Find vec a·vec b if |vec a|=10 , |vec b|=5 , |vec a×vec b|=16

Find $\vec a\cdot\vec b$ if $|\vec a|=10$, $|\vec b|=5$, $|\vec a\times\vec b|=16$.

Explanation of Method

The scalar product from the lengths and the magnitude of a×b.

Step 1: Computation

Step 2: Two signs

The area only fixes $\sin\varphi=\tfrac{16}{50}=0.32$. Both $\varphi\approx18.7^\circ$ (acute) and $\varphi\approx161.3^\circ$ (obtuse) are possible: $\vec a\cdot\vec b=\pm2\sqrt{561}\approx\pm47.37$.

$O$ $\vec a$ $\vec b_1$ $\vec b_2$ $h=1.6$

$\varphi\approx18.7^\circ:\ \vec a\cdot\vec b=+2\sqrt{561}$
$\varphi\approx161.3^\circ:\ \vec a\cdot\vec b=-2\sqrt{561}$

Both positions of $\vec b$ give the same area $16$ (same height $h=1.6$ over $|\vec a|=10$), but scalar products of opposite sign.

Result

\[\boxed{\,\vec a\cdot\vec b=\pm2\sqrt{561}\approx\pm47.37\,}\]