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If a⃗\vec{a} and b⃗\vec{b} are two non-zero vectors such that ∣a⃗×b⃗∣2+∣a⃗.b⃗∣2=100|\vec{a}\times\vec{b}|^2 + |\vec{a}.\vec{b}|^2 = 100 and ∣a⃗∣=5|\vec{a}| = 5 then the value of ∣b⃗∣|\vec{b}| is

Solution

✅ Correct Option: 1

The Lagrange identity gives ∣a⃗×b⃗∣2+∣a⃗⋅b⃗∣2=∣a⃗∣2∣b⃗∣2|\vec{a}\times\vec{b}|^2+|\vec{a}\cdot\vec{b}|^2=|\vec{a}|^2|\vec{b}|^2.

So 25∣b⃗∣2=10025|\vec{b}|^2=100, giving ∣b⃗∣2=4|\vec{b}|^2=4 and ∣b⃗∣=2|\vec{b}|=2.

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