Let $\vec{a}, \vec{b} , \vec{c} $ be three vectors in $\mathbb{R}^3 $ that are orthogonal to each other and have length 1. Show that for every differentiable $f:\mathbb{R}^3 \to \mathbb{R}$ and for every point $(x,y,z)$ we have:

$$ \Big( \frac{\partial f}{\partial \vec{a} } \Big)^2 +\Big( \frac{\partial f}{\partial \vec{b} } \Big)^2 +\Big( \frac{\partial f}{\partial \vec{c} } \Big)^2 = \Big( \frac{\partial f}{\partial {x} } \Big)^2 + \Big( \frac{\partial f}{\partial {y} } \Big)^2+\Big( \frac{\partial f}{\partial {z} } \Big)^2$$

The only possible argument I thought of is that the LHS represents the maximal value of the directional derivative, and it is obviously an invariant size under rotation. The problem is that this argument seems not very formal... Will you help me find a proper argument for this ?

Thanks a lot

  • $\begingroup$ How much linear algebra do you know? (Hint: there exists a linear transformation taking the $(a,b,c)$ vectors to the $(x,y,z)$ vectors. Call this linear transformation $A$. What you want to show is equivalent to showing $A^TA = I$ (why?)) $\endgroup$ Jan 17, 2014 at 9:30

1 Answer 1


Let $a=(a_1,a_2,a_3)$, $b=(b_1,b_2,b_3)$ and $c=(c_1,c_2,c_3)$. Then $$ \frac{\partial f}{\partial a}=a_1\frac{\partial f}{\partial x}+ a_2\frac{\partial f}{\partial y}+ a_3\frac{\partial f}{\partial z} $$ $$ \frac{\partial f}{\partial b}=b_1\frac{\partial f}{\partial x}+ b_2\frac{\partial f}{\partial y}+ b_3\frac{\partial f}{\partial z} $$ $$ \frac{\partial f}{\partial c}=c_1\frac{\partial f}{\partial x}+ c_2\frac{\partial f}{\partial y}+ c_3\frac{\partial f}{\partial z} $$ Thus $$ \nabla_{a,b,c}\ f=U\nabla_{x,y,z}\ f, $$ where $$ U=\left(\begin{matrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{matrix}\right) $$ Clearly $UU^T=I$, i.e., $U$ is orthogonal. Then \begin{align} \left|\frac{\partial f}{\partial a}\right|^2+\left|\frac{\partial f}{\partial b}\right|^2 +\left|\frac{\partial f}{\partial c}\right|^2&=\nabla_{a,b,c}\ f\cdot \nabla_{a,b,c}\ f =U\nabla_{x,y,z}\ f\cdot U\nabla_{x,y,z}\ f \\ &=\nabla_{a,b,c}\ f\cdot \nabla_{a,b,c}\ f =\nabla_{x,y,z}\ f\cdot U^TU\nabla_{x,y,z}\ f=\nabla_{x,y,z}\ f\cdot \nabla_{x,y,z}\ f\\ &=\left|\frac{\partial f}{\partial x}\right|^2+\left|\frac{\partial f}{\partial y}\right|^2 +\left|\frac{\partial f}{\partial z}\right|^2 \end{align}

  • $\begingroup$ Thanks for your detailed answer. There are a few things I can't understand: Can you please explain how did you use the chain rule in your first calculations ? I can't see how you got $a_1 , a_2 , a_3 $ as the coefficients $\endgroup$
    – criticism
    Jan 17, 2014 at 9:39
  • $\begingroup$ By definition, if $a$ is a nonzero vector, then $$\frac{\partial f}{\partial a}=\nabla f\cdot a.$$ $\endgroup$ Jan 17, 2014 at 9:45

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