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Home » Area per tile: 250 × 250 = 62,500 m². - ClickBalance

Area per tile: 250 × 250 = <<250*250=62500>>62,500 m². - ClickBalance

📅 May 18, 2026 👤 scraface

May 18, 2026

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📌 \|\mathbf{w}_\perp\|^2 = \|\mathbf{w}\|^2 - \|\mathbf{w}_\parallel\|^2 = 16 - \left(\frac{2}{3}\right)^2 \cdot 9 = 16 - 4 = 12.

📌 Now, \(\mathbf{u} \cdot \mathbf{w} = \mathbf{u} \cdot \mathbf{w}_\parallel + \mathbf{u} \cdot \mathbf{w}_\perp = \mathbf{u} \cdot \left(\frac{2}{3} \mathbf{v}\right) + \mathbf{u} \cdot \mathbf{w}_\perp = \frac{2}{3} (\mathbf{u} \cdot \mathbf{v}) + \mathbf{u} \cdot \mathbf{w}_\perp = \frac{2}{3}(1) + \mathbf{u} \cdot \mathbf{w}_\perp = \frac{2}{3} + \mathbf{u} \cdot \mathbf{w}_\perp.

📌 Since \(\|\mathbf{u} - \mathbf{u} \cdot \mathbf{v} \mathbf{v}\| \geq 0\), the projection of \(\mathbf{u}\) onto the plane perpendicular to \(\mathbf{v}\) satisfies:

📌 \|\mathbf{u}_\perp\|^2 = \|\mathbf{u}\|^2 - (\mathbf{u} \cdot \mathbf{v})^2 = 4 - 1 = 3.

📌 Let \(\mathbf{u}_\perp = \mathbf{u} - \frac{1}{9} \mathbf{v}\) (since \(\mathbf{v}\) has norm 3), so:

📌 \|\mathbf{u}_\perp\|^2 = \|\mathbf{u}\|^2 - (\mathbf{u} \cdot \mathbf{v})^2 / \|\mathbf{v}\|^2 \cdot

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To minimize the distance, minimize $\|\overrightarrow{CP}\|^2$:
D^2 = (-1 + 3t)^2 + (4 - 2t)^2 + (-5 + 3t)^2
= (1 - 6t + 9t^2) + (16 - 16t + 4t^2) + (25 - 30t + 9t^2) \
= 42 - 52t + 22t^2
Take derivative and set to zero:
rac{d}{dt}(42 - 52t + 22t^2) = -52 + 44t = 0 \Rightarrow t = rac{13}{...
Substitute $t = rac{13}{11}$ into $\mathbf{r}(t)$:
P = \langle 1 + 3\cdotrac{13}{11}, 3 - 2\cdotrac{13}{11}, -2 + 3\cdo...
angle = \left\langle rac{50}{11}, rac{7}{11}, rac{17}{11}
oxed{\left( rac{50}{11}, rac{7}{11}, rac{17}{11}

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