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The volume of the tetrahedron (in cubic units) formed by the plane $2 x+y+z=K$ and the coordinate planes is $\frac{2 V^3}{3}$, then $K: V=$
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Verified Answer
The correct answer is:
$2: 1$
Equation of given plane is
$2 x+y+z=K$
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 $O A B C$
$\begin{aligned} & =\frac{1}{6}[\mathbf{O A} \mathbf{O B} \mathbf{O C}]=\frac{1}{6}\left|\begin{array}{ccc}\frac{K}{2} & 0 & 0 \\ 0 & K & 0 \\ 0 & 0 & K\end{array}\right|=\frac{2 V^3}{3} \text { (given) } \\ & \Rightarrow \frac{1}{6} \frac{K^3}{2}=\frac{2 V^3}{3} \\ & \Rightarrow\left(\frac{K}{V}\right)^3=2^3 \Rightarrow K: V=2: 1\end{aligned}$
Hence, option (d) is correct.
$2 x+y+z=K$
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 $O A B C$
$\begin{aligned} & =\frac{1}{6}[\mathbf{O A} \mathbf{O B} \mathbf{O C}]=\frac{1}{6}\left|\begin{array}{ccc}\frac{K}{2} & 0 & 0 \\ 0 & K & 0 \\ 0 & 0 & K\end{array}\right|=\frac{2 V^3}{3} \text { (given) } \\ & \Rightarrow \frac{1}{6} \frac{K^3}{2}=\frac{2 V^3}{3} \\ & \Rightarrow\left(\frac{K}{V}\right)^3=2^3 \Rightarrow K: V=2: 1\end{aligned}$
Hence, option (d) is correct.
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