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:

(A), (B) and (D) only

The correct answer is Option (4) → (A), (B) and (D) only

Given: $\vec{p}$ and $\vec{q}$ are unit vectors and $|\vec{p} + \vec{q}| = \sqrt{2}$

Then:

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

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

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

Given: $|\vec{p} + \vec{q}| = \sqrt{2} \Rightarrow |\vec{p} + \vec{q}|^2 = 2$

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

So, $\vec{p}$ and $\vec{q}$ are orthogonal (perpendicular) unit vectors

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

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

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

$= 8(1) + 4(0) - 2(0) - 1 = 8 - 1 = 7$

Final Answers:

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

(B) ✔️ $\vec{p} \cdot \vec{q} = 0$ — Orthogonal

(C) ❌ Not collinear, since they are orthogonal

(D) ✔️ Value of dot product is 7