Find |vec a×vec b| if |vec a|=10 , |vec b|=5 , vec a·vec b=12

Explanation of Method

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

Step 1: Lagrange identity

From $|\vec a\times\vec b|=|\vec a||\vec b|\sin\varphi$, $\vec a\cdot\vec b=|\vec a||\vec b|\cos\varphi$ and $\sin^2\varphi+\cos^2\varphi=1$: $|\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2=|\vec a|^2|\vec b|^2$.

Step 2: Computation

\[ |\vec a\times\vec b|^2=100\cdot25-12^2=2500-144=2356=4\cdot589. \]

\[|\vec a\times\vec b|=2\sqrt{589}\approx48.54\]

(The cross product needs no sign discussion: a magnitude is non-negative. Here $\cos\varphi=\tfrac{12}{50}$, so $\varphi\approx76.1^\circ$.)

$\varphi$ $O$ $\vec a$ $\vec b$ $h$ $1.2$

$\mathrm{Pr}_{\vec a}\vec b=\tfrac{\vec a\cdot\vec b}{|\vec a|}=1.2$
$h=\tfrac{|\vec a\times\vec b|}{|\vec a|}\approx4.85$

Drawn to scale: $\cos\varphi=\tfrac{12}{50}$, $\varphi\approx76.1^\circ$. The area $|\vec a\times\vec b|=|\vec a|\,h\approx48.5$.

Result

\[\boxed{\,|\vec a\times\vec b|=2\sqrt{589}\approx48.54\,}\]