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4. Dados [tex]\(A(3, -4)\)[/tex] y [tex]\(B(-2, 1)\)[/tex], calcule el módulo de [tex]\(A\)[/tex] y [tex]\(B\)[/tex].

Sagot :

Para calcular el módulo (o magnitud) de un vector en el plano cartesiano, usamos la fórmula:

[tex]\[ \| \vec{V} \| = \sqrt{x^2 + y^2} \][/tex]

donde [tex]\(\| \vec{V} \|\)[/tex] es el módulo del vector [tex]\(\vec{V}\)[/tex], y [tex]\(x\)[/tex] y [tex]\(y\)[/tex] son las coordenadas del vector.

### Paso 1: Cálculo del módulo de [tex]\(A(3, -4)\)[/tex]

Primero, tomemos las coordenadas de [tex]\(A\)[/tex]:
- [tex]\(A_x = 3\)[/tex]
- [tex]\(A_y = -4\)[/tex]

Aplicamos la fórmula del módulo:

[tex]\[ \| A \| = \sqrt{A_x^2 + A_y^2} \][/tex]

Sustituyendo los valores:

[tex]\[ \| A \| = \sqrt{3^2 + (-4)^2} \][/tex]
[tex]\[ \| A \| = \sqrt{9 + 16} \][/tex]
[tex]\[ \| A \| = \sqrt{25} \][/tex]
[tex]\[ \| A \| = 5.0 \][/tex]

### Paso 2: Cálculo del módulo de [tex]\(B(-2, 1)\)[/tex]

Ahora, tomemos las coordenadas de [tex]\(B\)[/tex]:
- [tex]\(B_x = -2\)[/tex]
- [tex]\(B_y = 1\)[/tex]

Aplicamos la misma fórmula del módulo:

[tex]\[ \| B \| = \sqrt{B_x^2 + B_y^2} \][/tex]

Sustituyendo los valores:

[tex]\[ \| B \| = \sqrt{(-2)^2 + 1^2} \][/tex]
[tex]\[ \| B \| = \sqrt{4 + 1} \][/tex]
[tex]\[ \| B \| = \sqrt{5} \][/tex]
[tex]\[ \| B \| \approx 2.23606797749979 \][/tex]

### Resultados Finales

El módulo de [tex]\(A(3, -4)\)[/tex] es [tex]\(5.0\)[/tex].

El módulo de [tex]\(B(-2, 1)\)[/tex] es aproximadamente [tex]\(2.236\)[/tex].

Esto concluye el cálculo de los módulos de los puntos [tex]\(A\)[/tex] y [tex]\(B\)[/tex].
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