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Esercizi sulla Proiezione Ortogonale di un Vettore su uno Spazio Vettoriale
Introduzione
La proiezione ortogonale di un vettore \vec{v} v ⃗ \vec{v} v \{\vec{u_1}, \vec{u_2}, \ldots, \vec{u_n}\} { u 1 ⃗ , u 2 ⃗ , … , u n ⃗ } \{\vec{u_1}, \vec{u_2}, \ldots, \vec{u_n}\} { u 1 , u 2 , … , u n } \vec{v} v ⃗ \vec{v} v
La proiezione ortogonale di un vettore \vec{v} v ⃗ \vec{v} v \vec{u} u ⃗ \vec{u} u
\text{Proiezione}_{\vec{u}} \vec{v} = \frac{\vec{v} \cdot \vec{u}}{\vec{u} \cdot \vec{u}} \vec{u} Proiezione u ⃗ v ⃗ = v ⃗ ⋅ u ⃗ u ⃗ ⋅ u ⃗ u ⃗ \text{Proiezione}_{\vec{u}} \vec{v} = \frac{\vec{v} \cdot \vec{u}}{\vec{u} \cdot \vec{u}} \vec{u} Proiezione u v = u ⋅ u v ⋅ u u Esercizio 1: Proiezione di un Vettore su un Vettore
Problema: Trova la proiezione ortogonale del vettore \vec{v} = (3, 4) v ⃗ = ( 3 , 4 ) \vec{v} = (3, 4) v = ( 3 , 4 ) \vec{u} = (1, 2) u ⃗ = ( 1 , 2 ) \vec{u} = (1, 2) u = ( 1 , 2 )
Soluzione:
Calcoliamo il prodotto scalare \vec{v} \cdot \vec{u} v ⃗ ⋅ u ⃗ \vec{v} \cdot \vec{u} v ⋅ u
\vec{v} \cdot \vec{u} = 3 \cdot 1 + 4 \cdot 2 = 3 + 8 = 11 v ⃗ ⋅ u ⃗ = 3 ⋅ 1 + 4 ⋅ 2 = 3 + 8 = 11 \vec{v} \cdot \vec{u} = 3 \cdot 1 + 4 \cdot 2 = 3 + 8 = 11 v ⋅ u = 3 ⋅ 1 + 4 ⋅ 2 = 3 + 8 = 11
Calcoliamo il prodotto scalare \vec{u} \cdot \vec{u} u ⃗ ⋅ u ⃗ \vec{u} \cdot \vec{u} u ⋅ u
\vec{u} \cdot \vec{u} = 1 \cdot 1 + 2 \cdot 2 = 1 + 4 = 5 u ⃗ ⋅ u ⃗ = 1 ⋅ 1 + 2 ⋅ 2 = 1 + 4 = 5 \vec{u} \cdot \vec{u} = 1 \cdot 1 + 2 \cdot 2 = 1 + 4 = 5 u ⋅ u = 1 ⋅ 1 + 2 ⋅ 2 = 1 + 4 = 5
Ora possiamo calcolare la proiezione:
\text{Proiezione}_{\vec{u}} \vec{v} = \frac{11}{5} \vec{u} = \frac{11}{5} (1, 2) = \left( \frac{11}{5}, \frac{22}{5} \right) Proiezione u ⃗ v ⃗ = 11 5 u ⃗ = 11 5 ( 1 , 2 ) = ( 11 5 , 22 5 ) \text{Proiezione}_{\vec{u}} \vec{v} = \frac{11}{5} \vec{u} = \frac{11}{5} (1, 2) = \left( \frac{11}{5}, \frac{22}{5} \right) Proiezione u v = 5 11 u = 5 11 ( 1 , 2 ) = ( 5 11 , 5 22 )
Esercizio 2: Proiezione di un Vettore su uno Spazio Vettoriale
Problema: Trova la proiezione ortogonale del vettore \vec{v} = (1, 2, 3) v ⃗ = ( 1 , 2 , 3 ) \vec{v} = (1, 2, 3) v = ( 1 , 2 , 3 ) \vec{u_1} = (1, 0, 0) u 1 ⃗ = ( 1 , 0 , 0 ) \vec{u_1} = (1, 0, 0) u 1 = ( 1 , 0 , 0 ) \vec{u_2} = (0, 1, 0) u 2 ⃗ = ( 0 , 1 , 0 ) \vec{u_2} = (0, 1, 0) u 2 = ( 0 , 1 , 0 )
Soluzione:
Calcoliamo la proiezione di \vec{v} v ⃗ \vec{v} v \vec{u_1} u 1 ⃗ \vec{u_1} u 1
\text{Proiezione}_{\vec{u_1}} \vec{v} = \frac{\vec{v} \cdot \vec{u_1}}{\vec{u_1} \cdot \vec{u_1}} \vec{u_1} Proiezione u 1 ⃗ v ⃗ = v ⃗ ⋅ u 1 ⃗ u 1 ⃗ ⋅ u 1 ⃗ u 1 ⃗ \text{Proiezione}_{\vec{u_1}} \vec{v} = \frac{\vec{v} \cdot \vec{u_1}}{\vec{u_1} \cdot \vec{u_1}} \vec{u_1} Proiezione u 1 v = u 1 ⋅ u 1 v ⋅ u 1 u 1
\vec{v} \cdot \vec{u_1} = 1 \cdot 1 + 2 \cdot 0 + 3 \cdot 0 = 1 v ⃗ ⋅ u 1 ⃗ = 1 ⋅ 1 + 2 ⋅ 0 + 3 ⋅ 0 = 1 \vec{v} \cdot \vec{u_1} = 1 \cdot 1 + 2 \cdot 0 + 3 \cdot 0 = 1 v ⋅ u 1 = 1 ⋅ 1 + 2 ⋅ 0 + 3 ⋅ 0 = 1 \vec{u_1} \cdot \vec{u_1} = 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 = 1 u 1 ⃗ ⋅ u 1 ⃗ = 1 ⋅ 1 + 0 ⋅ 0 + 0 ⋅ 0 = 1 \vec{u_1} \cdot \vec{u_1} = 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 = 1 u 1 ⋅ u 1 = 1 ⋅ 1 + 0 ⋅ 0 + 0 ⋅ 0 = 1 Quindi:
\text{Proiezione}_{\vec{u_1}} \vec{v} = 1 \cdot (1, 0, 0) = (1, 0, 0) Proiezione u 1 ⃗ v ⃗ = 1 ⋅ ( 1 , 0 , 0 ) = ( 1 , 0 , 0 ) \text{Proiezione}_{\vec{u_1}} \vec{v} = 1 \cdot (1, 0, 0) = (1, 0, 0) Proiezione u 1 v = 1 ⋅ ( 1 , 0 , 0 ) = ( 1 , 0 , 0 )
Calcoliamo la proiezione di \vec{v} v ⃗ \vec{v} v \vec{u_2} u 2 ⃗ \vec{u_2} u 2
\text{Proiezione}_{\vec{u_2}} \vec{v} = \frac{\vec{v} \cdot \vec{u_2}}{\vec{u_2} \cdot \vec{u_2}} \vec{u_2} Proiezione u 2 ⃗ v ⃗ = v ⃗ ⋅ u 2 ⃗ u 2 ⃗ ⋅ u 2 ⃗ u 2 ⃗ \text{Proiezione}_{\vec{u_2}} \vec{v} = \frac{\vec{v} \cdot \vec{u_2}}{\vec{u_2} \cdot \vec{u_2}} \vec{u_2} Proiezione u 2 v = u 2 ⋅ u 2 v ⋅ u 2 u 2
Calcoliamo \vec{v} \cdot \vec{u_2} v ⃗ ⋅ u 2 ⃗ \vec{v} \cdot \vec{u_2} v ⋅ u 2
\vec{v} \cdot \vec{u_2} = 1 \cdot 0 + 2 \cdot 1 + 3 \cdot 0 = 2 v ⃗ ⋅ u 2 ⃗ = 1 ⋅ 0 + 2 ⋅ 1 + 3 ⋅ 0 = 2 \vec{v} \cdot \vec{u_2} = 1 \cdot 0 + 2 \cdot 1 + 3 \cdot 0 = 2 v ⋅ u 2 = 1 ⋅ 0 + 2 ⋅ 1 + 3 ⋅ 0 = 2
Calcoliamo \vec{u_2} \cdot \vec{u_2} u 2 ⃗ ⋅ u 2 ⃗ \vec{u_2} \cdot \vec{u_2} u 2 ⋅ u 2
\vec{u_2} \cdot \vec{u_2} = 0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0 = 1 u 2 ⃗ ⋅ u 2 ⃗ = 0 ⋅ 0 + 1 ⋅ 1 + 0 ⋅ 0 = 1 \vec{u_2} \cdot \vec{u_2} = 0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0 = 1 u 2 ⋅ u 2 = 0 ⋅ 0 + 1 ⋅ 1 + 0 ⋅ 0 = 1 \text{Proiezione}_{\vec{u_2}} \vec{v} = 2 \cdot (0, 1, 0) = (0, 2, 0) Proiezione u 2 ⃗ v ⃗ = 2 ⋅ ( 0 , 1 , 0 ) = ( 0 , 2 , 0 ) \text{Proiezione}_{\vec{u_2}} \vec{v} = 2 \cdot (0, 1, 0) = (0, 2, 0) Proiezione u 2 v = 2 ⋅ ( 0 , 1 , 0 ) = ( 0 , 2 , 0 )
La proiezione ortogonale di \vec{v} v ⃗ \vec{v} v \vec{u_1} u 1 ⃗ \vec{u_1} u 1 \vec{u_2} u 2 ⃗ \vec{u_2} u 2
\text{Proiezione}_{\text{piano}} \vec{v} = \text{Proiezione}_{\vec{u_1}} \vec{v} + \text{Proiezione}_{\vec{u_2}} \vec{v} Proiezione piano v ⃗ = Proiezione u 1 ⃗ v ⃗ + Proiezione u 2 ⃗ v ⃗ \text{Proiezione}_{\text{piano}} \vec{v} = \text{Proiezione}_{\vec{u_1}} \vec{v} + \text{Proiezione}_{\vec{u_2}} \vec{v} Proiezione piano v = Proiezione u 1 v + Proiezione u 2 v \text{Proiezione}_{\text{piano}} \vec{v} = (1, 0, 0) + (0, 2, 0) = (1, 2, 0) Proiezione piano v ⃗ = ( 1 , 0 , 0 ) + ( 0 , 2 , 0 ) = ( 1 , 2 , 0 ) \text{Proiezione}_{\text{piano}} \vec{v} = (1, 0, 0) + (0, 2, 0) = (1, 2, 0) Proiezione piano v = ( 1 , 0 , 0 ) + ( 0 , 2 , 0 ) = ( 1 , 2 , 0 )
Esercizio 3: Proiezione Ortogonale in uno Spazio di Dimensione Superiore
Problema: Trova la proiezione ortogonale del vettore \vec{v} = (2, 3, 4) v ⃗ = ( 2 , 3 , 4 ) \vec{v} = (2, 3, 4) v = ( 2 , 3 , 4 ) \vec{u_1} = (1, 1, 0) u 1 ⃗ = ( 1 , 1 , 0 ) \vec{u_1} = (1, 1, 0) u 1 = ( 1 , 1 , 0 ) \vec{u_2} = (0, 1, 1) u 2 ⃗ = ( 0 , 1 , 1 ) \vec{u_2} = (0, 1, 1) u 2 = ( 0 , 1 , 1 )
Soluzione:
Calcoliamo la proiezione di \vec{v} v ⃗ \vec{v} v \vec{u_1} u 1 ⃗ \vec{u_1} u 1
\text{Proiezione}_{\vec{u_1}} \vec{v} = \frac{\vec{v} \cdot \vec{u_1}}{\vec{u_1} \cdot \vec{u_1}} \vec{u_1} Proiezione u 1 ⃗ v ⃗ = v ⃗ ⋅ u 1 ⃗ u 1 ⃗ ⋅ u 1 ⃗ u 1 ⃗ \text{Proiezione}_{\vec{u_1}} \vec{v} = \frac{\vec{v} \cdot \vec{u_1}}{\vec{u_1} \cdot \vec{u_1}} \vec{u_1} Proiezione u 1 v = u 1 ⋅ u 1 v ⋅ u 1 u 1
Calcoliamo \vec{v} \cdot \vec{u_1} v ⃗ ⋅ u 1 ⃗ \vec{v} \cdot \vec{u_1} v ⋅ u 1
\vec{v} \cdot \vec{u_1} = 2 \cdot 1 + 3 \cdot 1 + 4 \cdot 0 = 2 + 3 + 0 = 5 v ⃗ ⋅ u 1 ⃗ = 2 ⋅ 1 + 3 ⋅ 1 + 4 ⋅ 0 = 2 + 3 + 0 = 5 \vec{v} \cdot \vec{u_1} = 2 \cdot 1 + 3 \cdot 1 + 4 \cdot 0 = 2 + 3 + 0 = 5 v ⋅ u 1 = 2 ⋅ 1 + 3 ⋅ 1 + 4 ⋅ 0 = 2 + 3 + 0 = 5
Calcoliamo \vec{u_1} \cdot \vec{u_1} u 1 ⃗ ⋅ u 1 ⃗ \vec{u_1} \cdot \vec{u_1} u 1 ⋅ u 1
\vec{u_1} \cdot \vec{u_1} = 1 \cdot 1 + 1 \cdot 1 + 0 \cdot 0 = 1 + 1 + 0 = 2 u 1 ⃗ ⋅ u 1 ⃗ = 1 ⋅ 1 + 1 ⋅ 1 + 0 ⋅ 0 = 1 + 1 + 0 = 2 \vec{u_1} \cdot \vec{u_1} = 1 \cdot 1 + 1 \cdot 1 + 0 \cdot 0 = 1 + 1 + 0 = 2 u 1 ⋅ u 1 = 1 ⋅ 1 + 1 ⋅ 1 + 0 ⋅ 0 = 1 + 1 + 0 = 2 \text{Proiezione}_{\vec{u_1}} \vec{v} = \frac{5}{2} (1, 1, 0) = \left( \frac{5}{2}, \frac{5}{2}, 0 \right) Proiezione u 1 ⃗ v ⃗ = 5 2 ( 1 , 1 , 0 ) = ( 5 2 , 5 2 , 0 ) \text{Proiezione}_{\vec{u_1}} \vec{v} = \frac{5}{2} (1, 1, 0) = \left( \frac{5}{2}, \frac{5}{2}, 0 \right) Proiezione u 1 v = 2 5 ( 1 , 1 , 0 ) = ( 2 5 , 2 5 , 0 )
Calcoliamo la proiezione di \vec{v} v ⃗ \vec{v} v \vec{u_2} u 2 ⃗ \vec{u_2} u 2
\text{Proiezione}_{\vec{u_2}} \vec{v} = \frac{\vec{v} \cdot \vec{u_2}}{\vec{u_2} \cdot \vec{u_2}} \vec{u_2} Proiezione u 2 ⃗ v ⃗ = v ⃗ ⋅ u 2 ⃗ u 2 ⃗ ⋅ u 2 ⃗ u 2 ⃗ \text{Proiezione}_{\vec{u_2}} \vec{v} = \frac{\vec{v} \cdot \vec{u_2}}{\vec{u_2} \cdot \vec{u_2}} \vec{u_2} Proiezione u 2 v = u 2 ⋅ u 2 v ⋅ u 2 u 2
Calcoliamo \vec{v} \cdot \vec{u_2} v ⃗ ⋅ u 2 ⃗ \vec{v} \cdot \vec{u_2} v ⋅ u 2
\vec{v} \cdot \vec{u_2} = 2 \cdot 0 + 3 \cdot 1 + 4 \cdot 1 = 0 + 3 + 4 = 7 v ⃗ ⋅ u 2 ⃗ = 2 ⋅ 0 + 3 ⋅ 1 + 4 ⋅ 1 = 0 + 3 + 4 = 7 \vec{v} \cdot \vec{u_2} = 2 \cdot 0 + 3 \cdot 1 + 4 \cdot 1 = 0 + 3 + 4 = 7 v ⋅ u 2 = 2 ⋅ 0 + 3 ⋅ 1 + 4 ⋅ 1 = 0 + 3 + 4 = 7
Calcoliamo \vec{u_2} \cdot \vec{u_2} u 2 ⃗ ⋅ u 2 ⃗ \vec{u_2} \cdot \vec{u_2} u 2 ⋅ u 2
\vec{u_2} \cdot \vec{u_2} = 0 \cdot 0 + 1 \cdot 1 + 1 \cdot 1 = 0 + 1 + 1 = 2 u 2 ⃗ ⋅ u 2 ⃗ = 0 ⋅ 0 + 1 ⋅ 1 + 1 ⋅ 1 = 0 + 1 + 1 = 2 \vec{u_2} \cdot \vec{u_2} = 0 \cdot 0 + 1 \cdot 1 + 1 \cdot 1 = 0 + 1 + 1 = 2 u 2 ⋅ u 2 = 0 ⋅ 0 + 1 ⋅ 1 + 1 ⋅ 1 = 0 + 1 + 1 = 2 \text{Proiezione}_{\vec{u_2}} \vec{v} = \frac{7}{2} (0, 1, 1) = \left( 0, \frac{7}{2}, \frac{7}{2} \right) Proiezione u 2 ⃗ v ⃗ = 7 2 ( 0 , 1 , 1 ) = ( 0 , 7 2 , 7 2 ) \text{Proiezione}_{\vec{u_2}} \vec{v} = \frac{7}{2} (0, 1, 1) = \left( 0, \frac{7}{2}, \frac{7}{2} \right) Proiezione u 2 v = 2 7 ( 0 , 1 , 1 ) = ( 0 , 2 7 , 2 7 )
La proiezione ortogonale di \vec{v} v ⃗ \vec{v} v \vec{u_1} u 1 ⃗ \vec{u_1} u 1 \vec{u_2} u 2 ⃗ \vec{u_2} u 2
\text{Proiezione}_{\text{piano}} \vec{v} = \text{Proiezione}_{\vec{u_1}} \vec{v} + \text{Proiezione}_{\vec{u_2}} \vec{v} Proiezione piano v ⃗ = Proiezione u 1 ⃗ v ⃗ + Proiezione u 2 ⃗ v ⃗ \text{Proiezione}_{\text{piano}} \vec{v} = \text{Proiezione}_{\vec{u_1}} \vec{v} + \text{Proiezione}_{\vec{u_2}} \vec{v} Proiezione piano v = Proiezione u 1 v + Proiezione u 2 v \text{Proiezione}_{\text{piano}} \vec{v} = \left( \frac{5}{2}, \frac{5}{2}, 0 \right) + \left( 0, \frac{7}{2}, \frac{7}{2} \right) = \left( \frac{5}{2}, \frac{5}{2} + \frac{7}{2}, \frac{7}{2} \right) = \left( \frac{5}{2}, 6, \frac{7}{2} \right) Proiezione piano v ⃗ = ( 5 2 , 5 2 , 0 ) + ( 0 , 7 2 , 7 2 ) = ( 5 2 , 5 2 + 7 2 , 7 2 ) = ( 5 2 , 6 , 7 2 ) \text{Proiezione}_{\text{piano}} \vec{v} = \left( \frac{5}{2}, \frac{5}{2}, 0 \right) + \left( 0, \frac{7}{2}, \frac{7}{2} \right) = \left( \frac{5}{2}, \frac{5}{2} + \frac{7}{2}, \frac{7}{2} \right) = \left( \frac{5}{2}, 6, \frac{7}{2} \right) Proiezione piano v = ( 2 5 , 2 5 , 0 ) + ( 0 , 2 7 , 2 7 ) = ( 2 5 , 2 5 + 2 7 , 2 7 ) = ( 2 5 , 6 , 2 7 )
English version
Exercises on Orthogonal Projection of a Vector on a Vector Space
Introduction
The orthogonal projection of a vector \vec{v} v ⃗ \vec{v} v \{\vec{u_1}, \vec{u_2}, \ldots, \vec{u_n}\} { u 1 ⃗ , u 2 ⃗ , … , u n ⃗ } \{\vec{u_1}, \vec{u_2}, \ldots, \vec{u_n}\} { u 1 , u 2 , … , u n } \vec{v} v ⃗ \vec{v} v
The orthogonal projection of a vector \vec{v} v ⃗ \vec{v} v \vec{u} u ⃗ \vec{u} u
\text{Projection}_{\vec{u}} \vec{v} = \frac{\vec{v} \cdot \vec{u}}{\vec{u} \cdot \vec{u}} \vec{u} Projection u ⃗ v ⃗ = v ⃗ ⋅ u ⃗ u ⃗ ⋅ u ⃗ u ⃗ \text{Projection}_{\vec{u}} \vec{v} = \frac{\vec{v} \cdot \vec{u}}{\vec{u} \cdot \vec{u}} \vec{u} Projection u v = u ⋅ u v ⋅ u u Exercise 1: Projection of a Vector onto a Vector
Problem: Find the orthogonal projection of the vector \vec{v} = (3, 4) v ⃗ = ( 3 , 4 ) \vec{v} = (3, 4) v = ( 3 , 4 ) \vec{u} = (1, 2) u ⃗ = ( 1 , 2 ) \vec{u} = (1, 2) u = ( 1 , 2 )
Solution:
Let's calculate the scalar product \vec{v} \cdot \vec{u} v ⃗ ⋅ u ⃗ \vec{v} \cdot \vec{u} v ⋅ u
\vec{v} \cdot \vec{u} = 3 \cdot 1 + 4 \cdot 2 = 3 + 8 = 11 v ⃗ ⋅ u ⃗ = 3 ⋅ 1 + 4 ⋅ 2 = 3 + 8 = 11 \vec{v} \cdot \vec{u} = 3 \cdot 1 + 4 \cdot 2 = 3 + 8 = 11 v ⋅ u = 3 ⋅ 1 + 4 ⋅ 2 = 3 + 8 = 11
Let's calculate the scalar product \vec{u} \cdot \vec{u} u ⃗ ⋅ u ⃗ \vec{u} \cdot \vec{u} u ⋅ u
\vec{u} \cdot \vec{u} = 1 \cdot 1 + 2 \cdot 2 = 1 + 4 = 5 u ⃗ ⋅ u ⃗ = 1 ⋅ 1 + 2 ⋅ 2 = 1 + 4 = 5 \vec{u} \cdot \vec{u} = 1 \cdot 1 + 2 \cdot 2 = 1 + 4 = 5 u ⋅ u = 1 ⋅ 1 + 2 ⋅ 2 = 1 + 4 = 5
Now we can calculate the projection:
\text{Projection}_{\vec{u}} \vec{v} = \frac{11}{5} \vec{u} = \frac{11}{5} (1, 2) = \left( \frac{11}{5}, \frac{22}{5} \right) Projection u ⃗ v ⃗ = 11 5 u ⃗ = 11 5 ( 1 , 2 ) = ( 11 5 , 22 5 ) \text{Projection}_{\vec{u}} \vec{v} = \frac{11}{5} \vec{u} = \frac{11}{5} (1, 2) = \left( \frac{11}{5}, \frac{22}{5} \right) Projection u v = 5 11 u = 5 11 ( 1 , 2 ) = ( 5 11 , 5 22 ) Exercise 2: Projection of a Vector on a Vector Space
Problem: Find the orthogonal projection of the vector \vec{v} = (1, 2, 3) v ⃗ = ( 1 , 2 , 3 ) \vec{v} = (1, 2, 3) v = ( 1 , 2 , 3 ) \vec{u_1} = (1, 0, 0) u 1 ⃗ = ( 1 , 0 , 0 ) \vec{u_1} = (1, 0, 0) u 1 = ( 1 , 0 , 0 ) \vec{u_2} = (0, 1, 0) u 2 ⃗ = ( 0 , 1 , 0 ) \vec{u_2} = (0, 1, 0) u 2 = ( 0 , 1 , 0 )
Solution: 1. We calculate the projection of \vec{v} v ⃗ \vec{v} v \vec{u_1} u 1 ⃗ \vec{u_1} u 1
\text{Projection}_{\vec{u_1}} \vec{v} = \frac{\vec{v} \cdot \vec{u_1}}{\vec{u_1} \cdot \vec{u_1}} \vec{u_1} Projection u 1 ⃗ v ⃗ = v ⃗ ⋅ u 1 ⃗ u 1 ⃗ ⋅ u 1 ⃗ u 1 ⃗ \text{Projection}_{\vec{u_1}} \vec{v} = \frac{\vec{v} \cdot \vec{u_1}}{\vec{u_1} \cdot \vec{u_1}} \vec{u_1} Projection u 1 v = u 1 ⋅ u 1 v ⋅ u 1 u 1
\vec{v} \cdot \vec{u_1} = 1 \cdot 1 + 2 \cdot 0 + 3 \cdot 0 = 1 v ⃗ ⋅ u 1 ⃗ = 1 ⋅ 1 + 2 ⋅ 0 + 3 ⋅ 0 = 1 \vec{v} \cdot \vec{u_1} = 1 \cdot 1 + 2 \cdot 0 + 3 \cdot 0 = 1 v ⋅ u 1 = 1 ⋅ 1 + 2 ⋅ 0 + 3 ⋅ 0 = 1 \vec{u_1} \cdot \vec{u_1} = 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 = 1 u 1 ⃗ ⋅ u 1 ⃗ = 1 ⋅ 1 + 0 ⋅ 0 + 0 ⋅ 0 = 1 \vec{u_1} \cdot \vec{u_1} = 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 = 1 u 1 ⋅ u 1 = 1 ⋅ 1 + 0 ⋅ 0 + 0 ⋅ 0 = 1 Therefore:
\text{Projection}_{\vec{u_1}} \vec{v} = 1 \cdot (1, 0, 0) = (1, 0, 0) Projection u 1 ⃗ v ⃗ = 1 ⋅ ( 1 , 0 , 0 ) = ( 1 , 0 , 0 ) \text{Projection}_{\vec{u_1}} \vec{v} = 1 \cdot (1, 0, 0) = (1, 0, 0) Projection u 1 v = 1 ⋅ ( 1 , 0 , 0 ) = ( 1 , 0 , 0 )
Let's calculate the projection of \vec{v} v ⃗ \vec{v} v \vec{u_2} u 2 ⃗ \vec{u_2} u 2
\text{Projection}_{\vec{u_2}} \vec{v} = \frac{\vec{v} \cdot \vec{u_2}}{\vec{u_2} \cdot \vec{u_2}} \vec{u_2} Projection u 2 ⃗ v ⃗ = v ⃗ ⋅ u 2 ⃗ u 2 ⃗ ⋅ u 2 ⃗ u 2 ⃗ \text{Projection}_{\vec{u_2}} \vec{v} = \frac{\vec{v} \cdot \vec{u_2}}{\vec{u_2} \cdot \vec{u_2}} \vec{u_2} Projection u 2 v = u 2 ⋅ u 2 v ⋅ u 2 u 2
Let's calculate \vec{v} \cdot \vec{u_2} v ⃗ ⋅ u 2 ⃗ \vec{v} \cdot \vec{u_2} v ⋅ u 2
\vec{v} \cdot \vec{u_2} = 1 \cdot 0 + 2 \cdot 1 + 3 \cdot 0 = 2 v ⃗ ⋅ u 2 ⃗ = 1 ⋅ 0 + 2 ⋅ 1 + 3 ⋅ 0 = 2 \vec{v} \cdot \vec{u_2} = 1 \cdot 0 + 2 \cdot 1 + 3 \cdot 0 = 2 v ⋅ u 2 = 1 ⋅ 0 + 2 ⋅ 1 + 3 ⋅ 0 = 2
Let's calculate \vec{u_2} \cdot \vec{u_2} u 2 ⃗ ⋅ u 2 ⃗ \vec{u_2} \cdot \vec{u_2} u 2 ⋅ u 2
\vec{u_2} \cdot \vec{u_2} = 0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0 = 1 u 2 ⃗ ⋅ u 2 ⃗ = 0 ⋅ 0 + 1 ⋅ 1 + 0 ⋅ 0 = 1 \vec{u_2} \cdot \vec{u_2} = 0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0 = 1 u 2 ⋅ u 2 = 0 ⋅ 0 + 1 ⋅ 1 + 0 ⋅ 0 = 1 \text{Projection}_{\vec{u_2}} \vec{v} = 2 \cdot (0, 1, 0) = (0, 2, 0) Projection u 2 ⃗ v ⃗ = 2 ⋅ ( 0 , 1 , 0 ) = ( 0 , 2 , 0 ) \text{Projection}_{\vec{u_2}} \vec{v} = 2 \cdot (0, 1, 0) = (0, 2, 0) Projection u 2 v = 2 ⋅ ( 0 , 1 , 0 ) = ( 0 , 2 , 0 )
The orthogonal projection of \vec{v} v ⃗ \vec{v} v \vec{u_1} u 1 ⃗ \vec{u_1} u 1 \vec{u_2} u 2 ⃗ \vec{u_2} u 2
\text{Projection}_{\text{plane}} \vec{v} = \text{Projection}_{\vec{u_1}} \vec{v} + \text{Projection}_{\vec{u_2}} \vec{v} Projection plane v ⃗ = Projection u 1 ⃗ v ⃗ + Projection u 2 ⃗ v ⃗ \text{Projection}_{\text{plane}} \vec{v} = \text{Projection}_{\vec{u_1}} \vec{v} + \text{Projection}_{\vec{u_2}} \vec{v} Projection plane v = Projection u 1 v + Projection u 2 v \text{Projection}_{\text{plane}} \vec{v} = (1, 0, 0) + (0, 2, 0) = (1, 2, 0) Projection plane v ⃗ = ( 1 , 0 , 0 ) + ( 0 , 2 , 0 ) = ( 1 , 2 , 0 ) \text{Projection}_{\text{plane}} \vec{v} = (1, 0, 0) + (0, 2, 0) = (1, 2, 0) Projection plane v = ( 1 , 0 , 0 ) + ( 0 , 2 , 0 ) = ( 1 , 2 , 0 ) Exercise 3: Orthogonal Projection in a Higher Dimensional Space
Problem: Find the orthogonal projection of the vector \vec{v} = (2, 3, 4) v ⃗ = ( 2 , 3 , 4 ) \vec{v} = (2, 3, 4) v = ( 2 , 3 , 4 ) \vec{u_1} = (1, 1, 0) u 1 ⃗ = ( 1 , 1 , 0 ) \vec{u_1} = (1, 1, 0) u 1 = ( 1 , 1 , 0 ) \vec{u_2} = (0, 1, 1) u 2 ⃗ = ( 0 , 1 , 1 ) \vec{u_2} = (0, 1, 1) u 2 = ( 0 , 1 , 1 )
Solution:
Let's calculate the projection of \vec{v} v ⃗ \vec{v} v \vec{u_1} u 1 ⃗ \vec{u_1} u 1
\text{Projection}_{\vec{u_1}} \vec{v} = \frac{\vec{v} \cdot \vec{u_1}}{\vec{u_1} \cdot \vec{u_1}} \vec{u_1} Projection u 1 ⃗ v ⃗ = v ⃗ ⋅ u 1 ⃗ u 1 ⃗ ⋅ u 1 ⃗ u 1 ⃗ \text{Projection}_{\vec{u_1}} \vec{v} = \frac{\vec{v} \cdot \vec{u_1}}{\vec{u_1} \cdot \vec{u_1}} \vec{u_1} Projection u 1 v = u 1 ⋅ u 1 v ⋅ u 1 u 1
Let's calculate \vec{v} \cdot \vec{u_1} v ⃗ ⋅ u 1 ⃗ \vec{v} \cdot \vec{u_1} v ⋅ u 1
\vec{v} \cdot \vec{u_1} = 2 \cdot 1 + 3 \cdot 1 + 4 \cdot 0 = 2 + 3 + 0 = 5 v ⃗ ⋅ u 1 ⃗ = 2 ⋅ 1 + 3 ⋅ 1 + 4 ⋅ 0 = 2 + 3 + 0 = 5 \vec{v} \cdot \vec{u_1} = 2 \cdot 1 + 3 \cdot 1 + 4 \cdot 0 = 2 + 3 + 0 = 5 v ⋅ u 1 = 2 ⋅ 1 + 3 ⋅ 1 + 4 ⋅ 0 = 2 + 3 + 0 = 5
Let's calculate \vec{u_1} \cdot \vec{u_1} u 1 ⃗ ⋅ u 1 ⃗ \vec{u_1} \cdot \vec{u_1} u 1 ⋅ u 1
\vec{u_1} \cdot \vec{u_1} = 1 \cdot 1 + 1 \cdot 1 + 0 \cdot 0 = 1 + 1 + 0 = 2 u 1 ⃗ ⋅ u 1 ⃗ = 1 ⋅ 1 + 1 ⋅ 1 + 0 ⋅ 0 = 1 + 1 + 0 = 2 \vec{u_1} \cdot \vec{u_1} = 1 \cdot 1 + 1 \cdot 1 + 0 \cdot 0 = 1 + 1 + 0 = 2 u 1 ⋅ u 1 = 1 ⋅ 1 + 1 ⋅ 1 + 0 ⋅ 0 = 1 + 1 + 0 = 2 \text{Projection}_{\vec{u_1}} \vec{v} = \frac{5}{2} (1, 1, 0) = \left( \frac{5}{2}, \frac{5}{2}, 0 \right) Projection u 1 ⃗ v ⃗ = 5 2 ( 1 , 1 , 0 ) = ( 5 2 , 5 2 , 0 ) \text{Projection}_{\vec{u_1}} \vec{v} = \frac{5}{2} (1, 1, 0) = \left( \frac{5}{2}, \frac{5}{2}, 0 \right) Projection u 1 v = 2 5 ( 1 , 1 , 0 ) = ( 2 5 , 2 5 , 0 )
Let's calculate the projection of \vec{v} v ⃗ \vec{v} v \vec{u_2} u 2 ⃗ \vec{u_2} u 2
\text{Projection}_{\vec{u_2}} \vec{v} = \frac{\vec{v} \cdot \vec{u_2}}{\vec{u_2} \cdot \vec{u_2}} \vec{u_2} Projection u 2 ⃗ v ⃗ = v ⃗ ⋅ u 2 ⃗ u 2 ⃗ ⋅ u 2 ⃗ u 2 ⃗ \text{Projection}_{\vec{u_2}} \vec{v} = \frac{\vec{v} \cdot \vec{u_2}}{\vec{u_2} \cdot \vec{u_2}} \vec{u_2} Projection u 2 v = u 2 ⋅ u 2 v ⋅ u 2 u 2
Let's calculate \vec{v} \cdot \vec{u_2} v ⃗ ⋅ u 2 ⃗ \vec{v} \cdot \vec{u_2} v ⋅ u 2
\vec{v} \cdot \vec{u_2} = 2 \cdot 0 + 3 \cdot 1 + 4 \cdot 1 = 0 + 3 + 4 = 7 v ⃗ ⋅ u 2 ⃗ = 2 ⋅ 0 + 3 ⋅ 1 + 4 ⋅ 1 = 0 + 3 + 4 = 7 \vec{v} \cdot \vec{u_2} = 2 \cdot 0 + 3 \cdot 1 + 4 \cdot 1 = 0 + 3 + 4 = 7 v ⋅ u 2 = 2 ⋅ 0 + 3 ⋅ 1 + 4 ⋅ 1 = 0 + 3 + 4 = 7
Let's calculate \vec{u_2} \cdot c{u_2} u 2 ⃗ ⋅ c u 2 \vec{u_2} \cdot c{u_2} u 2 ⋅ c u 2
\vec{u_2} \cdot \vec{u_2} = 0 \cdot 0 + 1 \cdot 1 + 1 \cdot 1 = 0 + 1 + 1 = 2 u 2 ⃗ ⋅ u 2 ⃗ = 0 ⋅ 0 + 1 ⋅ 1 + 1 ⋅ 1 = 0 + 1 + 1 = 2 \vec{u_2} \cdot \vec{u_2} = 0 \cdot 0 + 1 \cdot 1 + 1 \cdot 1 = 0 + 1 + 1 = 2 u 2 ⋅ u 2 = 0 ⋅ 0 + 1 ⋅ 1 + 1 ⋅ 1 = 0 + 1 + 1 = 2 \text{Projection}_{\vec{u_2}} \vec{v} = \frac{7}{2} (0, 1, 1) = \left( 0, \frac{7}{2}, \frac{7}{2} \right) Projection u 2 ⃗ v ⃗ = 7 2 ( 0 , 1 , 1 ) = ( 0 , 7 2 , 7 2 ) \text{Projection}_{\vec{u_2}} \vec{v} = \frac{7}{2} (0, 1, 1) = \left( 0, \frac{7}{2}, \frac{7}{2} \right) Projection u 2 v = 2 7 ( 0 , 1 , 1 ) = ( 0 , 2 7 , 2 7 )
The orthogonal projection of \vec{v} v ⃗ \vec{v} v \vec{u_1} u 1 ⃗ \vec{u_1} u 1 \vec{u_2} u 2 ⃗ \vec{u_2} u 2
\text{Projection}_{\text{plane}} \vec{v} = \text{Projection}_{\vec{u_1}} \vec{v} + \text{Projection}_{\vec{u_2}} \vec{v} Projection plane v ⃗ = Projection u 1 ⃗ v ⃗ + Projection u 2 ⃗ v ⃗ \text{Projection}_{\text{plane}} \vec{v} = \text{Projection}_{\vec{u_1}} \vec{v} + \text{Projection}_{\vec{u_2}} \vec{v} Projection plane v = Projection u 1 v + Projection u 2 v \text{Projection}_{\text{plane}} \vec{v} = \left( \frac{5}{2}, \frac{5}{2}, 0 \right) + \left( 0, \frac{7}{2}, \frac{7}{2} \right) = \left( \frac{5}{2}, \frac{5}{2} + \frac{7}{2}, \frac{7}{2} \right) = \left( \frac{5}{2}, 6, \frac{7}{2} \right) Projection plane v ⃗ = ( 5 2 , 5 2 , 0 ) + ( 0 , 7 2 , 7 2 ) = ( 5 2 , 5 2 + 7 2 , 7 2 ) = ( 5 2 , 6 , 7 2 ) \text{Projection}_{\text{plane}} \vec{v} = \left( \frac{5}{2}, \frac{5}{2}, 0 \right) + \left( 0, \frac{7}{2}, \frac{7}{2} \right) = \left( \frac{5}{2}, \frac{5}{2} + \frac{7}{2}, \frac{7}{2} \right) = \left( \frac{5}{2}, 6, \frac{7}{2} \right) Projection plane v = ( 2 5 , 2 5 , 0 ) + ( 0 , 2 7 , 2 7 ) = ( 2 5 , 2 5 + 2 7 , 2 7 ) = ( 2 5 , 6 , 2 7 )
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