I have a question concerning two planes and a given point of intersection in which I am asked to show that they do intersect at that point.
The sample method for answering the question goes as follows:
1) find normal vectors of the plane by implementing partial derivatives
2) substitute the point into the equations of the normal vectors
3) find the dot product of the two vectors and if it is zero then they intersect at that point
What I am having difficulty understanding is why the equations for the normal vectors are found using partial derivatives. Doesn't a first derivative give the slope of the tangent at that point and not the normal? In the example one of the planes is $z = 8 – x^3 – y^2$
so the normal vector is $(3x^2, 2y, 1)$. Why? Also, is the $1$ the derivative of $z$?
Though this is the sample way of answering the question I am wondering if there is a better way to do it, say using a system of equations or something?
Best Answer
The normal vector in your example is the gradient $(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})$. So, yes, the $1$ is the derivative of $z$.
If this is hard to understand, you can think of a 2D example. Let's say, $y=x^2$. The slope of the tangent line is $2x$, so the tangent vector is $(1,2x)$, $2x$ being $\frac{dy}{dx}$. This means when $x$ is increasing by $1$, $y$ is increasing by $\frac{dy}{dx}$.
The normal vector though, is $(2x, -1)$. They are perpendicular, that's why their dot product is $0$. They both involve the derivative $2x$, but in different component. You can easily draw the picture of $y=x^2$ to visualize its tangent and normal vectors.