Angle between two surfaces

Question

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.

Answer 1

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

Answer 2

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