Vector Projection and Orthogonalization Preview

Math Linear Algebra • Vectors and Basic Operations

Written by STEM Calculators Team Published February 21, 2026 Updated February 24, 2026

Compute the projection \(\mathrm{proj}_{\mathbf{W}}(\mathbf{V})=\left(\frac{\mathbf{V}\cdot\mathbf{W}}{\lVert\mathbf{W}\rVert^2}\right)\mathbf{W}\), the orthogonal component \(\mathbf{V}-\mathrm{proj}_{\mathbf{W}}(\mathbf{V})\), and preview a 2-vector Gram–Schmidt step.

Projection “shadow” + orthogonal component

2D: drag to pan • wheel to zoom • double-click to reset. 3D: drag to rotate • Shift+drag to pan • wheel to zoom • double-click to reset. Play animates \(\mathrm{proj}_{\mathbf{W}}(\mathbf{V})\) from 0 → final once and stops.

\(\mathbf{V}\) \(\mathbf{W}\) \(\mathrm{proj}_{\mathbf{W}}(\mathbf{V})\) \(\mathbf{V}-\mathrm{proj}\) (orth)

Enter vectors and click “Calculate”.

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Theory

Vector projection

The projection of \(\mathbf{V}\) onto \(\mathbf{W}\neq \mathbf{0}\) is the “shadow” of \(\mathbf{V}\) on the line spanned by \(\mathbf{W}\): \[ \mathrm{proj}_{\mathbf{W}}(\mathbf{V}) = \left(\frac{\mathbf{V}\cdot\mathbf{W}}{\lVert\mathbf{W}\rVert^2}\right)\mathbf{W}. \]

The scalar coefficient \[ \alpha=\frac{\mathbf{V}\cdot\mathbf{W}}{\lVert\mathbf{W}\rVert^2} \] tells you how much of \(\mathbf{W}\) you need to reach the projection point.

Orthogonal component

The remainder after subtracting the projection, \[ \mathbf{V}_\perp=\mathbf{V}-\mathrm{proj}_{\mathbf{W}}(\mathbf{V}), \] is orthogonal to \(\mathbf{W}\) (in exact arithmetic): \[ \mathbf{V}_\perp\cdot \mathbf{W} = 0. \]

Geometric meaning: \(\mathbf{V} = \mathbf{V}_{\parallel} + \mathbf{V}_{\perp}\), where \(\mathbf{V}_{\parallel}\) lies along \(\mathbf{W}\) and \(\mathbf{V}_{\perp}\) is perpendicular to that direction.

Gram–Schmidt preview (two vectors)

Given two vectors \(\mathbf{w}_1\) and \(\mathbf{v}_2\), one Gram–Schmidt step forms an orthogonal pair:

\(\mathbf{u}_1=\mathbf{w}_1\)
\(\mathbf{u}_2=\mathbf{v}_2-\mathrm{proj}_{\mathbf{u}_1}(\mathbf{v}_2)\)

If \(\mathbf{u}_1\neq \mathbf{0}\) and \(\mathbf{u}_2\neq \mathbf{0}\), normalize: \[ \mathbf{e}_1=\frac{\mathbf{u}_1}{\lVert\mathbf{u}_1\rVert},\quad \mathbf{e}_2=\frac{\mathbf{u}_2}{\lVert\mathbf{u}_2\rVert}. \]

If \(\mathbf{u}_2=\mathbf{0}\), then the second vector was dependent on the first (no new direction).

Least squares tease

Projection is the core idea behind least squares: projecting a vector onto a subspace gives the closest approximation inside that subspace. In full Gram–Schmidt, multiple projection subtractions build an orthonormal basis for that subspace.

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