Given vectors U(2,1,3) and V(4,-5,1), calculate the dot product, squared magnitude of U, magnitude of V, and the cosine of the angle between U and V.

Given vectors $\vec{U}(2,1,3)$ and $\vec{V}(4,-5,1)$.

• 1) Calculate $\vec{U} \cdot \vec{V}$ The dot product of two vectors $\vec{U} = (u_x, u_y, u_z)$ and $\vec{V} = (v_x, v_y, v_z)$ is $u_x v_x + u_y v_y + u_z v_z$.

  $$\vec{U} \cdot \vec{V} = (2)(4) + (1)(-5) + (3)(1) = 8 - 5 + 3 = \boxed{6}$$

• 2) Calculate $||\vec{U}||^2$ The squared magnitude of a vector $\vec{U}$ is $||\vec{U}||^2 = u_x^2 + u_y^2 + u_z^2$.

  $$||\vec{U}||^2 = 2^2 + 1^2 + 3^2 = 4 + 1 + 9 = \boxed{14}$$

• 3) Calculate $||\vec{V}||$ The magnitude of a vector $\vec{V}$ is $||\vec{V}|| = \sqrt{v_x^2 + v_y^2 + v_z^2}$.

  $$||\vec{V}|| = \sqrt{4^2 + (-5)^2 + 1^2} = \sqrt{16 + 25 + 1} = \sqrt{42} = \boxed{\sqrt{42}}$$

• 4) Calculate $\cos(\vec{U}, \vec{V})$ The cosine of the angle $\theta$ between two vectors is given by $\cos(\theta) = \frac{\vec{U} \cdot \vec{V}}{||\vec{U}|| \cdot ||\vec{V}||}$. First, find $||\vec{U}|| = \sqrt{||\vec{U}||^2} = \sqrt{14}$.

  $$\cos(\vec{U}, \vec{V}) = \frac{6}{\sqrt{14} \cdot \sqrt{42}} = \frac{6}{\sqrt{14 \cdot 42}} = \frac{6}{\sqrt{14 \cdot 3 \cdot 14}} = \frac{6}{\sqrt{14^2 \cdot 3}} = \frac{6}{14\sqrt{3}}$$

  Simplify the fraction and rationalize the denominator:

  $$\cos(\vec{U}, \vec{V}) = \frac{3}{7\sqrt{3}} = \frac{3\sqrt{3}}{7\sqrt{3}\sqrt{3}} = \frac{3\sqrt{3}}{7 \cdot 3} = \frac{\sqrt{3}}{7} = \boxed{\frac{\sqrt{3}}{7}}$$

• 5) Calculate $||\vec{U} + \vec{V}||$ First, calculate the vector sum $\vec{U} + \vec{V}$:

  $$\vec{U} + \vec{V} = (2+4, 1+(-5), 3+1) = (6, -4, 4)$$

  Now, calculate its magnitude:

  $$||\vec{U} + \vec{V}|| = \sqrt{6^2 + (-4)^2 + 4^2} = \sqrt{36 + 16 + 16} = \sqrt{68}$$

  Simplify the radical:

  $$\sqrt{68} = \sqrt{4 \cdot 17} = 2\sqrt{17} = \boxed{2\sqrt{17}}$$

Given points $A(-4,2,-1)$, $B(-1,-2,4)$, and $C(-2,3,1)$.

• 1) Calculate $\vec{AB}$ The vector $\vec{AB}$ is found by subtracting the coordinates of A from B.

  $$\vec{AB} = B - A = (-1 - (-4), -2 - 2, 4 - (-1)) = (3, -4, 5) = \boxed{(3, -4, 5)}$$

• 2) Calculate $\vec{AC}$ The vector $\vec{AC}$ is found by subtracting the coordinates of A from C.

  $$\vec{AC} = C - A = (-2 - (-4), 3 - 2, 1 - (-1)) = (2, 1, 2) = \boxed{(2, 1, 2)}$$

• 3) Calculate $\vec{AB} \cdot \vec{AC}$ Using the vectors calculated above:

  $$\vec{AB} \cdot \vec{AC} = (3)(2) + (-4)(1) + (5)(2) = 6 - 4 + 10 = \boxed{12}$$