ABC-is-any-triangle-C-B-A-are-respectively-middles-of-AB-AC-and-BC-we-suppose-that-AB-c-AC-b-BC-a-1-u-a-2-BC-b-2-C-A-c-2-AB-is-a-vector-Demonstrate-that-u

Question Number 77745 by mathocean1 last updated on 09/Jan/20

ABC is any triangle .

C^{'} . B^{'} . A^{'} are respectively middles

of [AB] . [AC] and [BC] .

we suppose that

AB = c AC = b BC = a .

1) \vec{u} = a^{2} B \vec{C} + b^{2} \vec{C A} + c^{2} A \vec{B} is a vector

Demonstrate that \vec{u} = (a^{2} - b^{2}) A \vec{C} + (c^{2} - a^{2}) A \vec{B} .

i h a v e d o n e i t .

2) D educt that \vec{u} is not a null vector .

Commented by mathocean1 last updated on 09/Jan/20

Please can you help me \dots