If a and b are vectors such that \(\|{a}\| = 4\), \(\|{b}\| = 4\), and \(\|{a+b}\| = 7\), then find \(\|{2a-3b}\|\).
x=||2a-3b||
\(7^2=4^2+4^2- 2*4*4*cos(A) \\ \Rightarrow cos(A)=-\dfrac{17}{32} \)
\(x^2 =8^2+12^2-2*8*12*cos(180^{\circ}-A )\\ x^2=208+192*cos(A) \)
\(x = \sqrt{106 }\)
