Question

Let
\(\vec{a}\) and \(\vec{b}\) be two vectors such that 
\(|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2,\  \vec{a} \cdot \vec{b} = 3\) and \(|\vec{a} \times \vec{b}|^2 = 75\)
Then \(|\vec{a}|^2\) is equal to _____.

Answer 1

Correct Answer: 14

\(\because |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2\)
or \(|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = |\vec{a}|^2 + 2|\vec{b}|^2\)
\(∴ |\vec{b}|^2=6 …(i)\)
Now,\(|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - \left(\vec{a} \cdot \vec{b}\right)^2\)
\(75=|\vec{a}|^2⋅6−9\)
\(∴ |\vec{a}|^2=14\)
So, the correct answer is 14.