Vector Algebra Problem Explained

The question asks us to evaluate the scalar triple product involving sums of vectors. We are given three non-coplanar vectors $\rm \vec{a}, \vec{b}, \vec{c}$. The expression to evaluate is:

$$ E = (\rm \vec{a}+\vec{b}+\vec{c}) \cdot[(\rm \vec{a}+\vec{b}) \times( \rm \vec{a}+\vec{c})] $$

The notation $\rm [\vec{a}\; \vec{b} \;\vec{c}]$ represents the scalar triple product $\rm \vec{a} \cdot (\vec{b} \times \vec{c})$. Since the vectors are non-coplanar, this value is non-zero.

Key Scalar Triple Product Properties

To solve this, we'll use the properties of the scalar triple product and vector operations:

• Cross Product Distributivity: The cross product distributes over vector addition.
• Zero Cross Product: The cross product of a vector with itself is the zero vector, i.e., $\rm \vec{u} \times \vec{u} = \vec{0}$.
• Anti-Commutativity: Swapping vectors in a cross product changes the sign, $\rm \vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$.
• Scalar Triple Product Definition: $\rm [\vec{u}\; \vec{v}\; \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$.
• Scalar Triple Product Properties:
  • If any two vectors are identical, the scalar triple product is zero: $\rm [\vec{u}\; \vec{u}\; \vec{w}] = 0$.
  • Swapping any two vectors changes the sign: $\rm [\vec{u}\; \vec{v}\; \vec{w}] = -[\vec{v}\; \vec{u}\; \vec{w}]$.
  • Cyclical permutation preserves the value: $\rm [\vec{u}\; \vec{v}\; \vec{w}] = [\vec{v}\; \vec{w}\; \vec{u}] = [\vec{w}\; \vec{u}\; \vec{v}]$.

Step-by-Step Calculation

Let's break down the calculation:

Step 1: Calculate the cross product term.

We need to find $(\rm \vec{a}+\vec{b}) \times (\rm \vec{a}+\vec{c})$. Using the distributive property:

$$ (\rm \vec{a}+\vec{b}) \times (\rm \vec{a}+\vec{c}) = (\rm \vec{a} \times \vec{a}) + (\rm \vec{a} \times \vec{c}) + (\vec{b} \times \rm \vec{a}) + (\vec{b} \times \vec{c}) $$

Using $\rm \vec{a} \times \vec{a} = \vec{0}$ and $\rm \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$:

$$ = \vec{0} + (\rm \vec{a} \times \vec{c}) - (\rm \vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) $$

$$ = \rm \vec{a} \times \vec{c} - \vec{a} \times \vec{b} + \vec{b} \times \vec{c} $$

Step 2: Compute the dot product.

Now, we take the dot product of $(\rm \vec{a}+\vec{b}+\vec{c})$ with the result from Step 1:

$$ E = (\rm \vec{a}+\vec{b}+\vec{c}) \cdot (\rm \vec{a} \times \vec{c} - \vec{a} \times \vec{b} + \vec{b} \times \vec{c}) $$

Distributing the dot product:

$$ E = \rm \vec{a} \cdot (\vec{a} \times \vec{c}) + \vec{a} \cdot (-\vec{a} \times \vec{b}) + \vec{a} \cdot (\vec{b} \times \vec{c}) $$

$$ \qquad + \rm \vec{b} \cdot (\vec{a} \times \vec{c}) + \vec{b} \cdot (-\vec{a} \times \vec{b}) + \vec{b} \cdot (\vec{b} \times \vec{c}) $$

$$ \qquad + \rm \vec{c} \cdot (\vec{a} \times \vec{c}) + \vec{c} \cdot (-\vec{a} \times \vec{b}) + \vec{c} \cdot (\vec{b} \times \vec{c}) $$

Step 3: Simplify using scalar triple product properties.

We apply the property that $\rm \vec{u} \cdot (\vec{v} \times \vec{w}) = 0$ if any two vectors are the same:

• $\rm \vec{a} \cdot (\vec{a} \times \vec{c}) = [\vec{a}\; \vec{a}\; \vec{c}] = 0$
• $\rm \vec{a} \cdot (-\vec{a} \times \vec{b}) = -[\vec{a}\; \vec{a}\; \vec{b}] = 0$
• $\rm \vec{b} \cdot (\vec{a} \times \vec{c}) = [\vec{b}\; \vec{a}\; \vec{c}]$
• $\rm \vec{b} \cdot (-\vec{a} \times \vec{b}) = -[\vec{b}\; \vec{a}\; \vec{b}] = 0$
• $\rm \vec{b} \cdot (\vec{b} \times \vec{c}) = [\vec{b}\; \vec{b}\; \vec{c}] = 0$
• $\rm \vec{c} \cdot (\vec{a} \times \vec{c}) = [\vec{c}\; \vec{a}\; \vec{c}] = 0$
• $\rm \vec{c} \cdot (-\vec{a} \times \vec{b}) = -[\vec{c}\; \vec{a}\; \vec{b}]$
• $\rm \vec{c} \cdot (\vec{b} \times \vec{c}) = [\vec{c}\; \vec{b}\; \vec{c}] = 0$

Substituting these back:

$$ E = 0 + 0 + \rm [\vec{a}\; \vec{b}\; \vec{c}] + [\vec{b}\; \vec{a}\; \vec{c}] + 0 + 0 + 0 - [\vec{c}\; \vec{a}\; \vec{b}] + 0 $$

$$ E = \rm [\vec{a}\; \vec{b}\; \vec{c}] + [\vec{b}\; \vec{a}\; \vec{c}] - [\vec{c}\; \vec{a}\; \vec{b}] $$

Now, apply permutation properties:

• $\rm [\vec{b}\; \vec{a}\; \vec{c}] = -[\vec{a}\; \vec{b}\; \vec{c}]$
• $\rm [\vec{c}\; \vec{a}\; \vec{b}] = [\vec{a}\; \vec{b}\; \vec{c}]$ (cyclical permutation)

Substituting these values:

$$ E = \rm [\vec{a}\; \vec{b}\; \vec{c}] + (-[\vec{a}\; \vec{b}\; \vec{c}]) - [\vec{a}\; \vec{b}\; \vec{c}] $$

$$ E = \rm [\vec{a}\; \vec{b}\; \vec{c}] - [\vec{a}\; \vec{b}\; \vec{c}] - [\vec{a}\; \vec{b}\; \vec{c}] $$

$$ E = -[\vec{a}\; \vec{b}\; \vec{c}] $$

Final Answer Derivation

The calculation shows that the expression equals $\rm -[\vec{a}\; \vec{b}\; \vec{c}]$. This matches option 4.