LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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$∥ v ∥_{2} d (w, R v) = ∣ \det [v w] ∣$ for $v \neq 0$ in $R^{2}$

Statement

For $v \neq 0$ , $∥ v ∥_{2} d (w, R v) = ∣ \det [v w] ∣$ .

The nearest point realizing the distance is the orthogonal projection of $w$ onto $R v$ .

Facts & Assumptions

Given: Vectors $v, w \in R^{2}$ with $v \neq 0$ , and the Euclidean base and height of Base and perpendicular height for a chosen side of a plane figure.

[L1]

For an orthonormal basis $(e_{i})$ of a subspace $W$ , the orthogonal projection is $P_{W} u = \sum_{i} ⟨ u, e_{i} ⟩ e_{i}$ (Orthogonal projection is linear, and an orthonormal basis $(e_{i})$ of $W$ gives $P_{W} v = \sum_{i} ⟨ v, e_{i} ⟩ e_{i}$ ).

[L2]

The vector $P_{W} u$ is the unique point of $W$ nearest to $u$ (The orthogonal projection is the unique nearest point in the subspace).

[L3]

For a real $2 \times 2$ matrix, the determinant is the signed permutation sum and $∣ \det A ∣$ is its ordinary absolute value (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L4]

Every nonnegative real has a unique nonnegative square root (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

Proof

technique · direct

1.1

Put $e : = v / ∥ v ∥_{2}$ . Then $(e)$ is an orthonormal basis of $R v$ , so [L1] and [L2] give $P_{R v} w = (⟨ w, v ⟩ / ∥ v ∥_{2}^{2}) v$ and $d (w, R v) = ∥ w - P_{R v} w ∥_{2}$ .

L1L2

2.1

Writing $v = (v_{1}, v_{2})$ and $w = (w_{1}, w_{2})$ , inner-product expansion of step 1.1 gives $∥ v ∥_{2}^{2} d (w, R v)^{2} = ∥ v ∥_{2}^{2} ∥ w ∥_{2}^{2} - ⟨ v, w ⟩^{2} = (v_{1} w_{2} - v_{2} w_{1})^{2} = \det [v w]^{2}$ .

step 1.1L3algebra

3.1

Both $∥ v ∥_{2} d (w, R v)$ and $∣ \det [v w] ∣$ are nonnegative, so equality of their squares in step 2.1 and [L4] give the claimed identity.

step 2.1L4algebra∎

Depends on

Used by

Dependency tree · two levels

27 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

W. F. Trench, Introduction to Real Analysis, §7.3 (standard reference, not scraped)
M. E. Taylor, Introduction to Analysis in Several Variables, Proposition 3.1.10 (standard reference, not scraped)

∥v∥2 d(w,Rv)=∣det⁡[v w]∣∥v∥2​d(w,Rv)=∣det[v w]∣ for v≠0v=0 in R2R2