Question

The volume of the tetrahedron (in cubic units) formed by the plane $2x+y+z=K$ and the coordinate planes is $\frac{2V^3}{3}$, then $K:V=$

• A. 1 : 2
• B. 1 : 6
• C. 4 : 3
• D. 2 : 1

Answer 1

Answer: (D)

Equation of given plane is
$2x+y+z=K$
Point of intersection of plane (i) with the coordinate axes is $A\left(\frac{K}{2},0,0\right),B(0,K,0)$ and $C(0,0,K)$.
Now, volume of the tetrahedron $OABC$
$=\frac{1}{6}[OAOBOC]=\frac{1}{6}\left|\begin{array}{ccc}\frac{K}{2}& 0& 0\\ 0& K& 0\\ 0& 0& K\end{array}\right|=\frac{2V^3}{3}($ given $)$
$\Rightarrow \frac{1}{6}\frac{K^3}{2}=\frac{2V^3}{3}$
$\Rightarrow {\left(\frac{K}{V}\right)}^3=2^3$
$\Rightarrow K:V=2:1$