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

$\because |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2$
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.