Given conditions: $\vec{a} \times \vec{b} = \vec{c}$

$\vec{a} \cdot \vec{c} = 2$

$\vec{b} \cdot \vec{c} = 1$

$|\vec{b}| = 1$

Since $\vec{c} = \vec{a} \times \vec{b}$,

we have: $|\vec{c}| = |\vec{a}||\vec{b}| \sin\theta = |\vec{a}|\sin\theta$ (because $|\vec{b}| = 1$)

Also: $\vec{a} \cdot \vec{c} = \vec{a} \cdot (\vec{a} \times \vec{b}) = 0$

But given: $\vec{a} \cdot \vec{c} = 2$

This is possible only if $\vec{c}$ is not perpendicular to $\vec{a}$, meaning $\vec{c}$ is not just $\vec{a} \times \vec{b}$ but also has a component along $\vec{a}$.

Use the identity: $(\vec{a} \times \vec{b}) \cdot \vec{c} = \det(\vec{a},\vec{b},\vec{c})$

But we need magnitudes. Take dot product of $\vec{b}$ with $\vec{c}$:

$\vec{b} \cdot \vec{c} = \vec{b} \cdot (\vec{a} \times \vec{b}) = 0$

But given $\vec{b} \cdot \vec{c} = 1$

So again $\vec{c}$ has components outside the perpendicular direction $\vec{c}$ is independent.

Use the vector triple product identity: $\vec{c} = \vec{a} \times \vec{b}$

Take magnitude squared:

$|\vec{c}|^2 = |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$

Since $|\vec{b}| = 1$: $|\vec{c}|^2 = |\vec{a}|^2 - (\vec{a} \cdot \vec{b})^2$

Now use the given dot products.

We know: $\vec{a} \cdot \vec{c} = 2$ $\vec{b} \cdot \vec{c} = 1$

Take magnitude squared of $\vec{c}$: $|\vec{c}|^2 = (\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b})$

But also: $|\vec{c}|^2 = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{c}) = 2 \cdot 1 = 2$

So: $|\vec{c}|^2 = 2$

Thus: $2 = |\vec{a}|^2 - (\vec{a} \cdot \vec{b})^2$

Also: $\vec{a} \cdot \vec{b} = \dfrac{\vec{a} \cdot \vec{c}}{|\vec{c}|} = \dfrac{2}{\sqrt{2}} = \sqrt{2}$

Therefore: $(\vec{a} \cdot \vec{b})^2 = 2$

Substitute: $2 = |\vec{a}|^2 - 2$

So: $|\vec{a}|^2 = 4$

Hence: $|\vec{a}| = 2$

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