Find the volume of the indicated region.
-$$\text { the tetrahedron cut off from the first octant by the plane } \frac { x } { 8 } + \frac { y } { 3 } + \frac { z } { 4 } = 1$$
A) 32
B) 24
C) 16
D) 48

Question

Quizplus · Accepted Answer

The tetrahedron is cut off from the first octant, which means that all of its vertices must be in the first octant. The plane intersects the $x$-axis at $(8,0,0)$, the $y$-axis at $(0,3,0)$, and the $z$-axis at $(0,0,4)$. These are the vertices of the tetrahedron.

To find the volume of the tetrahedron, we can use the formula $\frac{1}{3}Bh$, where $B$ is the area of the base and $h$ is the height of the tetrahedron. The base is a triangle with sides of length 8, 3, and 5 (which we can find using the Pythagorean theorem). The area of this triangle is $\frac{1}{2}(8)(3) = 12$.

To find the height of the tetrahedron, we note that the distance from the plane to the origin is $\frac{1}{\sqrt{\left(\frac{1}{8}ight)^2 + \left(\frac{1}{3}ight)^2 + \left(\frac{1}{4}ight)^2}} = \frac{24}{\sqrt{329}}$. Therefore, the height of the tetrahedron is this distance times $\frac{1}{4}$ (since the distance is the height of one of the four congruent triangles that make up the tetrahedron). So $h=\frac{6}{\sqrt{329}}$.

Putting this all together, we get that the volume of the tetrahedron is $\frac{1}{3}(12)\left(\frac{6}{\sqrt{329}}ight) = \boxed{	extbf{(C) }16}$.