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]