## Calculus (3rd Edition)

$$\langle-2\sqrt{2},-2\sqrt{2}\rangle.$$
Since the vector is in the direction $\langle-1,-1\rangle$, then it will be in the form $au=\langle-a,-a\rangle$. Now, since the length is $4$, then we have $$\sqrt{a^2+a^2}=\sqrt{2a^2}=4\Longrightarrow 2a^2=16\Longrightarrow a=\sqrt{8}=2\sqrt{2}.$$ Hence the required vector is given by the components $$\langle-2\sqrt{2},-2\sqrt{2}\rangle.$$