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5.- Normalizar el vector \( \quad a=\frac{1}{\sqrt{6}} i-\frac{1}{\sqrt{6}} j-\frac{1}{\sqrt{6}} k \)

Ask by Goodwin Fowler. in Mexico
Feb 25,2025

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Answer

El vector normalizado es \( \frac{1}{\sqrt{3}} \, \mathbf{i} - \frac{1}{\sqrt{3}} \, \mathbf{j} - \frac{1}{\sqrt{3}} \, \mathbf{k} \) o \( \frac{1}{\sqrt{3}} (\mathbf{i} - \mathbf{j} - \mathbf{k}) \).

Solution

Para normalizar un vector se divide el vector entre su magnitud. Dado   a = (1/√6) i − (1/√6) j − (1/√6) k, primero calculemos su norma:   ||a|| = √[(1/√6)² + (−1/√6)² + (−1/√6)²]      = √[1/6 + 1/6 + 1/6]      = √(3/6)      = √(1/2)      = 1/√2. La normalización consiste en dividir a entre ||a||:   a_normalizado = a / ||a|| = [(1/√6)/(1/√2)] i − [(1/√6)/(1/√2)] j − [(1/√6)/(1/√2)] k. Al simplificar, tenemos:   (1/√6) / (1/√2) = √2/√6 = √(2/6) = √(1/3) = 1/√3. Por lo tanto, el vector normalizado es:   a_normalizado = (1/√3) i − (1/√3) j − (1/√3) k  ó  a_normalizado = (1/√3)(i − j − k).

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The Deep Dive

Para normalizar el vector \( a = \frac{1}{\sqrt{6}} i - \frac{1}{\sqrt{6}} j - \frac{1}{\sqrt{6}} k \), primero debemos calcular su magnitud. La magnitud se calcula usando la fórmula \( \|a\| = \sqrt{x^2 + y^2 + z^2} \), donde \( x, y, z \) son las componentes del vector. En este caso, \( x = \frac{1}{\sqrt{6}}, y = -\frac{1}{\sqrt{6}}, z = -\frac{1}{\sqrt{6}} \). La magnitud \( \|a\| \) es \( \sqrt{\left(\frac{1}{\sqrt{6}}\right)^2 + \left(-\frac{1}{\sqrt{6}}\right)^2 + \left(-\frac{1}{\sqrt{6}}\right)^2} = \sqrt{\frac{1}{6} + \frac{1}{6} + \frac{1}{6}} = \sqrt{\frac{3}{6}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \). Una vez que tenemos la magnitud, podemos normalizar el vector dividiendo cada componente por la magnitud: \[ \hat{a} = \frac{1}{\|a\|} a = \frac{1}{\frac{1}{\sqrt{2}}} \left( \frac{1}{\sqrt{6}} i - \frac{1}{\sqrt{6}} j - \frac{1}{\sqrt{6}} k \right) = \sqrt{2} \left( \frac{1}{\sqrt{6}} i - \frac{1}{\sqrt{6}} j - \frac{1}{\sqrt{6}} k \right) = \left( \frac{\sqrt{2}}{\sqrt{6}} i - \frac{\sqrt{2}}{\sqrt{6}} j - \frac{\sqrt{2}}{\sqrt{6}} k \right). \] Por lo tanto, el vector normalizado es \( \hat{a} = \frac{\sqrt{2}}{\sqrt{6}} i - \frac{\sqrt{2}}{\sqrt{6}} j - \frac{\sqrt{2}}{\sqrt{6}} k \). Y ya está, ¡listo para ser usado!

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