Find the angle between vectors \(\vec{u} = (3,4,0)\) and \(\vec{v} = (2,1,-2)\) using the cross product.
Answer
The angle \(\theta\) between the vectors can be found using the formula \(\sin{\theta} = \frac{\|\vec{u} \times \vec{v}\|}{\|\vec{u}\|\|\vec{v}\|}\). Substituting the computed values, we get \(\sin{\theta} = \frac{\sqrt{89}}{5*3} = \frac{\sqrt{89}}{15}\). Thus, \(\theta = \arcsin{\left(\frac{\sqrt{89}}{15}\right)}\).
Step 1 ：First, calculate the cross product of \(\vec{u}\) and \(\vec{v}\). The cross product of \(\vec{u}\) and \(\vec{v}\) is given by \(\vec{u} \times \vec{v} = (u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1)\). Substituting the given values, we get \(\vec{u} \times \vec{v} = (4*(-2) - 0*1, 0*2 - 3*(-2), 3*1 - 4*2) = (-8,6,-5)\).
Step 2 ：The magnitude of a vector \(\vec{x} = (x_1, x_2, x_3)\) is given by \(\|\vec{x}\| = \sqrt{x_1^2 + x_2^2 + x_3^2}\). Applying this, we get \(\|\vec{u}\| = \sqrt{3^2 + 4^2 + 0^2} = 5\) and \(\|\vec{v}\| = \sqrt{2^2 + 1^2 + (-2)^2} = 3\). The magnitude of the cross product \(\|\vec{u} \times \vec{v}\| = \sqrt{(-8)^2 + 6^2 + (-5)^2} = \sqrt{89}\).
Step 3 ：The angle \(\theta\) between the vectors can be found using the formula \(\sin{\theta} = \frac{\|\vec{u} \times \vec{v}\|}{\|\vec{u}\|\|\vec{v}\|}\). Substituting the computed values, we get \(\sin{\theta} = \frac{\sqrt{89}}{5*3} = \frac{\sqrt{89}}{15}\). Thus, \(\theta = \arcsin{\left(\frac{\sqrt{89}}{15}\right)}\).
