The plane frac{x}{2}+frac{y}{3}+frac{z}{4}=1 | vedclass

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(D) The equation of the plane is $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$.The intercepts on the axes are $A(2, 0, 0)$,$B(0, 3, 0)$,and $C(0, 0, 4)$.The

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$\sqrt{61}$ sq. units

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(D) The equation of the plane is $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$.The intercepts on the axes are $A(2, 0, 0)$,$B(0, 3, 0)$,and $C(0, 0, 4)$.The vectors forming the sides of the triangle are $\vec{AB} = B - A = -2\hat{i} + 3\hat{j}$ and $\vec{AC} = C - A = -2\hat{i} + 4\hat{k}$.The cross product $\vec{AB} 	imes \vec{AC}$ is given by:$\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 3 & 0 \ -2 & 0 & 4 \end{vmatrix} = \hat{i}(12 - 0) - \hat{j}(-8 - 0) + \hat{k}(0 - (-6)) = 12\hat{i} + 8\hat{j} + 6\hat{k}$.The area of $	riangle ABC = \frac{1}{2} |\vec{AB} 	imes \vec{AC}|$.Area $= \frac{1}{2} \sqrt{12^2 + 8^2 + 6^2} = \frac{1}{2} \sqrt{144 + 64 + 36} = \frac{1}{2} \sqrt{244} = \frac{1}{2} \sqrt{4 	imes 61} = \sqrt{61}$ sq. units.

The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ cuts the $X$-axis at $A$,$Y$-axis at $B$,and $Z$-axis at $C$. The area of $\triangle ABC$ is:

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