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Se tienen 2 vectores cuyos módulos están en la misma razón que 5 y 2. Si cuando forman 53º su resultante mide 3√29u, ¿Cuánto mide el mayor vector?

Respuesta :

Answer:

|v2| = 3.27

Explanation:

You have two vector v1 and v2. The relation between the magnitudes of both vectors is given by:

[tex]\frac{|v_1|}{|v_2|}=\frac{5}{2}[/tex]

Furthermore, the projection (the dot product) of one vector on the other one is given by the following formula:

[tex]v_1 \cdot v_2 = |v_1||v_2|cos\theta[/tex]

The dot product between v1 and v2 is 3√29. If you multiply the right hand side of the last equation by |v2|/|v2| you obtain:

[tex]3\sqrt{29}=\frac{|v_1|}{|v_2|}|v_2|^2cos(53\°)[/tex]

you do |v2| the subject of the formula:

[tex]|v_2|=\sqrt{\frac{|v_2|}{|v_1|}\frac{3\sqrt{29}}{cos53\°}}\\\\|v_2|=\sqrt{\frac{2}{5}\frac{3\sqrt{29}}{cos53\°}}=3.27[/tex]

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