Doing an exercise a about directional derivatives, it was required to find the derivative of a given function $f(x,y,z)$ in the direction of the vector $ \vec{v}$ that forms with the coordinates axes $OX, OY, OZ$ the angles $\alpha, \beta, \gamma$, respectively. The book gives the following formula to solve the exercise:
$$\frac{df}{d\vec{v}}(x_o, y_o, z_o) = \vec{\nabla } f(x_o, y_o, z_o ) \cdot(\cos\alpha, \cos\beta, \cos\gamma)$$
In 2 variables (using sine and cosine of the angle $\alpha$ formed with $OX$), it is easy to see the formula using that $\sin^2 \alpha + \cos^2 \alpha = 1$. But, how can I show that $||(\cos\alpha, \cos\beta, \cos\gamma)|| = 1 $ using that $\alpha + \beta + \gamma = \pi /2$?
Maybe it is more a trigonometry problem than calculus.
Thanks!
Best Answer
This is indeed just a trigonometric problem in $\mathbb{R}^3$.
The cosine rule states that $$ |\mathbf{a}+\mathbf{b}|^2=|\mathbf{a}|^2+|\mathbf{a}|^2 +2|\mathbf{a}|\,|\mathbf{b}|\cos\theta, $$ or equivalently $$ \mathbf{a}\cdot\mathbf{b}=|\mathbf{a}|\,|\mathbf{b}|\cos\theta, $$ where $\theta$ is the angle formed by the vectors $\mathbf{a}$ and $\mathbf{b}$. In your case, this gives, assuming that $\mathbf{v}$ is a unit vector, $$ v_1=\mathbf{v}\cdot\mathbf{e}_1=\cos\alpha\quad\text{ etc.}, $$ and hence $$ \mathbf{v}=(\cos\alpha,\cos\beta,\cos\gamma). $$ As a consequence, $$ 1=|\mathbf{v}|=|(\cos\alpha,\cos\beta,\cos\gamma)| $$ as desired.