Question:
If three coterminous edges of a parallelopiped are represented by \(\vec{a}-\vec{b},\vec{b}-\vec{c}\) and \(\vec{c}-\vec{a}\) then its volume is
If three coterminous edges of a parallelopiped are represented by \(\vec{a}-\vec{b},\vec{b}-\vec{c}\) and \(\vec{c}-\vec{a}\) then its volume is
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The three edge vectors add up to zero, so they are coplanar and their scalar triple product vanishes.
Updated On: Nov 12, 2026
- \([\vec{a}\ \vec{b}\ \vec{c}]\)
- \([\vec{a}\ \vec{b}\ \vec{c}]^{2}\)
- \(2[\vec{a}\ \vec{b}\ \vec{c}]\)
- \(0\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
The volume is the scalar triple product \([\vec{a}-\vec{b}\ \ \vec{b}-\vec{c}\ \ \vec{c}-\vec{a}]\).
Step 2: Look for dependence
Adding the three vectors gives \((\vec{a}-\vec{b})+(\vec{b}-\vec{c})+(\vec{c}-\vec{a})=\vec{0}\).
Step 3: Conclude
So the vectors are linearly dependent, hence coplanar, and the triple product is 0.
Final Answer:
The volume is 0. \[ \boxed{\text{(D) }0} \]
The volume is the scalar triple product \([\vec{a}-\vec{b}\ \ \vec{b}-\vec{c}\ \ \vec{c}-\vec{a}]\).
Step 2: Look for dependence
Adding the three vectors gives \((\vec{a}-\vec{b})+(\vec{b}-\vec{c})+(\vec{c}-\vec{a})=\vec{0}\).
Step 3: Conclude
So the vectors are linearly dependent, hence coplanar, and the triple product is 0.
Final Answer:
The volume is 0. \[ \boxed{\text{(D) }0} \]
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