Thus, a⃗⋅c⃗=−10+2+8=0 ⟹ θ=cos−1(0)=90∘\vec a\cdot \vec c=-10+2+8=0\implies \theta=\cos^{-1}(0)=90^\circa⋅c=−10+2+8=0⟹θ=cos−1(0)=90∘
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