\( \vec{a}=3 i+4 j+11 k \) \( \vec{b}=6 i+7 j+9 k \) find the area of vectors

To find the area formed by the vectors \( \vec{a} \) and \( \vec{b} \), we can use the cross product. The magnitude of the cross product \( |\vec{a} \times \vec{b}| \) gives the area of the parallelogram formed by the two vectors. First, calculate the cross product: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 11 \\ 6 & 7 & 9 \end{vmatrix} \] Calculating this determinant, we find: \[ \vec{a} \times \vec{b} = \hat{i}(4 \cdot 9 - 11 \cdot 7) - \hat{j}(3 \cdot 9 - 11 \cdot 6) + \hat{k}(3 \cdot 7 - 4 \cdot 6) \] This results in: \[ \vec{a} \times \vec{b} = \hat{i}(-53) - \hat{j}(-33) + \hat{k}(-10) = -53 \hat{i} + 33 \hat{j} - 10 \hat{k} \] Now calculate the magnitude: \[ |\vec{a} \times \vec{b}| = \sqrt{(-53)^2 + 33^2 + (-10)^2} = \sqrt{2809 + 1089 + 100} = \sqrt{3998} \] Thus, the area is: \[ \text{Area} = |\vec{a} \times \vec{b}| = \sqrt{3998} \approx 63.24 \] So, the area of the parallelogram formed by the vectors \( \vec{a} \) and \( \vec{b} \) is approximately 63.24 square units. Imagine two vectors as arrows originating from a common point! The area they span looks like a trampoline stretched between them. The larger your arrows (vectors), the bigger your trampoline (area) will be! So remember to measure your vectors right to get the best bounce! If you're intrigued by the practical applications of vector areas, think about physics. In fields such as robotics or game design, understanding how to manipulate areas defined by vectors can help control movements, design paths, or even calculate the forces acting on shapes in motion! Exploring these concepts can enhance your abilities in simulations or animations.