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

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(C) The equation of the plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b,$ and $c$ are the intercepts on the $x, y,$ and $z

Vedclass · Accepted Answer

$\sqrt{61}$

Vedclass · Answer

(C) The equation of the plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b,$ and $c$ are the intercepts on the $x, y,$ and $z$ axes respectively.Comparing this with the given equation $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$,we get $a = 2, b = 3,$ and $c = 4$.The coordinates of the points where the plane cuts the axes are $A(2, 0, 0), B(0, 3, 0),$ and $C(0, 0, 4)$.The area of a triangle with vertices $(a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ is given by the formula $	ext{Area} = \frac{1}{2} \sqrt{a^2b^2 + b^2c^2 + c^2a^2}$.Substituting the values $a=2, b=3, c=4$:$	ext{Area} = \frac{1}{2} \sqrt{(2^2 	imes 3^2) + (3^2 	imes 4^2) + (4^2 	imes 2^2)}$$	ext{Area} = \frac{1}{2} \sqrt{(4 	imes 9) + (9 	imes 16) + (16 	imes 4)}$$	ext{Area} = \frac{1}{2} \sqrt{36 + 144 + 64}$$	ext{Area} = \frac{1}{2} \sqrt{244}$$	ext{Area} = \frac{1}{2} \sqrt{4 	imes 61} = \frac{1}{2} 	imes 2 \sqrt{61} = \sqrt{61}$.

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

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