my question is:
Consider a temperature function in space given by
$$R(x,y,z) = \frac{1}{(x-1)^2 + (y-1)^2 + (z-3)^2 + 1}$$
- A bird is located at the point $(-1, 1, 3)$ and wants to minimize radiation at the maximum speed. In which direction should the bird fly, considering the negative gradient of the temperature function?
- A person is restricted in their movement and is confined to the plane $z = y - x + 1$. They aim to minimize radiation at the maximum speed. In which direction should the person move within the plane?
the first question is somehow easy as i solved it by finding the negative gradient of the function at the given point. so the answer should be $(\frac{-4}{25},0,0)$.
but my problem is with the second question, I know that if the move is restricted to a surface or a plane, the direction of steepest descent isn't just given by the negative gradient of the function (like flying), but by the projection of the negative gradient onto the tangent plane of the surface at the given point but the thing i still don't know how should i solve it.
You're right. You are one step away from the final answer, which is to find the projection of vector $\vec{a}=(-\frac{4}{25},0,0)$ onto the plane $x-y+z=1$. Firstly, for the plane $Ax+By+Cz=1$, the direction of its normal vector is $(A,B,C)$, which is $\vec{b}=(1,-1,1)$ in this case. The orthogonal component of vector $\vec{a}$ relative to $\vec{b}$ is $\vec{b}-|\vec{a}|<\vec{a},\vec{b}>$. Therefore, the final answer is $(-\frac{4}{25}-\frac{4\sqrt{3}}{75},\frac{4\sqrt{3}}{75},-\frac{4\sqrt{3}}{75})$.