Vectors and Norms

Question

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

I couldn't figure out how to start off this problem. I attempted to use $$\cos \theta = \frac{a\cdot b}{\|a\|\cdot \|b\|}$$ But I still don't know what to do. Could someone nudge me in the right direction?

Thank you!

Answer 1

Hint: $$\|a\mathbf{x}+b\mathbf{y}\|^2 = (a\mathbf{x}+b\mathbf{y})\cdot(a\mathbf{x}+b\mathbf{y})$$

Answer 2

$a^2+b^2+2ab\cos\theta=49$ and $a^2=16,b^2=25$ thus $ab\cos\theta=4$

$\|2 {a} - 3 {b}\|=\sqrt{4a^2+9b^2-12ab\cos\theta}=\sqrt{64+225-12\times 4}=\sqrt{241}$

Answer 3

hint use the square.

$$\|u+v\|^2=<u+v,u+v> $$ $$=\|u\|^2+\|v\|^2+2 <u,v> $$

Answer 4

Hint: $a.b = \frac{1}{2}(\|a+b\|^2 - \| a\|^2 - \|b\|^b)$

Now Just expand $\|2a-3b\|^2$