Question

If $\vec{a}$, $\vec{b}$, $\vec{c}$ are mutually perpendicular vectors having magnitudes $1$, $2$, $3$ respectively, then $[\vec{a} + \vec{b} + \vec{c} \quad \vec{b} - \vec{a} \quad \vec{c}] =$

• A. $12$
• B. $18$
• C. $0$
• D. $6$

Hint:

You can solve this quickly using a determinant format of coefficients! Form a matrix of the component multipliers of $\vec{a}, \vec{b}, \vec{c}$: $$\begin{vmatrix} 1 & 1 & 1
-1 & 1 & 0
0 & 0 & 1 \end{vmatrix} \times [\vec{a} \quad \vec{b} \quad \vec{c}]$$ Expanding the determinant along the third row gives $1(1 - (-1)) = 2$. Multiply this directly by the base volume ($1 \times 2 \times 3 = 6$) to get $2 \times 6 = 12$ in a few seconds!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the evaluation of a scalar triple product involving linear combinations of three mutually perpendicular spatial vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$.

Step 2: Key Formula or Approach:
The scalar triple product $[\vec{x} \quad \vec{y} \quad \vec{z}]$ can be expanded algebraically as $\vec{x} \cdot (\vec{y} \times \vec{z})$. Since the vectors are mutually perpendicular: $$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$$ The scalar triple product of the base vectors forms an orthogonal box of volume: $$[\vec{a} \quad \vec{b} \quad \vec{c}] = |\vec{a}| \cdot |\vec{b}| \cdot |\vec{c}|$$

Step 3: Detailed Explanation:
Let's expand the scalar triple product structure linearly: $$[\vec{a} + \vec{b} + \vec{c} \quad \vec{b} - \vec{a} \quad \vec{c}] = (\vec{a} + \vec{b} + \vec{c}) \cdot [(\vec{b} - \vec{a}) \times \vec{c}]$$ Distribute the cross product operator inside the bracket: $$= (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{b} \times \vec{c} - \vec{a} \times \vec{c})$$ Now, expand the scalar dot product across these terms: $$= \vec{a}\cdot(\vec{b}\times\vec{c}) - \vec{a}\cdot(\vec{a}\times\vec{c}) + \vec{b}\cdot(\vec{b}\times\vec{c}) - \vec{b}\cdot(\vec{a}\times\vec{c}) + \vec{c}\cdot(\vec{b}\times\vec{c}) - \vec{c}\cdot(\vec{a}\times\vec{c})$$ Recall that any scalar triple product containing a repeated vector is automatically zero (as the cross product is perpendicular to both component vectors): $$\vec{a}\cdot(\vec{a}\times\vec{c}) = 0, \quad \vec{b}\cdot(\vec{b}\times\vec{c}) = 0, \quad \vec{c}\cdot(\vec{b}\times\vec{c}) = 0, \quad \vec{c}\cdot(\vec{a}\times\vec{c}) = 0$$ This leaves only two surviving non-zero triple product expressions: $$= [\vec{a} \quad \vec{b} \quad \vec{c}] - [\vec{b} \quad \vec{a} \quad \vec{c}]$$ Using the cyclic permutation property, flipping two adjacent vectors changes the algebraic sign: $$-[\vec{b} \quad \vec{a} \quad \vec{c}] = +[\vec{a} \quad \vec{b} \quad \vec{c}]$$ Thus, the expression simplifies to: $$= 2 [\vec{a} \quad \vec{b} \quad \vec{c}]$$ Since $\vec{a}$, $\vec{b}$, $\vec{c}$ are mutually perpendicular, their scalar triple product is equal to the product of their magnitudes: $$[\vec{a} \quad \vec{b} \quad \vec{c}] = 1 \times 2 \times 3 = 6$$ Substitute this back to reach our final calculation: $$\text{Value} = 2 \times 6 = 12$$

Step 4: Final Answer:
The scalar triple product evaluates to $12$, which matches option (A).