The area of a triangle with adjacent sides represented by vectors OA and OB is given by:
\text{Area} = \frac{1}{2} |\vec{OA} \times \vec{OB}|$ 1. **Find the cross product $\vec{OA} \times \vec{OB}$:**\vec{OA} \times \vec{OB} = i^1−3j^2−2k^31= \hat{i}(2 - (-6)) - \hat{j}(1 - (-9)) + \hat{k}(-2 - (-6))= 8\hat{i} - 10\hat{j} + 4\hat{k}2.∗∗Calculatethemagnitude:∗∗|\vec{OA} \times \vec{OB}| = \sqrt{8^2 + (-10)^2 + 4^2} = \sqrt{64 + 100 + 16} = \sqrt{180}3.∗∗CalculatetheArea:∗∗\text{Area} = \frac{1}{2} \sqrt{180} = \sqrt{\frac{180}{4}} = \sqrt{45}\text{ sq. units}$$