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

Question

Dr. Abubacker · Accepted Answer

(c)
Plane meets axes at $A(2,\ 0,\ 0),\ B(0,\ 3,\ 0)$ and $C(0,\ 0,\ 6)$
Then area of $\mathrm{\Delta}ABC$ is
$\frac{1}{2}\left| \overrightarrow{AB} 	imes \overrightarrow{AC} ight|$
$= \frac{1}{2}\left| \left( - 2\widehat{i} + 3\widehat{j} ight) 	imes \left( - 2\widehat{i} + 6\widehat{j} ight) ight|$
$= 3\sqrt{14}\ \ \$sq units

Q. If the plane $\frac{x}{2} + \frac{y}{3} + \frac{\mathcal{z}}{6} = 1$ cuts the axes of coordinates at points $A,\ B$ and $C$, then find the area of then triangle $ABC$

Explanation

Reviewed by Dr. Abubacker

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