So now, we know what is partial derivative. So let us move on towards the actual equation of tangent line. So what we do know now, we know two straight lines, tangent lines in two cross sections with two streams, x equal to a and y equal to b, which are definitely into our tangent plane. So is it enough to actually write an equation of this plane? Well, it turns out that it is. Firstly, we need to revisit some basic things about the equation of a plane. As we start with a scalar equation of a plane, which is as following. It is scalar product or just as a result of multiplication and summation of two basic essentials. So we assume that we are trying to characterize arbitrary point of our plan x, y, z, and so we are going to do so by two scenes. One, given a point of our plane for example x_0, y_0 and z_0, and a vector which is orthogonal for all our plane. In other words, it is orthogonal for any vector that license a plane actually. So what we're going to say is that scalar equation is basically the idea that these two vectors are orthogonal, or in other manner, we are going to multiply our coordinates of the orthogonal vector and the vector which is definitely lying in the plane, which is vector between point x, y, z, our arbitrary point and point x_0, y_0 and z_0. That's our equation of a plane. It's not actually the only thing that we can define our plane, but it's quite convenient and it's usually used in many many occasions. So another thing I'm going to remind you is the concept of scalar product, which is a mapping from two vectors into real numbers and it works pretty much as we stated above. It's just a summation of multiplied coordinates of two vectors. It is pretty much what we have done in our scalar equation plane, and it is your responsibility to know but let us reiterate is, scalar project has its geometrical meaning, which is basically a product of the lengths of two vectors multiplied by the cosine of an angle between them. You also should consider that in two dimensional cases, multidimensional cases and spaces of three, four and other dimensions, that's basically a definition of an angle. So it is not quite an easy way to roll with the scalar product here. It's extremely complicated actually. But it produces some other 3D scene here because what we're going to say looking at our multiplication by the cosine function is that the orthogonal case is actually characterized by the value of a scalar product if two vectors are orthogonal, which means that they have an angle of half of Pi. So cosine of half of a Pi is zero, so multiplication by the cosine function results into zero as a scalar project result. So basically, that's how one can define that two vectors are orthogonal. So how are we going to use it. We are going to use it in the following manner. First of all, we know the point that definitely belongs to our tangent plane. It is a point a,b,f from a, b. That's what we are expecting, that's the point of our approximation. So we need to find the orthogonal vector to all our tangent planes here, but we do know we do know two vectors which has directional for our tangent lines which we have mentioned whilst we were defining our partial derivatives. They are directional for our tangent lines. In other way, we do know that one of the variables during the calculation of partial derivatives were fixed. For example, whilst we were defining as the partial derivative y were fixed, thus means that the y-coordinate of the directional vector of this tangent line equals to zero. The same applies for the y case and x-coordinate in this case. So we have actually two vectors. 1,0 partial derivative towards x, and 0,1 partial derivative derivative towards y. Which results into 2x as definitely lying in our tangent plane. So yes, we do know it. We need to construct another vector, a third vector which is orthogonal to this two, and it's quite easy. Just let us revisit it. It is a vector which is partial derivative towards x, partial derivative towards y and minus one. Let's say that whilst we can calculate this scalar product with the first vector here and the second vector here, we are almost definitely going to get zero. Well, it is easy because we multiply 1 by some partial derivative, and then minus 1 by very much partial derivative answers in summation we get zero. So now, we can write our scalar equation of a plane and with just an easy transformation, because it is nice to have that tangent variable on the left side of the equation, and to the right-hand, they are going to just write this down. So our equation of the tangent plane is our tangent variable equals the function at this very point, and then partial derivatives was multiplied by the change of arguments x and y respectively. So let us for example look at the case of previously discussed function, square root from absolute value of product x and y. We have actually already stated that that the point 0,0 both of our partial derivatives are actually equals to zero. Okay so I am going to just write it down. Okay, that's fine. Let us proceed with an actual equation. We need to write z equals to. First of all, the value of our function at point 0,0. Well, the absolute value of zero is zero, so it was quite easy. Then we need to write our partial derivative towards x multiplied by x minus 0 plus partial derivative towards y multiplied by y minus 0. Well, that is quite an easy one. So the thing is that we are going to call our tangent plane, these right points for function square root from multiplication of x and y is horizontal plane. The planes that coincides with 0 x, y plane of the three-dimensional space. Okay, and that's nice. That's a fine way to define our tangent plane.
