کافیست قرار دهیم:$A= a_{1}\overrightarrow{i}+a_{2} \overrightarrow{j} +a_{3} \overrightarrow{k} $ و $B= b_{1} \overrightarrow{i}+b_{2} \overrightarrow{j} +b_{3} \overrightarrow{k} $و$C= c_{1} \overrightarrow{i}+c_{2}\overrightarrow{j} +c_{3} \overrightarrow{k} $
وطرفین را حساب کنیم. اولا:$A.C=a_{1} c_{1}+a_{2} c_{2} +a_{3} c_{3}$ و :$A.B=a_{1} b_{1}+a_{2} b_{2} +a_{3} b_{3}$
پس داریم:
$$B(A.C)=(a_{1} c_{1}+a_{2} c_{2} +a_{3} c_{3}) b_{1} \overrightarrow{i}+(a_{1} c_{1}+a_{2} c_{2} +a_{3} c_{3})b_{2} \overrightarrow{j} +(a_{1} c_{1}+a_{2} c_{2} +a_{3} c_{3})b_{3} \overrightarrow{k} $$
و همچنین:
$$C(A.B)=(a_{1} b_{1}+a_{2} b_{2} +a_{3} b_{3}) c_{1}\overrightarrow{i}+(a_{1} b_{1}+a_{2} b_{2} +a_{3} b_{3})c_{2} \overrightarrow{j} +$$
$$(a_{1} b_{1}+a_{2} b_{2} +a_{3} b_{3})c_{3} \overrightarrow{k} $$
لذا داریم:
$$B(A.C)-C(A.B)=(a_{1} c_{1}b_{1}+a_{2} c_{2} b_{1}+a_{3} c_{3}b_{1}-a_{1} b_{1}c_{1}-a_{2} b_{2} c_{1}-a_{3} b_{3}c_{1})\overrightarrow{i}+$$
$$(a_{1} c_{1}b_{2}+a_{2} c_{2} b_{2}+a_{3} c_{3}b_{2}-a_{1} b_{1}c_{2}-a_{2} b_{2} c_{2}-a_{3} b_{3}c_{2})\overrightarrow{j} +$$
$$(a_{1} c_{1}b_{3}+a_{2} c_{2} b_{3}+a_{3} c_{3}b_{3}-a_{1} b_{1}c_{3}-a_{2} b_{2} c_{3}-a_{3} b_{3}c_{3}) \overrightarrow{k} $$
پس داریم:
$$B(A.C)-C(A.B)=(a_{2} b_{1}c_{2}+a_{3} b_{1}c_{3}-a_{2} b_{2} c_{1}-a_{3} b_{3}c_{1})\overrightarrow{i}+$$
$$(a_{1} b_{2}c_{1}+a_{3} b_{2}c_{3}-a_{1} b_{1}c_{2}-a_{3} b_{3}c_{2})\overrightarrow{j} +$$
$$(a_{1} b_{3}c_{1}+a_{2} b_{3}c_{2}-a_{1} b_{1}c_{3}-a_{2} b_{2} c_{3}) \overrightarrow{k} $$
از طرف دیگر داریم:
$$B \times C= \begin{vmatrix}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k} \\b_{1} & b_{2}& b_{3} \\c_{1} & c_{2}& c_{3}\end{vmatrix} =$$
$$(b_{2}c_{3}-b_{3}c_{2})\overrightarrow{i}-(b_{1}c_{3}-b_{3}c_{1})\overrightarrow{j} +(b_{1}c_{2}-b_{2}c_{1}) \overrightarrow{k} $$
وهمچنین:
$$A\times (B \times C)= \begin{vmatrix}\overrightarrow{i} & \overrightarrow{j}& \overrightarrow{k} \\a_{1} & a_{2}& a_{3} \\b_{2}c_{3}-b_{3}c_{2}& b_{3}c_{1}-b_{1}c_{3}& b_{1}c_{2}-b_{2}c_{1}\end{vmatrix} =$$
$$(a_{2}(b_{1}c_{2}-b_{2}c_{1})-a_{3}(b_{3}c_{1}-b_{1}c_{3}))\overrightarrow{i} -(a_{1}(b_{1}c_{2}-b_{2}c_{1})-a_{3}(b_{2}c_{3}-b_{3}c_{2}))\overrightarrow{j} +$$
$$(a_{1}(b_{3}c_{1}-b_{1}c_{3})-a_{2}(b_{2}c_{3}-b_{3}c_{2}))) \overrightarrow{k} $$
یعنی داریم:
$$A\times (B \times C)=(a_{2}b_{1}c_{2}-a_{2}b_{2}c_{1}+a_{3}b_{1}c_{3}-a_{3}b_{3}c_{1}))\overrightarrow{i}$$
$$ +(-a_{1}b_{1}c_{2}+a_{1}b_{2}c_{1}+a_{3}b_{2}c_{3}-a_{3}b_{3}c_{2})\overrightarrow{j} +$$
$$(-a_{1}b_{1}c_{3}+a_{1}b_{3}c_{1}-a_{2}b_{2}c_{3}+a_{2}b_{3}c_{2}) \overrightarrow{k} $$
با مقایسه طرفین برابری نتیجه می شود