If |a×b|^2+(a·b)^2=400 and |a|=5 , then |b| is equal to

The correct answer is Option (4) → 4

Let \(\theta\) be the angle between \(\vec{a}\) and \(\vec{b}\). Then

    \(|\vec{a}\times\vec{b}| = |\vec{a}|\,|\vec{b}|\sin\theta\)
    \(\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta\)

Squaring and adding, with \(\sin^2\theta+\cos^2\theta = 1\), gives Lagrange’s identity:

    \(|\vec{a}\times\vec{b}|^{2}+(\vec{a}\cdot\vec{b})^{2} = |\vec{a}|^{2}|\vec{b}|^{2}\)

So

    \(25\,|\vec{b}|^{2} = 400\)
    \(|\vec{b}|^{2} = 16\)
    \(|\vec{b}| = 4\)