Find the norm of a linear combination of vectors, given their norms

Question

If a and b are vectors such that $\|a\| = 4$, $\|b\| = 5$, and $\|a + b\| = 7$, then find $\|2a-3b\|$.

So I first squared both sides and then got $ab = -44$. What do I do now?

Answer 1

First compute A=angle between a and b using law of cosines $$\Vert a+b\Vert^2=\Vert a\Vert^2 +\Vert b\Vert^2 + 2\Vert a\Vert \Vert b\Vert cos(a,b)$$ 7x7 =4x4 + 5x5 + 2x4x5 cosA ,
cosA=1/5 ,

Norm of $$\Vert a-b\Vert^2=\Vert a\Vert^2 +\Vert b\Vert^2 - 2\Vert(a)\Vert b\Vert cos(a,b)$$ $$\Vert(a-b)\Vert^2 =16 +25 -2.4.5.1/5=33$$ Checking $$ 4 \Vert a\Vert \Vert b\Vert=\Vert a+b\Vert^2 -\Vert a-b\Vert^2$$ 4x4x5x1/5=16=49-33=16 , it's OK

Again use law of cosines $$\Vert 2a +(-3b)\Vert^2 =\Vert2a\Vert^2 + \Vert 3b\Vert^2 -2\Vert 2a\Vert.\Vert3b\Vert cos A$$ $$\Vert 2a +(-3b)\Vert^2=4.16 + 9.25 -2.8.15.1/5 $$ $$\Vert 2a -3b \Vert^2 =241$$ , $$\Vert (2a -3b)\Vert=15.52$$