I have no idea where they came up with the fact that $a=\hat{i}* \hat{v}$ $/$ $|i|$ or $b=\hat{j}* \hat{v}$ $/$ $|j|$ etc. Can I can get proof explanation?
I don't see this equality:
Best Answer
Note that $\hat{i}$, $\hat{j}$ and $\hat{k}$ are orthonormal. Consequently, $\hat{i}\cdot\hat{j}=\hat{i}\cdot\hat{k}=\hat{j}\cdot\hat{k}=0$, and $\hat{i}\cdot\hat{i}=\hat{j}\cdot\hat{j}=\hat{k}\cdot\hat{k}=1$. Therefore, if $\mathbf{v} = a\hat{i}+b\hat{j}+c\hat{k}$ then $\hat{i}\cdot\mathbf{v} = a \hat{i}\cdot\hat{i}+ b \hat{i}\cdot\hat{j} + c \hat{i}\cdot\hat{k}=a$. Similarly, $\hat{j}\cdot\mathbf{v}=b$, and $\hat{k}\cdot\mathbf{v}=c$.
The gradient $\nabla f$ gives you the normal to the surface (your hill). To see in which direction the ball would roll, you have to project a downward pointing vector (representing gravitational attraction to the earth) on the plane normal to $\nabla f$ (which corresponds to substracting the normal force applied on the ball by the hill).
He wants to solve $p=x^2+y^2$.
If there is a solution, $p$ is odd so one of the squares has to be even. So instead, look at $p=x^2+(2y)^2=x^2+4yy$.
The clever bit is to look at $p=x^2+4yz$ instead of $p=x^2+4yy$. Most solutions come in pairs, because $x^2+4yz=x^2+4zy$. Only if $y=z$ does a solution not have a partner.
He counts the number of solutions, and because of the other formula finds there is an odd number.
So they don't all come in pairs, so there is a solution with $y=z$ - that is, $p=x^2+(2y)^2$
Best Answer
Note that $\hat{i}$, $\hat{j}$ and $\hat{k}$ are orthonormal. Consequently, $\hat{i}\cdot\hat{j}=\hat{i}\cdot\hat{k}=\hat{j}\cdot\hat{k}=0$, and $\hat{i}\cdot\hat{i}=\hat{j}\cdot\hat{j}=\hat{k}\cdot\hat{k}=1$. Therefore, if $\mathbf{v} = a\hat{i}+b\hat{j}+c\hat{k}$ then $\hat{i}\cdot\mathbf{v} = a \hat{i}\cdot\hat{i}+ b \hat{i}\cdot\hat{j} + c \hat{i}\cdot\hat{k}=a$. Similarly, $\hat{j}\cdot\mathbf{v}=b$, and $\hat{k}\cdot\mathbf{v}=c$.