we do dot product to find the angle between the surfaces? Jun 6, 2019 at 19:50