Question

hi. anyone can help me prove it

Answer 1

For the second one, don't know what u can assume

If [abc] = a.(bxc) then

[abc]d = [dbc]a + [adc]b + [abd]c is a well known identity.

So if [abc] not 0 (not coplanar) then can divide by it ->

d = RHS/ [abc] which is of the form D = aA + bB +cC