Prove $\vec a [\vec b\ \vec c\ \vec d] + \vec c[\vec a\ \vec b\ \vec d]=\vec b[\vec a\ \vec c\ \vec d] + \vec d[\vec a \ \vec b \ \vec c]$

Question

Consider the non zero vectors $ \vec a, \vec b,\vec c$ and $\vec d$ such that no three of which are coplanar then prove that $$\vec a [\vec b\ \vec c\ \vec d] + \vec c[\vec a\ \vec b\ \vec d]=\vec b[\vec a\ \vec c\ \vec d] + \vec d[\vec a \ \vec b \ \vec c]$$ Hence prove that $ \vec a, \vec b,\vec c$ and $\vec d$ represent the position vectors of the vertices of a plane quadrilateral if and only if $$\frac{[\vec b\ \vec c\ \vec d]+[\vec a\ \vec b\ \vec d]}{[\vec a\ \vec c\ \vec d]+[\vec a \ \vec b \ \vec c]} =1$$

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WHere do I begin ?

Answer 1

Write each vector as a sum of components in the form $\vec{a} = a_i\hat{i} + a_j\hat{j} + a_k\hat{k}$, and observe that

$\vec{a}[\vec{b}, \vec{c}, \vec{d}] + \vec{c}[\vec{a}, \vec{b}, \vec{d}] = a_i[\vec{b}, \vec{c}, \vec{d}]\hat{i} + a_j[\vec{b}, \vec{c}, \vec{d}]\hat{j} + a_k[\vec{b}, \vec{c}, \vec{d}]\hat{k} + c_i[\vec{a}, \vec{b}, \vec{d}]\hat{i} + c_j[\vec{a}, \vec{b}, \vec{d}]\hat{j} + c_k[\vec{a}, \vec{b}, \vec{d}]\hat{k} = (a_i[\vec{b}, \vec{c}, \vec{d}] + c_i[\vec{a}, \vec{b}, \vec{d}])\hat{i} + (a_j[\vec{b}, \vec{c}, \vec{d}] + c_j[\vec{a}, \vec{b}, \vec{d}])\hat{j} + (a_k[\vec{b}, \vec{c}, \vec{d}] + c_k[\vec{a}, \vec{b}, \vec{d}])\hat{k}$

It remains to be shown that $a_x[\vec{b}, \vec{c}, \vec{d}] + c_x[\vec{a}, \vec{b}, \vec{d}] = b_x[\vec{a}, \vec{c}, \vec{d}] + d_x[\vec{a}, \vec{b}, \vec{c}]$ for $x \in \{1,2,3\}$. This is cumbersome yet manageable.

Answer 2

Let $\vec{a} = x \vec{b}+ y \vec{c}+ z \vec{d}$.

Then 'dotting' with $\vec{c} \times \vec{d}$ on both sides we get

$[\vec{a} \ \ \vec{c} \ \ \vec{d}] = x [\vec{b} \ \ \vec{c} \ \ \vec{d}]$

Thus we get $[\vec{b} \ \ \vec{c} \ \ \vec{d}]\vec{a}=[\vec{a} \ \ \vec{c} \ \ \vec{d}]\vec{b}+ [\vec{a} \ \ \vec{d} \ \ \vec{b}]\vec{c}+[\vec{a} \ \ \vec{b} \ \ \vec{c}]\vec{d} $

which using the properties of the triple product yield upon rearranging that

$[\vec{b} \ \ \vec{c} \ \ \vec{d}]\vec{a}+ [\vec{a} \ \ \vec{b} \ \ \vec{d}]\vec{c}=[\vec{a} \ \ \vec{c} \ \ \vec{d}]\vec{b}+[\vec{a} \ \ \vec{b} \ \ \vec{c}]\vec{d} $

Answer 3

Since these are 3-vectors, any four of them must be linearly-dependent. This linear dependence is preserved by the mapping $(v_x,v_y,v_z)\mapsto (v_x,v_y,v_z,v_x)$, yielding four linearly-dependent 4-vectors. In terms of determinants, this means

$$\begin{vmatrix} a_x & b_x & c_x & d_x \\ a_y & b_y & c_y & d_y \\ a_z & b_z & c_z & d_z \\ a_x & b_x & c_x & d_x\end{vmatrix}=0.$$ Expanding along the last row and employing the [] notation for scalar triple product, we obtain

$$a_x\!\left[\vec{b}\;\vec{c}\;\vec{d}\right]-b_x\!\left[\vec{a}\;\vec{c}\;\vec{d}\right]+c_x\!\left[\vec{a}\;\vec{b}\;\vec{d}\right]-d_x\!\left[\vec{a}\;\vec{b}\;\vec{c}\right]=0$$ which rearranges to $$a_x\!\left[\vec{b}\;\vec{c}\;\vec{d}\right]+c_x\!\left[\vec{a}\;\vec{b}\;\vec{d}\right]=b_x\!\left[\vec{a}\;\vec{c}\;\vec{d}\right]+d_x\!\left[\vec{a}\;\vec{b}\;\vec{c}\right].$$ Analogous arguments apply to the $y$-,$z$-components, which yields the desired vector equation. (This approach generalizes immediately to $n+1$ vectors in $\mathbb{R}^n$, with $n$-by-$n$ determinants replacing scalar triple products.)