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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  3-D Geometry

If the plane $\large\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ cuts the coordinate axis in $A,B,C$ respectively, then the area of the $\Delta\:ABC$ is ?

$\begin{array}{1 1} \sqrt {21} \\ \sqrt {41} \\ \sqrt {61} \\ None\;of\;the\;above \end{array}$

1 Answer

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  • Area of $\Delta\:ABC =\large\frac{1}{2}$$|\overrightarrow {AB}\times\overrightarrow {AC}|$
Given equation of the plane is $\large\frac{x}{2}+\frac{y}{3}+\frac{z}{4}$$=1$
This plane cuts the coordinate axis at the points $A(2,0,0)\:\:B(0,3,0)\:\:and\:\:C(0,0,4)$
$\Rightarrow\:\overrightarrow {AB}=(-2,3,0)\:\:\:and\:\:\:\overrightarrow {AC}=(-2,0,4)$
$\Rightarrow\:\overrightarrow {AB}\times\overrightarrow {AC}=\left |\begin {array}{ccc} \hat i & \hat j & \hat k \\ -2 & 3 & 0\\ -2 & 0 & 4 \end {array} \right|=(12,8,6)$
$\Rightarrow\:|\overrightarrow {AB}\times\overrightarrow {AC}|=\sqrt {144+64+36}=\sqrt {244}$
$\therefore$ Area of the $\Delta\:ABC=\large\frac{\sqrt {244}}{2}=$$\sqrt {61}$
answered Jan 3, 2014 by rvidyagovindarajan_1
edited Sep 29, 2014 by sharmaaparna1
 

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