Find \(|\vec x|\) if \((\vec x - \vec a) \cdot (\vec x + \vec a) = 12\) and \(\vec a\) is a unit vector ?

CONCEPT:

• \(\vec a \cdot \vec a = |\vec a|^2\)
• \(\vec a \cdot \vec b = \vec b \cdot \vec a\)
• If \(\vec a\) is a unit vector then \(|\vec a| = 1\)

CALCULATION:

Given: \((\vec x - \vec a) \cdot (\vec x + \vec a) = 12\) and \(\vec a\) is a unit vector

⇒ \((\vec x - \vec a) \cdot (\vec x + \vec a) = |\vec x|^2 + \vec x \cdot \vec a - \vec a \cdot \vec x - |\vec a|^2 = 12\)

As we know that, \(\vec a \cdot \vec b = \vec b \cdot \vec a\)

⇒ \((\vec x - \vec a) \cdot (\vec x + \vec a) = |\vec x|^2 - |\vec a|^2 = 12\)

As we know that, if \(\vec a\) is a unit vector then \(|\vec a| = 1\)

⇒ \(|\vec x|^2 = 13 \Rightarrow |\vec x| = \sqrt {13}\)

Hence, correct option is 2.