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:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (B) and (D) only
Squaring the given magnitude, with \(|\vec{p}| = |\vec{q}| = 1\):
\( \displaystyle |\vec{p} + \vec{q}|^2 = |\vec{p}|^2 + |\vec{q}|^2 + 2\,\vec{p} \cdot \vec{q} \)
\( \displaystyle 2 = 1 + 1 + 2\,\vec{p} \cdot \vec{q} \)
\( \displaystyle \vec{p} \cdot \vec{q} = 0 \)
So \(|\vec{p}| = |\vec{q}| = 1\) and \(\vec{p} \cdot \vec{q} = 0\).
Checking the statements
(A) \(|\vec{p}| = |\vec{q}| = 1\) — Correct. \(\vec{p}\) and \(\vec{q}\) are given to be unit vectors.
(B) \(\vec{p}\) and \(\vec{q}\) are orthogonal vectors — Correct. \(\vec{p} \cdot \vec{q} = 0\), so they are perpendicular.
(C) \(\vec{p}\) and \(\vec{q}\) are collinear vectors — Incorrect. Collinear unit vectors have \(\vec{p} \cdot \vec{q} = \pm 1\), not 0.
(D) \((4\vec{p} - \vec{q}).(2\vec{p} + \vec{q}) = 7\) — Correct. Expanding:
\( \displaystyle (4\vec{p} - \vec{q}) \cdot (2\vec{p} + \vec{q}) = 8|\vec{p}|^2 + 4\,\vec{p} \cdot \vec{q} - 2\,\vec{q} \cdot \vec{p} - |\vec{q}|^2 \)
\( \displaystyle = 8 + 0 - 0 - 1 = 7 \)