If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}| = 10, |\vec{b}| = 2$ and $\vec{a} \cdot \vec{b} = 12$, then $|\vec{a} \times \vec{b}|$ is equal to

Given $|\vec{a}| = 10$, $|\vec{b}| = 2$, and $\vec{a} \cdot \vec{b} = 12$.

Using the dot product formula $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$:

$12 = 10 \times 2 \times \cos\theta$

$12 = 20\cos\theta$

$\cos\theta = \frac{12}{20}$

$\cos\theta = \frac{3}{5}$

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Using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$:

$\sin^2\theta = 1 - \cos^2\theta$

$\sin^2\theta = 1 - \left(\frac{3}{5}\right)^2$

$\sin^2\theta = 1 - \frac{9}{25}$

$\sin^2\theta = \frac{16}{25}$

$\sin\theta = \frac{4}{5}$

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Using the cross product magnitude formula $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$:

$|\vec{a} \times \vec{b}| = 10 \times 2 \times \frac{4}{5}$

$|\vec{a} \times \vec{b}| = 20 \times \frac{4}{5}$

$|\vec{a} \times \vec{b}| = \frac{80}{5}$

$|\vec{a} \times \vec{b}| = 16$

Therefore, $|\vec{a} \times \vec{b}| = 16$.