Given $|\vec a| = 3, |\vec b| = 5$ and $|\vec a+\vec b| = 7$. Determine $|\vec a-\vec b|$.

Question

I can't seem to understand this question at all. It does not make sense to me.

The question is

Given $\left|\vec a\right| = 3, \left|\vec b\right| = 5$ and $\left|\vec a+\vec b\right| = 7$. Determine $\left|\vec a-\vec b\right|$.

I have tried finding $\left|\vec a+\vec b\right|$ using cosine rule such that $\left|\vec a+\vec b\right| = 7 = 3^2 + 5^2 - 2\cosθ$

Which failed as I clearly am unable to picture this question correctly in my head. If someone could explain this question (or maybe help me sketch it) that's be very helpful, thanks in advance.

Answer 1

You are thinking in the right direction, but have used the wrong formula. The right ones are: $$|\vec a\pm\vec b|^2=|\vec a|^2+|\vec b|^2\pm2|\vec a||\vec b|\cos\theta$$ Since you have $|\vec a+\vec b|$, you can find $\cos\theta$ and then plug in the value to find $|\vec a-\vec b|$

Answer 2

HINT

If you are in, for example, $\mathbb{R}^2$, write $a = (a_x,a_y), b = (b_x,b_y)$ then $$ \begin{split} |a+b|^2 &= a_x^2 + b_x^2 + 2a_x b_x + a_y^2 + b_y^2 + 2a_yb_y\\ |a-b|^2 &= a_x^2 + b_x^2 - 2a_x b_x + a_y^2 + b_y^2 - 2a_yb_y \\ |a+b|^2 + |a-b|^2 &= 2a_x^2 + 2b_x^2 + 2a_y^2 + 2b_y^2 \\ &= 2|a|^2 + 2|b|^2 \end{split} $$ Can you finish and generalize for $\mathbb{R}^n$?