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Home » Can COUNTIF Work Without Blank Cells? Try This Surprising Hacks! - AIKO, infinite ways to autonomy.

Can COUNTIF Work Without Blank Cells? Try This Surprising Hacks! - AIKO, infinite ways to autonomy.

📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 Question: Find $ y $ such that the vectors $ \begin{pmatrix} 2 \\ y \\ -1 \end{pmatrix} $ and $ \begin{pmatrix} 3 \\ -2 \\ y \end{pmatrix} $ are orthogonal.

📌 Solution: Two vectors are orthogonal if their dot product is zero. Compute:

📌 2(3) + y(-2) + (-1)(y) = 6 - 2y - y = 6 - 3y = 0.

📌 Solving $ 6 - 3y = 0 $ gives $ y = 2 $. Final answer: $ \boxed{2} $.

📌 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:

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