In rhombus ABCD, the diagonals AC = 6 and BD = 8 are known. Find the length of the BA-BC vector.

Let the values (modules) of the diagonals AC = 6 and BD = 8 in the rhombus of AВСD.

Let us determine the magnitude (modulus) of the vector ВA – BC. To solve the AВСD rhombus diagram, we will find this vector, taking into account the vector rule: -BC = – (- BC) = СВ. That is, we get that the sought vector BA – BC = BA + CB = CB + BA = CA.

Here we used the vector permutation rule: BA + CB = CB + BA, and the rule for summing two or more vectors, as a result we got the CA vector corresponding to the beginning of the first CB vector and the end of the second BA vector, which corresponds to the CA vector, or the module of the CA diagonal = 6.