Question

The volume of the tetrahedron bounded by the planes \(x=1\), \(y=2\), \(z=3\) and \(12x+8y+6z=70\) is rounded off to one decimal place.

Hint:

For a tetrahedron cut from coordinate planes by \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\), the volume is \(\frac{abc}{6}\).

Answer 1

Correct Answer: 4

Step 1: Move the corner to the origin.
Let $X=x-1$, $Y=y-2$, $Z=z-3$. Then the planes $x=1,y=2,z=3$ become $X=0,Y=0,Z=0$.

Step 2: Shift the slanted plane.
Substituting into $12x+8y+6z=70$ gives $12X+8Y+6Z+46=70$, that is $12X+8Y+6Z=24$.

Step 3: Intercept form.
\[ \frac{X}{2}+\frac{Y}{3}+\frac{Z}{4}=1 \]
So the intercepts are $2,3,4$.

Step 4: Volume formula.
A tetrahedron cut off near the origin with intercepts $a,b,c$ has volume $\frac16 abc$.

Step 5: Plug in.
\[ V=\frac16\cdot 2\cdot 3\cdot 4=4 \]
\[ \boxed{4.0} \]