Question:
If the plane \(2x+3y+z = 6\) cuts coordinate axes at A, B and C, then the volume of tetrahedron OABC (where O is the origin) is ......cubic units.
If the plane \(2x+3y+z = 6\) cuts coordinate axes at A, B and C, then the volume of tetrahedron OABC (where O is the origin) is ......cubic units.
Show Hint
Find the intercepts and use volume = one sixth of the product of the intercepts.
Updated On: Oct 1, 2026
- \(30\)
- \(6\)
- \(36\)
- \(180\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
A plane meets the axes at \(A(a,0,0)\), \(B(0,b,0)\), \(C(0,0,c)\). The tetrahedron \(OABC\) has volume \(\frac16|abc|\).
Step 2: Key Formula or Approach:
Write \(2x + 3y + z = 6\) as \(\frac{x}{3} + \frac{y}{2} + \frac{z}{6} = 1\).
Step 3: Detailed Explanation:
Intercepts: \(a = 3\), \(b = 2\), \(c = 6\).
\[ V = \frac16\times 3\times 2\times 6 = 6\ \text{cubic units} \]
The value \(36\) is the product \(abc\), so it forgets the factor \(\frac16\). The value \(30\) and \(180\) come from wrong intercepts.
Final Answer:
The volume is \(6\) cubic units, option (B). \[ \boxed{6} \]
A plane meets the axes at \(A(a,0,0)\), \(B(0,b,0)\), \(C(0,0,c)\). The tetrahedron \(OABC\) has volume \(\frac16|abc|\).
Step 2: Key Formula or Approach:
Write \(2x + 3y + z = 6\) as \(\frac{x}{3} + \frac{y}{2} + \frac{z}{6} = 1\).
Step 3: Detailed Explanation:
Intercepts: \(a = 3\), \(b = 2\), \(c = 6\).
\[ V = \frac16\times 3\times 2\times 6 = 6\ \text{cubic units} \]
The value \(36\) is the product \(abc\), so it forgets the factor \(\frac16\). The value \(30\) and \(180\) come from wrong intercepts.
Final Answer:
The volume is \(6\) cubic units, option (B). \[ \boxed{6} \]
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