Following the examples shown in the textbook I was able to d

Following the examples shown in the textbook I was able to do the following:

However at this point I am at a loss as far as how to proceed, the only example in the textbook is as follows:

parrallelogram formed by (3a+b) and (a-b) if: ||a||= 1; ||b||= 3; and the angle formed by (a,b)= 30 degrees

they then show

=6 square units

I don\'t know how they jumped from one point to another, as this was the only example given in the textbook of how such a problem is to be tackled. without knowing the steps inbetween, I can\'t find a way to solve this question.

Solution

Area = || (a-b) x (2a - 3b) ||

= || a x 2a - 3a x b - b x2a + b x3b ||

Now a x a = ; b xb =0 ; ax b = -b x a

So, = || a x 2a - 3a x b - b x2a + b x3b ||

= || -3a x b +a x b ||

= || -2a x b ||

= 2||a||*||b||*sin60

= 2*2*sqrt3*sqrt3/2

= 2*3 = 6 units