If $\vec{p}$ and $\vec{q}$ are two unit vectors such that $|\vec{p} + \vec{q}| = \sqrt{2}$, then which of the following are correct?

(A) $|\vec{p}| = |\vec{q}| = 1$

(B) $\vec{p}$ and $\vec{q}$ are orthogonal vectors

(C) $\vec{p}$ and $\vec{q}$ are collinear vectors

(D) $(4\vec{p} - \vec{q}).(2\vec{p} + \vec{q}) = 7$

Choose the correct answer from the options given below:

Given $\vec{p}$ and $\vec{q}$ are unit vectors, so $|\vec{p}| = |\vec{q}| = 1$.

Also given $|\vec{p} + \vec{q}| = \sqrt{2}$.

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$|\vec{p} + \vec{q}|^2 = (\sqrt{2})^2$

$(\vec{p} + \vec{q}) \cdot (\vec{p} + \vec{q}) = 2$

$\vec{p} \cdot \vec{p} + \vec{p} \cdot \vec{q} + \vec{q} \cdot \vec{p} + \vec{q} \cdot \vec{q} = 2$

$\vec{p} \cdot \vec{p} + 2\vec{p} \cdot \vec{q} + \vec{q} \cdot \vec{q} = 2$

Since $\vec{p}$ and $\vec{q}$ are unit vectors:

$1 + 2\vec{p} \cdot \vec{q} + 1 = 2$

$2 + 2\vec{p} \cdot \vec{q} = 2$

$2\vec{p} \cdot \vec{q} = 0$

$\vec{p} \cdot \vec{q} = 0$

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Checking statement (A): $|\vec{p}| = |\vec{q}| = 1$

This is true by the given condition that $\vec{p}$ and $\vec{q}$ are unit vectors.

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Checking statement (B): $\vec{p}$ and $\vec{q}$ are orthogonal vectors

Since $\vec{p} \cdot \vec{q} = 0$, the vectors are orthogonal.

This is true.

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Checking statement (C): $\vec{p}$ and $\vec{q}$ are collinear vectors

Collinear vectors are parallel. Since $\vec{p}$ and $\vec{q}$ are orthogonal (perpendicular), they cannot be collinear.

This is false.

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Checking statement (D): $(4\vec{p} - \vec{q}) \cdot (2\vec{p} + \vec{q}) = 7$

$(4\vec{p} - \vec{q}) \cdot (2\vec{p} + \vec{q})$

$= 4\vec{p} \cdot 2\vec{p} + 4\vec{p} \cdot \vec{q} - \vec{q} \cdot 2\vec{p} - \vec{q} \cdot \vec{q}$

$= 8(\vec{p} \cdot \vec{p}) + 4(\vec{p} \cdot \vec{q}) - 2(\vec{p} \cdot \vec{q}) - (\vec{q} \cdot \vec{q})$

$= 8(\vec{p} \cdot \vec{p}) + 2(\vec{p} \cdot \vec{q}) - (\vec{q} \cdot \vec{q})$

$= 8(1) + 2(0) - 1$

$= 8 + 0 - 1$

$= 7$

This is true.

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Statements (A), (B), and (D) are correct.