Double vector product
Literature: Collection of problems in mathematics. Part 1. Edited by A.V. Efimov, B.P. Demidovich.
The double vector product is defined as the vector $[a, [b, c]]$, where $[a, b]$ denotes the vector product of vectors $a$ and $b$.
2.123*. The vector $[a, [b, c]]$ is called the double vector product. Prove that the equality $$[a, [b, c]]=b(a, c)-c(a, b).$$ holds true.
Solution.
Let $a=(a_x, a_y, a_z), b=(b_x, b_y, b_z), c=(c_x, c_y, c_z).$ Then
$$[b, c]=\begin{vmatrix}i&j&k\\b_x&b_y&b_z\\c_x&c_y&c_z\end{vmatrix}=i(b_yc_z-b_zc_y)-j(b_xc_z-b_zc_x)+k(b_xc_y-b_yc_x).$$
$$[a, [b,c]]=\begin{vmatrix}i&j&k\\a_x&a_y&a_z\\b_yc_z-b_zc_y&-b_xc_z+b_zc_x&b_xc_y-b_yc_x\end{vmatrix}=$$ $$=i(a_y(b_xc_y-b_yc_x)-a_z(b_zc_x-b_xc_z))-$$ $$-j(a_x(b_xc_y-b_yc_x)-a_z(b_yc_z-b_zc_y))+$$ $$+k(a_x(b_zc_x-b_xc_z)-a_y(b_yc_z-b_zc_y))=$$ $$=i(a_yb_xc_y-a_yb_yc_x+a_zb_xc_z-a_zb_zc_x)-$$ $$-j(a_xb_xc_y-a_xb_yc_x-a_zb_yc_z+a_zb_zc_y)+$$ $$+k(a_xb_zc_x-a_xb_xc_z-a_yb_yc_z+a_yb_zc_y).$$
Next, let's find the vector $b(a, c)-c(a, b):$
$$b(a, c)=(b_x, b_y, b_z)(a_xc_x+a_yc_y+a_zc_z)=$$ $$=(a_xb_xc_x+a_yb_xc_y+a_zb_xc_z, a_xb_yc_x+a_yb_yc_y+a_zb_yc_z, $$ $$a_xb_zc_x+a_yb_zc_y+a_zb_zc_z);$$
$$c(a, b)=(c_x, c_y, c_z)(a_xb_x+a_yb_y+a_zb_z)=$$ $$=(a_xb_xc_x+a_yb_yc_x+a_zb_zc_x, a_xb_xc_y+a_yb_yc_y+a_zb_zc_y, $$ $$a_xb_xc_z+a_yb_yc_z+a_zb_zc_z);$$
Hence,
$$b(a, c)-c(a, b)=i(a_yb_xc_y-a_yb_yc_x+a_zb_xc_z-a_zb_zc_x)-$$ $$-j(a_xb_xc_y-a_xb_yc_x-a_zb_yc_z+a_zb_zc_y)+$$ $$+k(a_xb_zc_x-a_xb_xc_z-a_yb_yc_z+a_yb_zc_y)=[a, [b, c]].$$
Which is what needed to be proved.
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