Consider P(1, 2, -3),Q(-2, 1, -4),and R(3, 4, | vedclass

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(D) The vector $\vec{B}$ is defined as $\vec{B} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$,where $A_x, A_y, A_z$ are the areas of the projections of $

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$\frac{9}{2}$

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(D) The vector $\vec{B}$ is defined as $\vec{B} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$,where $A_x, A_y, A_z$ are the areas of the projections of $\Delta PQR$ on the $yz, zx, xy$ planes respectively. By the property of vector areas,$\vec{B} = \frac{1}{2} (\vec{PQ} 	imes \vec{PR})$. Given $P(1, 2, -3)$,$Q(-2, 1, -4)$,and $R(3, 4, -2)$,we have: $\vec{PQ} = (-2-1)\hat{i} + (1-2)\hat{j} + (-4 - (-3))\hat{k} = -3\hat{i} - \hat{j} - \hat{k}$. $\vec{PR} = (3-1)\hat{i} + (4-2)\hat{j} + (-2 - (-3))\hat{k} = 2\hat{i} + 2\hat{j} + \hat{k}$. Now,$\vec{PQ} 	imes \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & -1 & -1 \ 2 & 2 & 1 \end{vmatrix} = \hat{i}(-1 - (-2)) - \hat{j}(-3 - (-2)) + \hat{k}(-6 - (-2)) = \hat{i} + \hat{j} - 4\hat{k}$. Thus,$\vec{B} = \frac{1}{2} (\hat{i} + \hat{j} - 4\hat{k})$. Then,$|\vec{B}|^2 = \left( \frac{1}{2} ight)^2 (1^2 + 1^2 + (-4)^2) = \frac{1}{4} (1 + 1 + 16) = \frac{18}{4} = \frac{9}{2}$.

Consider $P(1, 2, -3)$,$Q(-2, 1, -4)$,and $R(3, 4, -2)$. Let $\vec{B} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$,where $A_x, A_y$,and $A_z$ are the projections of the area of triangle $PQR$ on the $yz, zx$,and $xy$ planes,respectively. Then,the value of $|\vec{B}|^2$ is:

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