Math Problem Statement

Si \( \vec{a} = (3,5) \) y \( \vec{b} = (6,1) \), representa gráficamente y determina la \text{Proy}_{\vec{b}}\vec{a} y la \text{Comp}_{\vec{b}^{\perp}}\vec{a}.

Solution

The problem asks to find both the projection of vector a\vec{a} onto vector b\vec{b} (Proyba\text{Proy}_{\vec{b}}\vec{a}) and the component of a\vec{a} perpendicular to b\vec{b} (Compba\text{Comp}_{\vec{b}^{\perp}}\vec{a}).

Given:

  • a=(3,5)\vec{a} = (3, 5)
  • b=(6,1)\vec{b} = (6, 1)

Step 1: Find the projection of a\vec{a} onto b\vec{b}

The formula for the projection of a\vec{a} onto b\vec{b} is: Proyba=abbbb\text{Proy}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{\vec{b} \cdot \vec{b}} \vec{b}

First, calculate the dot products:

  1. ab=3×6+5×1=18+5=23\vec{a} \cdot \vec{b} = 3 \times 6 + 5 \times 1 = 18 + 5 = 23.
  2. bb=6×6+1×1=36+1=37\vec{b} \cdot \vec{b} = 6 \times 6 + 1 \times 1 = 36 + 1 = 37.

Now, substitute these into the formula: Proyba=2337b=2337(6,1)=(13837,2337)\text{Proy}_{\vec{b}}\vec{a} = \frac{23}{37} \vec{b} = \frac{23}{37} (6, 1) = \left(\frac{138}{37}, \frac{23}{37}\right)

Simplifying the fractions: Proyba(3.73,0.62)\text{Proy}_{\vec{b}}\vec{a} \approx (3.73, 0.62)

Step 2: Find the component of a\vec{a} perpendicular to b\vec{b}

The component of a\vec{a} perpendicular to b\vec{b} is given by: Compba=aProyba\text{Comp}_{\vec{b}^{\perp}}\vec{a} = \vec{a} - \text{Proy}_{\vec{b}}\vec{a}

Substitute a=(3,5)\vec{a} = (3, 5) and Proyba(3.73,0.62)\text{Proy}_{\vec{b}}\vec{a} \approx (3.73, 0.62): Compba=(3,5)(3.73,0.62)=(0.73,4.38)\text{Comp}_{\vec{b}^{\perp}}\vec{a} = (3, 5) - (3.73, 0.62) = (-0.73, 4.38)

Summary

  • The projection of a\vec{a} onto b\vec{b} is approximately (3.73,0.62)(3.73, 0.62).
  • The component of a\vec{a} perpendicular to b\vec{b} is approximately (0.73,4.38)(-0.73, 4.38).

Would you like a more detailed explanation, or do you have any questions about these steps?

Relative Questions

  1. What is the geometric interpretation of the vector projection?
  2. How do the dot product and magnitude of vectors relate to their projections?
  3. Can the projection of a vector onto another be longer than the original vector?
  4. What happens to the perpendicular component if the vectors are parallel?
  5. How can we find the angle between two vectors using their dot product?

Tip

When working with vector projections, remember that the projection lies in the direction of the vector onto which you are projecting, while the perpendicular component is orthogonal to this direction.

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Math Problem Analysis

Mathematical Concepts

Vector Projections
Perpendicular Components
Dot Products

Formulas

Projection of vector a onto b: \text{Proy}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{\vec{b} \cdot \vec{b}} \vec{b}
Component perpendicular to b: \text{Comp}_{\vec{b}^{\perp}}\vec{a} = \vec{a} - \text{Proy}_{\vec{b}}\vec{a}

Theorems

Dot Product
Orthogonal Decomposition Theorem

Suitable Grade Level

Grades 10-12 (High School)