If the plane frac{x}{2}+frac{y}{3}+frac{z}{6} | vedclass

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Question

(B) The equation of the plane is $\frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1$. The plane cuts the coordinate axes at points $A, B, C$. Setting $y=0,

Vedclass · Accepted Answer

$3 \sqrt{14}$ sq. units

Vedclass · Answer

(B) The equation of the plane is $\frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1$. The plane cuts the coordinate axes at points $A, B, C$. Setting $y=0, z=0$,we get $x=2$,so $A = (2, 0, 0)$. Setting $x=0, z=0$,we get $y=3$,so $B = (0, 3, 0)$. Setting $x=0, y=0$,we get $z=6$,so $C = (0, 0, 6)$. The vectors representing the sides are $\vec{AB} = B - A = (-2, 3, 0)$ and $\vec{AC} = C - A = (-2, 0, 6)$. The area of triangle $ABC$ is given by $\frac{1}{2} |\vec{AB} 	imes \vec{AC}|$. Calculating the cross product: $\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 3 & 0 \ -2 & 0 & 6 \end{vmatrix} = \hat{i}(18 - 0) - \hat{j}(-12 - 0) + \hat{k}(0 - (-6)) = 18\hat{i} + 12\hat{j} + 6\hat{k}$. The magnitude is $\sqrt{18^2 + 12^2 + 6^2} = \sqrt{324 + 144 + 36} = \sqrt{504} = \sqrt{36 	imes 14} = 6\sqrt{14}$. Thus,the area is $\frac{1}{2} 	imes 6\sqrt{14} = 3\sqrt{14}$ sq. units.

Point	Plane
$(-6, 0, 0)$	$2x - 3y + 6z - 2 = 0$

If the plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1$ cuts the coordinate axes at points $A, B, C$ respectively,then the area of the triangle $ABC$ is

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$(-6, 0, 0)$ $2x - 3y + 6z - 2 = 0$

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