Given that vec a + 2vec b + 3vec c = vec O→a+2→b+3→c=→O
Now, Taking cross product or **vector product ** with vec a →a on both sides:-
(vec a + 2vec b + 3vec c ) xx vec a = vec O xx vec a(→a+2→b+3→c)×→a=→O×→a
Since cross product is distributive and cross product with vec O→O OR null vector is vec O→O itself,
=> vec a xx vec a + 2(vec b xx vec a) + 3(vec c xx vec a) = vec O⇒→a×→a+2(→b×→a)+3(→c×→a)=→O
because the cross product of a vector with itself is null vector or vec O therefore vec a xx vec a = vec O
=> 2(vec b xx vec a) + 3(vec c xx vec a) = vec O ------------------- 1.
Similarly, taking cross product with vec b and vecc on both sides:-
(vec a xx vec b) + 3(vec c xx vec b) = vec O ------------ 2.
(vec a xx vec c) + 2(vec b xx vec c) = vec O ------------ 3.
Adding 1., 2. and 3.
2(vec b xx vec a) + 3(vec c xx vec a) + (vec a xx vec b) + 3(vec c xx vec b) + (vec a xx vec c) + 2(vec b xx vec c) = vec O
Now for any two vectors vec x and vec y, color(red)(vec x xx vec y = - vec y xx vec x).
Also, sum of any vector with null vector is the vector itself.
=> (vec a xx vec b) - 2(vec a xx vec b) + 2(vec b xx vec c) - 3(vec b xx vec c) + 3(vec c xx vec a) - (vec c xx vec a) = vec O
=> -(vec a xx vec b) - (vec b xx vec c) - (vec c xx vec a) + 3(vec c xx vec a) = vec O
=> -(vec a xx vec b) - (vec b xx vec c) - (vec c xx vec a) = - 3(vec c xx vec a)
Taking product on both sides with -1,
=> vec a xx vec b + vec b xx vec c + vec c xx vec a = 3(vec c xx vec a)
