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: Find the Angle:
\(\vec P\cdot\vec Q=PQ\cos\theta\) and \(|\vec P\times\vec Q|=PQ\sin\theta\). Setting them equal gives \(\tan\theta=1\), so \(\theta=45^{\circ}\).

Step 2: Resultant Formula:
\[ R=\sqrt{P^2+Q^2+2PQ\cos\theta} \]

Step 3: Substitute:
With \(\cos45^{\circ}=\dfrac1{\sqrt2}\), \(2PQ\cos45^{\circ}=\sqrt2\,PQ\). So
\[ R=\sqrt{P^2+Q^2+\sqrt2\,PQ} \]

Step 4: Check the Other Options:
Option (A) corresponds to \(\theta=90^{\circ}\). Option (C) has \(\tfrac{PQ}{\sqrt2}\), which would come from forgetting the factor 2 in \(2PQ\cos\theta\). Option (B) is not a valid form. So (D) is correct.

Final Answer:
The magnitude is \(\sqrt{P^2+Q^2+\sqrt2PQ}\), option (D). \[ \boxed{\text{(D) } \sqrt{P^2+Q^2+\sqrt{2}\,PQ}} \]