The plane x2- y3- z4-1 cuts the axes in A-B-C then the area of the ABC in sq-units is

Given : x2+ y3+ z4=1 So, intercept on x,y,z axes are a=2, b=3, c=4 respectively. ⇒ Area of triangle =√(_x^2+_y^2+_z^2) =√((ab)^24+(bc)^24+(ca)^24) =12√(36+144 ...

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Byju's Answer

Standard XII

Mathematics

Auxiliary Circle of Ellipse

The plane x2+...

Question

The plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$ cuts the axes in $A, B, C$ then the area of the $△ A B C$ (in $sq.units$ ) is

\sqrt{59}

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\sqrt{61}

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\sqrt{63}

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\sqrt{65}

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Solution

The correct option is B $\sqrt{61}$
Given :
$\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$
So, intercept on $x, y, z$ axes are $a = 2, b = 3, c = 4$ respectively.
$\Rightarrow$ Area of triangle $= \sqrt{△_{x}^{2} + △_{y}^{2} + △_{z}^{2}} = \sqrt{\frac{(a b)^{2}}{4} + \frac{(b c)^{2}}{4} + \frac{(c a)^{2}}{4}} = \frac{1}{2} \sqrt{36 + 144 + 64} = \sqrt{61}$

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The plane x2+y3+z4=1 cuts the axes in A,B,C then the area of the △ABC (in sq.units) is

The correct option is B √61Given : x2+y3+z4=1 So, intercept on x,y,z axes are a=2,b=3,c=4 respectively. ⇒ Area of triangle =√△2x+△2y+△2z=√(ab)24+(bc)24+(ca)24=12√36+144+64=√61

The plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$ cuts the axes in $A, B, C$ then the area of the $△ A B C$ (in $sq.units$ ) is