Question

At what angle should the two forces \(2 \overrightarrow{\mathrm{P}}\) and \(\sqrt{2} \overrightarrow{\mathrm{P}}\) act so that the resultant force is \(\sqrt{10} \overrightarrow{\mathrm{P}}\) ?

• A. \(\mathrm{cos}^{-1}(\frac{1}{\sqrt{2}})\)
• B. \(\mathrm{cos}^{-1}(\frac{1}{2})\)
• C. \(\mathrm{cos}^{-1}(\frac{\sqrt{3}}{2})\)
• D. \(\mathrm{cos}^{-1}(\frac{1}{\sqrt{3}})\)

Hint:

Use the magnitude of the resultant of two vectors with the cosine rule.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Resultant: \(R^2 = A^2 + B^2 + 2AB\cos\theta\).

Step 2: Substitute
\(A = 2P\), \(B = \sqrt2 P\), \(R = \sqrt{10}P\):
\(10P^2 = 4P^2 + 2P^2 + 2(2P)(\sqrt2P)\cos\theta\)

Step 3: Solve
\(4 = 4\sqrt2\cos\theta\), so \(\cos\theta = \dfrac{1}{\sqrt2}\).

Step 4: Result
\(\theta = \cos^{-1}\left(\dfrac{1}{\sqrt2}\right)\), option A.

Final Answer:
The angle is the inverse cosine of 1 over root 2. \[ \boxed{\text{(A) }\cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right)} \]