Let vec{a} = 2hat{i} + 3hat{j} + 3hat{k} and | vedclass

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(C) The area $A$ of a triangle with adjacent sides $\vec{u}$ and $\vec{v}$ is given by $A = \frac{1}{2} |\vec{u} 	imes \vec{v}|$.Here, $\vec{u} = 2\v

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$1800$

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(C) The area $A$ of a triangle with adjacent sides $\vec{u}$ and $\vec{v}$ is given by $A = \frac{1}{2} |\vec{u} 	imes \vec{v}|$.Here, $\vec{u} = 2\vec{a} + 3\vec{b}$ and $\vec{v} = \vec{a} - \vec{b}$.Calculating the cross product: $\vec{u} 	imes \vec{v} = (2\vec{a} + 3\vec{b}) 	imes (\vec{a} - \vec{b}) = 2(\vec{a} 	imes \vec{a}) - 2(\vec{a} 	imes \vec{b}) + 3(\vec{b} 	imes \vec{a}) - 3(\vec{b} 	imes \vec{b})$.Since $\vec{a} 	imes \vec{a} = 0$ and $\vec{b} 	imes \vec{b} = 0$, and $\vec{b} 	imes \vec{a} = -(\vec{a} 	imes \vec{b})$, we get:$\vec{u} 	imes \vec{v} = 0 - 2(\vec{a} 	imes \vec{b}) - 3(\vec{a} 	imes \vec{b}) - 0 = -5(\vec{a} 	imes \vec{b})$.Now, $\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 3 \ 6 & 3 & 3 \end{vmatrix} = \hat{i}(9-9) - \hat{j}(6-18) + \hat{k}(6-18) = 0\hat{i} + 12\hat{j} - 12\hat{k}$.Magnitude $|\vec{a} 	imes \vec{b}| = \sqrt{0^2 + 12^2 + (-12)^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2}$.Thus, $A = \frac{1}{2} |-5(\vec{a} 	imes \vec{b})| = \frac{5}{2} |\vec{a} 	imes \vec{b}| = \frac{5}{2} 	imes 12\sqrt{2} = 30\sqrt{2}$.The square of the area is $A^2 = (30\sqrt{2})^2 = 900 	imes 2 = 1800$.

Let $\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}$ and $\vec{b} = 6\hat{i} + 3\hat{j} + 3\hat{k}$. Then the square of the area of the triangle with adjacent sides determined by the vectors $(2\vec{a} + 3\vec{b})$ and $(\vec{a} - \vec{b})$ is :

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