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(D) We are given the relation $|\vec{a} 	imes \vec{b}|^2 = k^2 - (\vec{a} \cdot \vec{b})^2$. Rearranging this, we get $k^2 = |\vec{a} 	imes \vec{b}|

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$k=7$ and $	heta$ is arbitrary

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(D) We are given the relation $|\vec{a} 	imes \vec{b}|^2 = k^2 - (\vec{a} \cdot \vec{b})^2$. Rearranging this, we get $k^2 = |\vec{a} 	imes \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2$. Using the definitions of the cross product and dot product, we know that $|\vec{a} 	imes \vec{b}| = |\vec{a}||\vec{b}| \sin 	heta$ and $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos 	heta$. Substituting these into the equation: $k^2 = (|\vec{a}||\vec{b}| \sin 	heta)^2 + (|\vec{a}||\vec{b}| \cos 	heta)^2$ $k^2 = |\vec{a}|^2 |\vec{b}|^2 (\sin^2 	heta + \cos^2 	heta)$ Since $\sin^2 	heta + \cos^2 	heta = 1$, we have $k^2 = |\vec{a}|^2 |\vec{b}|^2$. Given $|\vec{a}|=7$ and $|\vec{b}|=1$, we get $k^2 = (7)^2 	imes (1)^2 = 49$. Therefore, $k = 7$. Since the equation holds for any $	heta$, $	heta$ can be any value.

If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b}$ such that $|\vec{a}|=7$, $|\vec{b}|=1$ and $|\vec{a} \times \vec{b}|^2 = k^2 - (\vec{a} \cdot \vec{b})^2$, then the values of $k$ and $\theta$ are

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