Vector Algebra

Let a, b be vectors in R^3, c be a constant.
(a) |a x b|=|a|*|b|*sin(pi/2)=2*2*1=4
(b) 2a+(c-3)b // ca-b if and only if [2a+(c-3)b] x (ca-b) = 0 (vector)
0=[2a+(c-3)b] x (c a-b)
=2c axa -2axb+ c(c-3) bxa - (c-3)bxb (distributive law)
= 0 +2 bxa + c(c-3) bxa - 0 ( a//a iff axa=0, and axb=-bxa)
=(c^2-3c+2) bxa
so c^2-3c+2=0, c=1 or 2

Note: when c=1, 2a+(c-3)b=2a-2b , ca-b=a-b, (2a-2b) // (a-b). Right!
when c=2, 2a+(c-3)b= 2a-b, ca-b=2a-b, (2a-b)//(2a-b). Right!