The area of the triangle formed by plane $2 x + 3 y + 6 z + 9 = 0$ with y-axis, z-axis is?(in Sq. units)

9

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\frac{9}{2}

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\frac{9}{4}

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None

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Solution

The correct option is C $\frac{9}{4}$
The equation of plane is $2 x + 3 y + 6 z = - 9$
$\Rightarrow \frac{x}{\frac{- 9}{2}} + \frac{y}{- 3} + \frac{z}{\frac{- 3}{2}} = 1$
So, the x,y,z intercepts are $\frac{- 9}{2}, - 3, \frac{- 3}{2}$
respectively.
Area of the triangle formed with y - axis, z - axis
$= \frac{1}{2} \times (y - i n t e r c e p t) \times (z - i n t e r c e p t)$
$= \frac{1}{2} \times - 3 \times \frac{3}{2} = \frac{9}{4} s q . u n i t s$

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