The area of the triangle with vertices (1,2,0 | vedclass

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Question

(B) Let the vertices be $A(1,2,0)$,$B(1,0,2)$,and $C(0,3,1)$.The vectors representing the sides are $\vec{AB} = (1-1)\hat{i} + (0-2)\hat{j} + (2-0)\ha

Vedclass · Accepted Answer

$\sqrt{6}$ sq. units

Vedclass · Answer

(B) Let the vertices be $A(1,2,0)$,$B(1,0,2)$,and $C(0,3,1)$.The vectors representing the sides are $\vec{AB} = (1-1)\hat{i} + (0-2)\hat{j} + (2-0)\hat{k} = -2\hat{j} + 2\hat{k}$ and $\vec{AC} = (0-1)\hat{i} + (3-2)\hat{j} + (1-0)\hat{k} = -\hat{i} + \hat{j} + \hat{k}$.The area of $	riangle ABC$ is given by $\frac{1}{2} |\vec{AB} 	imes \vec{AC}|$.Calculating the cross product: $\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & -2 & 2 \ -1 & 1 & 1 \end{vmatrix} = \hat{i}(-2-2) - \hat{j}(0 - (-2)) + \hat{k}(0-2) = -4\hat{i} - 2\hat{j} - 2\hat{k}$.The magnitude is $|\vec{AB} 	imes \vec{AC}| = \sqrt{(-4)^2 + (-2)^2 + (-2)^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6}$.Therefore,the area is $\frac{1}{2} 	imes 2\sqrt{6} = \sqrt{6}$ sq. units.

The area of the triangle with vertices $(1,2,0)$,$(1,0,2)$ and $(0,3,1)$ is.

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