The volume of the tetrahedron (in cubic units | vedclass

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(D) The equation of the given plane is $2x + y + z = K$. Dividing by $K$, we get the intercept form: $\frac{x}{K/2} + \frac{y}{K} + \frac{z}{K} = 1$.

Vedclass · Accepted Answer

$2:1$

Vedclass · Answer

(D) The equation of the given plane is $2x + y + z = K$. Dividing by $K$, we get the intercept form: $\frac{x}{K/2} + \frac{y}{K} + \frac{z}{K} = 1$. The vertices of the tetrahedron formed by the plane and the coordinate planes are $O(0, 0, 0)$, $A(K/2, 0, 0)$, $B(0, K, 0)$, and $C(0, 0, K)$. The volume of the tetrahedron $OABC$ is given by the formula $V_{tet} = \frac{1}{6} |x_A y_B z_C| = \frac{1}{6} 	imes \frac{K}{2} 	imes K 	imes K = \frac{K^3}{12}$. Given that the volume is $\frac{2V^3}{3}$, we equate the two expressions: $\frac{K^3}{12} = \frac{2V^3}{3}$. Multiplying both sides by $12$, we get $K^3 = 8V^3$. Taking the cube root on both sides, we get $K = 2V$, which implies $\frac{K}{V} = \frac{2}{1}$. Thus, $K:V = 2:1$. Therefore, option $D$ is correct.

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 $K:V =$

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