I need more examples and solution under dot multiplication and cross multiplication in application of vector.

The dot product of two vectors $\vec{u} = u_x\mathbf{i} + u_y\mathbf{j} + u_z\mathbf{k}$ and $\vec{v} = v_x\mathbf{i} + v_y\mathbf{j} + v_z\mathbf{k}$ is given by $\vec{u} \cdot \vec{v} = u_xv_x + u_yv_y + u_zv_z$.

$$\vec{u} \cdot \vec{v} = (3)(1) + (2)(-4) + (-1)(2)$$ $$\vec{u} \cdot \vec{v} = 3 - 8 - 2$$ $$\vec{u} \cdot \vec{v} = -7$$

The magnitude of a vector $\vec{u} = u_x\mathbf{i} + u_y\mathbf{j} + u_z\mathbf{k}$ is $|\vec{u}| = \sqrt{u_x^2 + u_y^2 + u_z^2}$.

$$|\vec{u}| = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$$ $$|\vec{v}| = \sqrt{1^2 + (-4)^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21}$$

The formula relating the dot product, magnitudes, and angle is $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta$.
Rearranging for $\cos \theta$:

$$\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$$ $$\cos \theta = \frac{-7}{\sqrt{14} \sqrt{21}}$$ $$\cos \theta = \frac{-7}{\sqrt{294}}$$ $$\cos \theta = \frac{-7}{7\sqrt{6}}$$ $$\cos \theta = \frac{-1}{\sqrt{6}}$$

Now, find $\theta$:

$$\theta = \arccos\left(\frac{-1}{\sqrt{6}}\right)$$ $$\theta \approx \arccos(-0.4082)$$ $$\theta \approx \boxed{\mathbf{114.09^\circ}}$$

Cross Product Application: Finding a Perpendicular Vector and Area of a Parallelogram

Problem: Given vectors $\vec{a} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}$ and $\vec{b} = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k}$, find a vector perpendicular to both $\vec{a}$ and $\vec{b}$, and then find the area of the parallelogram formed by these two vectors.

The cross product of two vectors $\vec{a} = a_x\mathbf{i} + a_y\mathbf{j} + a_z\mathbf{k}$ and $\vec{b} = b_x\mathbf{i} + b_y\mathbf{j} + b_z\mathbf{k}$ is given by:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix}$$ $$\vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix}$$ $$\vec{a} \times \vec{b} = \mathbf{i}((2)(2) - (-1)(-1)) - \mathbf{j}((1)(2) - (-1)(3)) + \mathbf{k}((1)(-1) - (2)(3))$$ $$\vec{a} \times \vec{b} = \mathbf{i}(4 - 1) - \mathbf{j}(2 - (-3)) + \mathbf{k}(-1 - 6)$$ $$\vec{a} \times \vec{b} = 3\mathbf{i} - 5\mathbf{j} - 7\mathbf{k}$$

A vector perpendicular to both $\vec{a}$ and $\vec{b}$ is $\boxed{\mathbf{3\mathbf{i} - 5\mathbf{j} - 7\mathbf{k}}}$.

The area of the parallelogram formed by $\vec{a}$ and $\vec{b}$ is equal to the magnitude of their cross product, $|\vec{a} \times \vec{b}|$.

$$|\vec{a} \times \vec{b}| = |3\mathbf{i} - 5\mathbf{j} - 7\mathbf{k}|$$ $$|\vec{a} \times \vec{b}| = \sqrt{3^2 + (-5)^2 + (-7)^2}$$ $$|\vec{a} \times \vec{b}| = \sqrt{9 + 25 + 49}$$ $$|\vec{a} \times \vec{b}| = \sqrt{83}$$

The area of the parallelogram is $\boxed{\mathbf{\sqrt{83} \text{ square units}}}$.