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Home » \( t = 2 \pm \sqrt14 \) - Crosslake

\( t = 2 \pm \sqrt14 \) - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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Direction vector: \(\mathbf{d} = egin{pmatrix} 3 \ 1 \end{pmatrix}\)
Point to project: \(\mathbf{p} = egin{pmatrix} 5 \ 4 \end{pmatrix}\)
Vector from point on line to \(\mathbf{p}\):
\mathbf{v} = \mathbf{p} - \mathbf{r}_0 = egin{pmatrix} 5 - 1 \ 4 - (-...
Project \(\mathbf{v}\) onto \(\mathbf{d}\):
ext{proj}_{\mathbf{d}} \mathbf{v} = \left( rac{\mathbf{v} \cdot \math...
ight) \mathbf{d}
Compute dot products:
\mathbf{v} \cdot \mathbf{d} = 4 \cdot 3 + 6 \cdot 1 = 12 + 6 = 18
\mathbf{d} \cdot \mathbf{d} = 3^2 + 1^2 = 9 + 1 = 10

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