Question

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

• A. $18$ sq. units
• B. $6$ sq. units
• C. $3\sqrt{14}$ sq. units
• D. $4\sqrt{14}$ sq. units

Answer 1

Answer: (B)

$6$ sq. units

Plane cut x-axis, when $y,z=0$ $\Rightarrow x=2$

$\Rightarrow A=(2,0,0)$

Similarly plane cuts y-axis at $y=3$

$\Rightarrow B=(0,3,0)$

Similarly plane cuts z-axis at $z=6$

$\Rightarrow C=(0,0,6)$

$\Rightarrow area=|\frac{A_x(B_y-C_y)+B_x(C_y-A_y)+C_x(A_y-B_y)}{2}|$

$\Rightarrow area=|\frac{2(3-6)+0+0}{2}|$

$\Rightarrow area=6$