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

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(A) Step $1$: Identify the coordinates of the points $A, B, C$ where the plane intersects the axes.For $x$-axis, $y=0, z=0 \implies \frac{x}{2} = 1 \i

Vedclass · Accepted Answer

$\sqrt{61}$ sq. units

Vedclass · Answer

(A) Step $1$: Identify the coordinates of the points $A, B, C$ where the plane intersects the axes.For $x$-axis, $y=0, z=0 \implies \frac{x}{2} = 1 \implies x=2$. So, $A = (2, 0, 0)$.For $y$-axis, $x=0, z=0 \implies \frac{y}{3} = 1 \implies y=3$. So, $B = (0, 3, 0)$.For $z$-axis, $x=0, y=0 \implies \frac{z}{4} = 1 \implies z=4$. So, $C = (0, 0, 4)$.Step $2$: Find vectors $\vec{AB}$ and $\vec{AC}$.$\vec{AB} = B - A = (0-2, 3-0, 0-0) = (-2, 3, 0)$.$\vec{AC} = C - A = (0-2, 0-0, 4-0) = (-2, 0, 4)$.Step $3$: Calculate the cross product $\vec{AB} 	imes \vec{AC}$.$\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 3 & 0 \ -2 & 0 & 4 \end{vmatrix} = \hat{i}(12-0) - \hat{j}(-8-0) + \hat{k}(0 - (-6)) = 12\hat{i} + 8\hat{j} + 6\hat{k}$.Step $4$: Calculate the area of triangle $ABC = \frac{1}{2} |\vec{AB} 	imes \vec{AC}|$.Area $= \frac{1}{2} \sqrt{12^2 + 8^2 + 6^2} = \frac{1}{2} \sqrt{144 + 64 + 36} = \frac{1}{2} \sqrt{244} = \frac{1}{2} \sqrt{4 	imes 61} = \frac{2\sqrt{61}}{2} = \sqrt{61}$ sq. units.

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

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