Please excuse the non-specific title, this is a rather long problem.

So on our last exam in multivariable calculus, our professor gave us a very lengthy vector manipulation problem as a bonus. Seeing as it's no longer worth any points to me, I was wondering if someone could help me understand how to solve it.

So let $ \vec v_1,\vec v_2,\vec v_3 \in \Bbb R^3$, such that $\vec v_1 \cdot (\vec v_2 \times \vec v_3) \neq 0$

Now define

$$k_1 = \frac{\vec v_2 \times \vec v_3}{\vec v_1 \cdot (\vec v_2 \times \vec v_3)}, k_2 = \frac{\vec v_3 \times \vec v_1}{\vec v_1 \cdot (\vec v_2 \times \vec v_3)}, k_3 = \frac{\vec v_1 \times \vec v_32}{\vec v_1 \cdot (\vec v_2 \times \vec v_3)}$$

We must show:

i) Each $k_i$ is perpendicular to the corresponding $v_j$ where $j \neq i$;

ii) $\vec k_1 \cdot (\vec k_2 \times \vec k_3) = \frac{1}{\vec v_1 \cdot (\vec v_2 \times \vec v_3)}$

The second part seems somewhat straightforward to solve through brute force, but I'm wondering if there's a simpler way. However, I have no idea how to approach the first part; I have no notion of vector division and that seems to be what it's approaching.

  • 1
    $\begingroup$ The denominator in each of those expressions is just a number (the triple scalar product), so each $k_i$ is just a scalar multiple of the cross product in the numerator, and you know that a cross product of two vectors is orthogonal to both. $\endgroup$ Mar 31, 2015 at 4:29
  • $\begingroup$ So each $k_i$ would resolve to some scalar times the cross product of $v_j$ and $v_k$, where $i \neq j \neq k$? $\endgroup$ Mar 31, 2015 at 4:42
  • $\begingroup$ I mean, each $k_i$ is, by definition, some scalar times the cross product. $\endgroup$ Mar 31, 2015 at 4:43
  • $\begingroup$ Oh, I did not see that somehow. That completely clarifies part i. Thank you. $\endgroup$ Mar 31, 2015 at 4:44
  • $\begingroup$ I think it's mostly that the notation is confusing. You just see a lot of vectors and operations on those vectors, but in fact the dot product is a very different operation from the cross product because it returns a scalar rather than a vector. It's confusing to use notation that makes them seem so similar. $\endgroup$ Mar 31, 2015 at 4:52

2 Answers 2


Here is a clean way to solve this problem without any brute force, although it might involve more linear algebra than you've seen. Combine the following facts:

  1. The triple scalar product $v_1 \cdot (v_2 \times v_3)$ is the determinant of the $3 \times 3$ matrix $V$ with columns $v_1, v_2, v_3$. It is also the determinant of the $3 \times 3$ matrix with rows $v_1, v_2, v_3$.
  2. $k_1, k_2, k_3$ are the rows of the inverse matrix $V^{-1}$.
  3. The determinant of the inverse of a matrix is the inverse of its determinant, so $\det(V^{-1}) = \det(V)^{-1}$.
  • $\begingroup$ Wow, it makes perfect sense when you frame it like that. I guess I was just bogged out by the sheer number of vectors thrown at me... but anyway, I see exactly what you're getting at with statement 3, and I see how to use that to prove the second part of the question. Thank you so much, man. Go bears. $\endgroup$ Mar 31, 2015 at 5:00

$\vec k_i \cdot \vec v_j = \frac{(\vec v_j \times \vec v_k) \cdot (\vec v_j)}{\vec v_1 \cdot (\vec v_2 \times \vec v_3 )} =0 $

since the numerator contains the inner product of $\vec v_j$ with a vector orthogonal to $\vec v_j$. We may replace $j$ with $k$ to complete the proof.

For ii) we have

$$\begin{align} \vec k_1 \cdot (\vec k_2 \times\vec k_3)&=\frac{(\vec v_2 \times \vec v_3)\cdot (\vec v_3 \times \vec v_1)\times (\vec v_1 \times \vec v_2)}{\vec v_1 \cdot (\vec v_2 \times \vec v_3)}\\\\ &=\frac{(\vec v_2 \times \vec v_3)\cdot \vec v_1[(\vec v_3 \times \vec v_1)\cdot \vec v_2]}{\vec v_1 \cdot (\vec v_2 \times \vec v_3)}\\\\ &=\frac{ [\vec v_1 \cdot(\vec v_2 \times \vec v_3)][(\vec v_3 \times \vec v_1)\cdot \vec v_2]}{\vec v_1 \cdot (\vec v_2 \times \vec v_3)}\\\\ &=\vec v_2 \cdot (\vec v_3 \times \vec v_1)\\\\ &=\vec v_1 \cdot (\vec v_2 \times \vec v_3) \end{align}$$


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