Given v= 2I-J and w=2I + 3J, find the angle between v and w.

To find the angle between two vectors, you can use the dot product formula:

If the angle between two vectors is θ, then the dot product of the two vectors is given by:

{\vec{v} \cdot \vec{w} = |\vec{v}| \cdot |\vec{w}| \cdot \cos(\theta)}

Given that \vec{v} = 2\vec{I} - \vec{J} and \vec{w} = 2\vec{I} + 3\vec{J} , the dot product of v and w is:

{\vec{v} \cdot \vec{w} = (2\vec{I} - \vec{J}) \cdot (2\vec{I} + 3\vec{J})}

{\vec{v} \cdot \vec{w} = 2 \cdot 2 + (-1) \cdot 3}

{\vec{v} \cdot \vec{w} = 4 - 3}

{\vec{v} \cdot \vec{w} = 1}

Now, to find the magnitude of v and w:

|\vec{v}| = \sqrt{(2)^2 + (-1)^2}

|\vec{v}| = \sqrt{4 + 1}

|\vec{v}| = \sqrt{5}

|\vec{w}| = \sqrt{(2)^2 + 3^2}

|\vec{w}| = \sqrt{4 + 9}

|\vec{w}| = \sqrt{13}

Substitute these magnitudes and the calculated dot product back into the formula:

{\cos(\theta) = \frac{\vec{v} \cdot \vec{w}}{|\vec{v}| \cdot |\vec{w}|}}

{\cos(\theta) = \frac{1}{\sqrt{5} \cdot \sqrt{13}}}

{\cos(\theta) = \frac{1}{\sqrt{65}}}

\theta=\cos^{-1}\left(\frac{1}{\sqrt{65}}\approx82.87^{\circ}\right.

\boxed{\theta\approx82.87^{\circ}}