The directional derivative of $f(x,y)$ at $(1,2)$ in the direction of $\vec a =\vec i + \vec j$ is $2\sqrt{2}$.
We also know that the directional derivative of $f(x,y)$ at $(1,2)$ in the direction of $\vec b = -2\vec j$ is $-3$.
What is the derivative of $f$ at $(1,2)$ in the direction of $\vec u=-\vec i -2\vec j$
What I did:
We know at $(1,2)$ : $\frac{\langle \nabla f,\vec a\rangle}{|\vec a|}=\frac{\langle \nabla f,\vec i +\vec j\rangle}{|\vec i+\vec j|}=\frac{\langle \nabla f,\vec i\rangle}{|\vec i +\vec j|}+\frac{\langle \nabla f,\vec j\rangle}{|\vec i +\vec j|}=2\sqrt{2}$
We also know that at (1,2): $\frac{\langle \nabla f,\vec b\rangle}{|\vec b|}=\frac{\langle \nabla f,-2\vec j\rangle}{|-2\vec j|}=\frac{-2}{2}\frac{\langle \nabla f,\vec j\rangle }{|j|}=-\frac{\langle \nabla f,\vec j\rangle }{|j|}=-3$
So overall we have:
$\frac{\langle \nabla f,\vec i\rangle}{|\vec i +\vec j|}+\frac{\langle \nabla f,\vec j\rangle}{|\vec i +\vec j|}=2\sqrt{2}$ and $\frac{\langle \nabla f,\vec j\rangle }{|j|}=3$
And from this, without any knowledge of what $\vec i$ or $\vec j$ are, we are supposed to find $\frac{\langle \nabla f,-\vec i -2\vec j\rangle}{|-\vec i -2\vec j|}$
I dont Immediately see how can this be done. $\vec i$ and $\vec j$ could be anything, the results we know don't help us solve what we want.
I'd appreciate any insight.
Just write $\vec u=-\vec a+\frac12 \vec b$. Then,
$\nabla f \cdot \vec u=\nabla f \cdot (-\vec a+\frac12 \vec b)=-(\nabla f \cdot \vec a)+\frac12 (\nabla f \cdot \vec b)=-(2\sqrt{2})(\sqrt{2})+\frac12 (-3)(2)=-7$
Now just divide by $|\vec u|=\sqrt{5}$ and find the directional derivative is $-7\sqrt{5}/5$.