First slide

Planes in 3D

Question

If the areas of triangles formed by a plane with the positive $X, Y : Y, Z : Z, X$ axes respectively are 12, 9, 6 sq. unit respectively then the equation of the plane is

Moderate

A

$\frac{x}{4} + \frac{y}{6} + \frac{z}{3} = 1$
B

$\frac{x}{6} + \frac{y}{3} + \frac{z}{4} = 1$
C

$\frac{x}{3} + \frac{y}{4} + \frac{z}{6} = 1$
D

$\frac{x}{3} + \frac{y}{6} + \frac{z}{4} = 1$

Solution

If $a, b, c$ are the intercepts of the required plane then $\frac{1}{2} a b = 12, \frac{1}{2} b c = 9, \frac{1}{2} c a = 6$
$\Rightarrow a b = 24, b c = 18, c a = 12$
$a^{2} = \frac{a b a c}{b c} = \frac{24 \times 12}{18} = 16 \Rightarrow a = 4$
$a b = 24, a c = 12 \Rightarrow b = 6, c = 3$
The equation of the plane is $\frac{x}{4} + \frac{y}{6} + \frac{z}{3} = 1$

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