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benefits of lemon and cucumber water - Network4

📅 May 19, 2026 👤 scraface

May 19, 2026

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📌 Question: If $ \mathbf{a}, \mathbf{b}, \mathbf{c} $ are unit vectors with $ \mathbf{a} \cdot \mathbf{b} = \frac{1}{2} $, $ \mathbf{b} \cdot \mathbf{c} = \frac{\sqrt{3}}{2} $, find the maximum value of $ \mathbf{a} \cdot \mathbf{c} $.

📌 Solution: Let $ \theta $ be the angle between $ \mathbf{a} $ and $ \mathbf{b} $, so $ \cos\theta = \frac{1}{2} \Rightarrow \theta = 60^\circ $. Let $ \phi $ be the angle between $ \mathbf{b} $ and $ \mathbf{c} $, so $ \cos\phi = \frac{\sqrt{3}}{2} \Rightarrow \phi = 30^\circ $. To maximize $ \mathbf{a} \cdot \mathbf{c} = \cos(\alpha) $, where $ \alpha $ is the angle between $ \mathbf{a} $ and $ \mathbf{c} $, arrange $ \mathbf{a}, \mathbf{b}, \mathbf{c} $ in a plane. The maximum occurs when $ \mathbf{a} $ and $ \mathbf{c} $ are aligned, but constrained by their angles relative to $ \mathbf{b} $. The minimum angle between $ \mathbf{a} $ and $ \mathbf{c} $ is $ 60^\circ - 30^\circ = 30^\circ $, so $ \cos(30^\circ) = \frac{\sqrt{3}}{2} $. However, if they are aligned, $ \alpha = 0^\circ $, but this requires $ \theta = \phi = 0^\circ $, which contradicts the given dot products. Instead, use the cosine law for angles: $ \cos\alpha \leq \cos(60^\circ - 30^\circ) = \cos(30^\circ) = \frac{\sqrt{3}}{2} $. Thus, the maximum is $ \boxed{\frac{\sqrt{3}}{2}} $.

📌 Question: Find the vector $ \mathbf{v} $ such that $ \mathbf{v} \times \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} -4 \\ 6 \\ 2 \end{pmatrix} $.

📌 Solution: Let $ \mathbf{v} = \begin{pmatrix} a \\ b \\ c \end{pmatrix} $. The cross product is:

📌 \mathbf{v} \times \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} b(2) - c(0) \\ c(1) - a(2) \\ a(0) - b(1) \end{pmatrix} = \begin{pmatrix} 2b \\ c - 2a \\ -b \end{pmatrix} = \begin{pmatrix} -4 \\ 6 \\ 2 \end{pmatrix}.

📌 This gives the system:

📌 $ 2b = -4 \Rightarrow b = -2 $,

📌 $ c - 2a = 6 $,

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