The plane frac{x}{2} + frac{y}{3} + frac{z}{4 | vedclass

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(C) The plane equation is $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$. Setting $y=0, z=0$ gives $x=2$, so $A = (2, 0, 0)$. Setting $x=0, z=0$ gives

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$\sqrt{61}$

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(C) The plane equation is $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$. Setting $y=0, z=0$ gives $x=2$, so $A = (2, 0, 0)$. Setting $x=0, z=0$ gives $y=3$, so $B = (0, 3, 0)$. Setting $x=0, y=0$ gives $z=4$, so $C = (0, 0, 4)$. The vectors $\vec{AB} = B - A = (-2, 3, 0)$ and $\vec{AC} = C - A = (-2, 0, 4)$. The cross product $\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 3 & 0 \ -2 & 0 & 4 \end{vmatrix} = \hat{i}(12 - 0) - \hat{j}(-8 - 0) + \hat{k}(0 - (-6)) = 12\hat{i} + 8\hat{j} + 6\hat{k}$. The area of triangle $ABC$ is $\frac{1}{2} |\vec{AB} 	imes \vec{AC}| = \frac{1}{2} \sqrt{12^2 + 8^2 + 6^2} = \frac{1}{2} \sqrt{144 + 64 + 36} = \frac{1}{2} \sqrt{244} = \frac{1}{2} \sqrt{4 	imes 61} = \sqrt{61}$.

The plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$ cuts the axes at the points $A, B, C$. Find the area of triangle $ABC$.

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