The area of the triangle formed by the vector | vedclass

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(D) The area of a triangle formed by two vectors $\vec{a}$ and $\vec{b}$ as adjacent sides is given by the formula: $	ext{Area} = \frac{1}{2} |\vec{a

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$\frac{\sqrt{14}}{2}$ sq. units

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(D) The area of a triangle formed by two vectors $\vec{a}$ and $\vec{b}$ as adjacent sides is given by the formula: $	ext{Area} = \frac{1}{2} |\vec{a} 	imes \vec{b}|$.First,calculate the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -1 \ 1 & 1 & 1 \end{vmatrix}$$= \hat{i}(1 - (-1)) - \hat{j}(2 - (-1)) + \hat{k}(2 - 1)$$= \hat{i}(2) - \hat{j}(3) + \hat{k}(1) = 2\hat{i} - 3\hat{j} + \hat{k}$.Next,calculate the magnitude of the cross product:$|\vec{a} 	imes \vec{b}| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.Finally,the area of the triangle is:$	ext{Area} = \frac{1}{2} 	imes \sqrt{14} = \frac{\sqrt{14}}{2}$ sq. units.

The area of the triangle formed by the vectors $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ is:

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