00:01
In this problem, we're trying to find a point on the surface, the given surface here, that is perpendicular to the line that is defined parametrically right here.
00:15
So essentially, if the plane is perpendicular to the given line, then what we know is that the normal vector is parallel to the slope vector of our line.
00:27
Okay.
00:28
So in other words, if we call the surface f, and we call our line g, then the gradient of f would be equal to the gradient of g.
00:46
And that's pretty much what we're trying to show.
00:50
So first let's rewrite our surface as a function of three variables.
00:59
So f of x, y, z would be equal to 8 minus 3 x squared minus 2.
01:09
Square minus b and our gradient here we take the partial derivatives is a gradient of x of f so with if we take the derivative with respect to x that would be negative x with respect to y with respect to y that would be negative for y and with respect to z and it simply be negative one now, to get the gradient of g, what we can recall, our parametric equations here, note that we get those by taking x of 0 and the partial derivative of x at the given point multiplied by t.
02:11
So based on this format, we can see that the gradient of the line, which we're calling g, would be the coefficient of my of the t variable...