Question

\( (\vec{a}+2\vec{b}-\vec{c}) \cdot ((\vec{a}-\vec{b}) \times (\vec{a}-\vec{b}-\vec{c})) = \)

• A. \( [\vec{a}\vec{b}\vec{c}] \)
• B. \( 3[\vec{a}\vec{b}\vec{c}] \)
• C. \( [\vec{a}\vec{b}\vec{c}]^2 \)
• D. \( 2[\vec{a}\vec{b}\vec{c}] \) Correct Answer

Hint:

Scalar triple product properties: \( [\vec{X}\vec{Y}\vec{Z}] = \vec{X} \cdot (\vec{Y} \times \vec{Z}) \). If any two vectors in a scalar triple product are identical, the product is 0 (e.g., \( [\vec{X}\vec{X}\vec{Y}]=0 \)). Cyclic permutation maintains the value: \( [\vec{X}\vec{Y}\vec{Z}] = [\vec{Y}\vec{Z}\vec{X}] = [\vec{Z}\vec{X}\vec{Y}] \). Anti-cyclic permutation negates the value: \( [\vec{X}\vec{Z}\vec{Y}] = -[\vec{X}\vec{Y}\vec{Z}] \). Cross product property: \( \vec{U} \times (\vec{V}-\vec{W}) = (\vec{U} \times \vec{V}) - (\vec{U} \times \vec{W}) \). And \( \vec{U} \times \vec{U} = \vec{0} \).

Answer 1

The Correct Option is B

Step 1: Expand the cross product term.
Let \( \vec{u} = \vec{a}-\vec{b} \) and \( \vec{v} = \vec{a}-\vec{b}-\vec{c} \).
We need to calculate \( \vec{u} \times \vec{v} \).
Note that \( \vec{v} = \vec{u} - \vec{c} \).
So, \( (\vec{a}-\vec{b}) \times (\vec{a}-\vec{b}-\vec{c}) = \vec{u} \times (\vec{u}-\vec{c}) = (\vec{u} \times \vec{u}) - (\vec{u} \times \vec{c}) \).
Since \( \vec{u} \times \vec{u} = \vec{0} \), this simplifies to \( -(\vec{u} \times \vec{c}) = \vec{c} \times \vec{u} \).
So, \( (\vec{a}-\vec{b}) \times (\vec{a}-\vec{b}-\vec{c}) = \vec{c} \times (\vec{a}-\vec{b}) = (\vec{c} \times \vec{a}) - (\vec{c} \times \vec{b}) \).

Step 2: Calculate the scalar triple product.
The expression is \( (\vec{a}+2\vec{b}-\vec{c}) \cdot [(\vec{c} \times \vec{a}) - (\vec{c} \times \vec{b})] \).
This is \( (\vec{a}+2\vec{b}-\vec{c}) \cdot (\vec{c} \times \vec{a}) - (\vec{a}+2\vec{b}-\vec{c}) \cdot (\vec{c} \times \vec{b}) \).
Term 1: \( (\vec{a}+2\vec{b}-\vec{c}) \cdot (\vec{c} \times \vec{a}) \) Using properties of scalar triple product: \( \vec{X} \cdot (\vec{Y} \times \vec{Z}) = [\vec{X}\vec{Y}\vec{Z}] \).
So, \( \vec{a} \cdot (\vec{c} \times \vec{a}) + 2\vec{b} \cdot (\vec{c} \times \vec{a}) - \vec{c} \cdot (\vec{c} \times \vec{a}) \).
\( [\vec{a}\vec{c}\vec{a}] = 0 \) because two vectors are identical.
\( [\vec{c}\vec{c}\vec{a}] = 0 \) because two vectors are identical.
So Term 1 is \( 2\vec{b} \cdot (\vec{c} \times \vec{a}) = 2[\vec{b}\vec{c}\vec{a}] \).
We know \( [\vec{b}\vec{c}\vec{a}] = [\vec{a}\vec{b}\vec{c}] \) (cyclic permutation).
So Term 1 = \( 2[\vec{a}\vec{b}\vec{c}] \).
Term 2: \( (\vec{a}+2\vec{b}-\vec{c}) \cdot (\vec{c} \times \vec{b}) \) This is \( \vec{a} \cdot (\vec{c} \times \vec{b}) + 2\vec{b} \cdot (\vec{c} \times \vec{b}) - \vec{c} \cdot (\vec{c} \times \vec{b}) \).
\( [\vec{b}\vec{c}\vec{b}] = 0 \).
\( [\vec{c}\vec{c}\vec{b}] = 0 \).
So Term 2 is \( \vec{a} \cdot (\vec{c} \times \vec{b}) = [\vec{a}\vec{c}\vec{b}] \).
We know \( [\vec{a}\vec{c}\vec{b}] = -[\vec{a}\vec{b}\vec{c}] \) (anti-cyclic permutation).
So Term 2 = \( -[\vec{a}\vec{b}\vec{c}] \).

Step 3: Combine Term 1 and Term 2.
The original expression is Term 1 - Term 2.
\( 2[\vec{a}\vec{b}\vec{c}] - (-[\vec{a}\vec{b}\vec{c}]) = 2[\vec{a}\vec{b}\vec{c}] + [\vec{a}\vec{b}\vec{c}] = 3[\vec{a}\vec{b}\vec{c}] \).
This matches option (2).