If the plane frac{x}{3}+frac{y}{2}-frac{z}{4} | vedclass

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(C) The equation of the plane is $\frac{x}{3}+\frac{y}{2}-\frac{z}{4}=1$. To find the points where the plane cuts the coordinate axes,we set the other

Vedclass · Accepted Answer

$\sqrt{61}$ sq. units

Vedclass · Answer

(C) The equation of the plane is $\frac{x}{3}+\frac{y}{2}-\frac{z}{4}=1$. To find the points where the plane cuts the coordinate axes,we set the other two variables to zero. For the $x$-axis,set $y=0$ and $z=0$: $\frac{x}{3}=1 \implies x=3$. So,$A = (3, 0, 0)$. For the $y$-axis,set $x=0$ and $z=0$: $\frac{y}{2}=1 \implies y=2$. So,$B = (0, 2, 0)$. For the $z$-axis,set $x=0$ and $y=0$: $-\frac{z}{4}=1 \implies z=-4$. So,$C = (0, 0, -4)$. The area of a triangle with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $\frac{1}{2} |\vec{AB} 	imes \vec{AC}|$. $\vec{AB} = B - A = (-3, 2, 0)$ and $\vec{AC} = C - A = (-3, 0, -4)$. $\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & 2 & 0 \ -3 & 0 & -4 \end{vmatrix} = \hat{i}(-8 - 0) - \hat{j}(12 - 0) + \hat{k}(0 - (-6)) = -8\hat{i} - 12\hat{j} + 6\hat{k}$. The magnitude is $\sqrt{(-8)^2 + (-12)^2 + 6^2} = \sqrt{64 + 144 + 36} = \sqrt{244} = \sqrt{4 	imes 61} = 2\sqrt{61}$. Area $= \frac{1}{2} 	imes 2\sqrt{61} = \sqrt{61}$ sq. units.

If the plane $\frac{x}{3}+\frac{y}{2}-\frac{z}{4}=1$ cuts the coordinate axes at points $A, B$,and $C$,then the area of the triangle $ABC$ is

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