If $|\vec a|=12$ and $|\vec b|=4\sqrt 3$ and $\vec b.\vec c=24$ , then find $|(\vec a \times \vec b) +(\vec c \times \vec a)|$ Let $\Delta PQR$ be a triangle. Let $\vec a = \vec {QR}, \vec b = \vec {RP}$ and $\vec c = \vec {PQ}$. If $|\vec a|=12$ and $|\vec b|=4\sqrt 3$ and $\vec b.\vec c=24$ , then find $|(\vec a \times \vec b) +(\vec c \times \vec a)|$
Actually this is an easy problem but I realised it only later. I am looking for a variety of solutions. Thank you .
I think the simplest solution is by first using the triangle equality.
 A: Hint: $|\vec{a}\times \vec{b} + \vec{c}\times \vec{a}|= |\vec{a}\times \vec{b}-\vec{a}\times \vec{c}|=|\vec{a}\times(\vec{b}-\vec{c})|$. But first use the law of cosine to find $|\vec{a}|^2 = |\vec{b}|^2+|\vec{c}|^2+ 2bc\cos(P)=|\vec{b}|^2+|\vec{c}|^2+2\vec{b}\cdot \vec{c}\Rightarrow 144 = 48+ |\vec{c}|^2 +48\Rightarrow |\vec{c}| = 4\sqrt{3}$. Also, using the identity: $|\vec{b}-\vec{c}|^2+|\vec{b}+\vec{c}|^2 = 2(|\vec{b}|^2+|\vec{c}|^2)\Rightarrow |\vec{b}-\vec{c}|^2 + |\vec{a}|^2 = 2(|\vec{b}|^2+|\vec{c}|^2)$. From this you can find $|\vec{b}-\vec{c}|$,and the angle between $\vec{a}$,and $\vec{b}-\vec{c}$ using the parallelogram $PQRS$. You should be done in few steps. Also $\triangle PQR$ is now isosceles.
A: One solution which provides a variety of results in the process of solution is as following:
By the triangle law of vectors, observe that $$\vec {PQ}+\vec {QR}+\vec {RP} =0$$
i.e $$\vec c+\vec a+\vec b =0$$
i.e $$\vec c = -\vec a -\vec b$$
Then again, we have $$\vec b.\vec c=24$$ 
or, $$2\vec b.\vec c=48$$
or, by using the cosine rule, $$b^2+c^2-a^2=-48$$
or, $$48+c^2-144=-48$$
or, $$|\vec c|^2=48$$
or, $$|\vec c|=4\sqrt3$$
So by using cosine rule and the values of $|\vec a|$,$|\vec b|$ and $|\vec c|$, you can calculate the sines and cosines of all $3$ angles.
Also note that $\vec a \times \vec b$ and $\vec c \times \vec a$ are coplanar. So they can be added algebraically.
Hence, $$|\vec a \times \vec b|+|\vec c \times \vec a|$$
$$=||\vec a |\cdot|\vec b|\cdot \sin (180^\circ-R) \cdot \hat n+ |\vec c|\cdot|\vec a |\cdot \sin (180^\circ-P) \cdot \hat n|$$
$$=||\vec a |\cdot|\vec b|\cdot \sin R + |\vec c|\cdot|\vec a |\cdot \sin P|\cdot|\hat n|$$
$$=|12\cdot4\sqrt3\cdot \sin R + 4\sqrt3\cdot 12 \cdot \sin P|\cdot 1$$
Now $$\sin^2 R = 1-\cos ^2R = 1-\frac{b^2+c^2-a^2}{2bc}$$
and $$\sin^2 P = 1-\cos ^2P = 1-\frac{a^2+c^2-b^2}{2ac}$$
Hope you can calculate the rest on your own.
