If \(\vec a \:and\: \vec b\) are two vectors such that \(|\vec a + \vec b|= |\vec a - \vec b|=4,\) then which one of the following is correct?

CONCEPT:

The scalar product of two vectors \(\vec a \ and \ \vec b \)is given by \(\vec a \cdot \;\vec b = \left| {\vec a} \right| \times \left| {\vec b} \right|\cos θ \)

If the vectors \(\vec a \ and \ \vec b \) are perpendicular then \(\vec a \cdot \;\vec b = 0\)

CALCULATION:

Given: \(\vec a \:and\: \vec b\) are two vectors such that \(|\vec a + \vec b|= |\vec a - \vec b|=4\)

⇒ \(|\vec a + \vec b|^2= |\vec a - \vec b|^2\)

⇒ \((\vec a + \vec b) \cdot (\vec a + \vec b) = (\vec a - \vec b) \cdot (\vec a - \vec b)\)

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

∵ \(\vec a \cdot \vec b = \vec b \cdot \vec a\)

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

⇒ \(4\vec a \cdot \vec b = 0\)

⇒ \(\vec a \cdot \vec b = 0\)

So, \(\vec a \) must be perpendicular to \(\vec b.\)

Hence, correct option is 3.