Question

For two vectors \(\overset{⃗}{P}\) and \(\overset{⃗}{Q}\), \(\overset{⃗}{P}\cdot \overset{⃗}{Q} = |\overset{⃗}{P}\times \overset{⃗}{Q}|\)
The magnitude of \(\overset{⃗}{R} = \overset{⃗}{P}+\overset{⃗}{Q}\) is (\(cos45^{\circ} = \frac{1}{\sqrt{2}}\)) ?

• A. \(\sqrt{P^2+Q^2}\)
• B. \(\frac{\sqrt{P^2+Q^2}}{\sqrt{2}}\)
• C. \(\sqrt{P^2+Q^2+\frac{PQ}{\sqrt{2}}}\)
• D. \(\sqrt{P^2+Q^2+\sqrt{2}\,PQ}\)

Hint:

Equal dot and cross magnitudes mean the angle between the vectors is 45 degrees.

Answer 1

The Correct Option is D

Step 1: Components:
Take $\vec P$ along the x-axis. With the angle $\theta$ between them, $\vec Q=(Q\cos\theta,\,Q\sin\theta)$. The condition $PQ\cos\theta=PQ\sin\theta$ gives $\theta=45^{\circ}$.

Step 2: Add and Square:
$\vec R=(P+Q/\sqrt2,\ Q/\sqrt2)$. $R^2=P^2+\sqrt2PQ+\dfrac{Q^2}2+\dfrac{Q^2}2=P^2+Q^2+\sqrt2PQ$.

Step 3: Answer:
$R=\sqrt{P^2+Q^2+\sqrt2PQ}$. Option (D).

Final Answer:
Option (D). \[ \boxed{\text{(D) } \sqrt{P^2+Q^2+\sqrt{2}\,PQ}} \]