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If a⃗,b⃗\vec{a},\vec{b} and c⃗\vec{c} are three vectors such that ∣a⃗∣=12\lvert \vec{a} \rvert=\frac{1}{\sqrt{2}}, ∣b⃗∣=13\lvert \vec{b} \rvert=\frac{1}{\sqrt{3}}, ∣c⃗∣=16\lvert \vec{c} \rvert=\frac{1}{\sqrt{6}}, ∣a⃗+b⃗+c⃗∣=1\lvert \vec{a}+\vec{b}+\vec{c} \rvert=1 and c⃗=λ(a⃗×b⃗)\vec{c}=\lambda(\vec{a}\times\vec{b}). The angle between a⃗\vec{a} and b⃗\vec{b} is

Solution

✅ Correct Option: 1

Since c⃗\vec c is along a⃗×b⃗\vec a\times\vec b, we get b⃗⋅c⃗=c⃗⋅a⃗=0\vec b\cdot\vec c=\vec c\cdot\vec a=0.

Expanding, 1=12+13+16+2 a⃗⋅b⃗=1+2 a⃗⋅b⃗1=\frac12+\frac13+\frac16+2\,\vec a\cdot\vec b=1+2\,\vec a\cdot\vec b.

So a⃗⋅b⃗=0\vec a\cdot\vec b=0 and the angle is π2\frac{\pi}{2}.

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