Question

Resultant of two vectors \(\overset{⃗}{P}\) and \(\overset{⃗}{Q}\) is of magnitude A. If \(\overset{⃗}{Q}\) is reversed, then the resultant is of magnitude B. The value of \(A^2+B^2\) is

• A. \(P^2+Q^2\)
• B. \(P^2-Q^2\)
• C. \(2(P^2+Q^2)\)
• D. \(2(P^2-Q^2)\)

Hint:

Write \(A^2\) and \(B^2\) with the cosine law and add.

Answer 1

The Correct Option is C

Step 1: Use the vector sum and difference
$A=|\vec P+\vec Q|$ and $B=|\vec P-\vec Q|$.

Step 2: Apply the identity
$|\vec P+\vec Q|^2+|\vec P-\vec Q|^2=2|\vec P|^2+2|\vec Q|^2$, since the dot product terms cancel. This is option (C).

Final Answer:
The sum is $2(P^2+Q^2)$, option (C). \[ \boxed{2(P^2+Q^2)} \]