Vector Algebra
Vector Algebra
nta_abhyas_2025
Grade 12
Question:
Given $|\vec{a}| = 1$, $|\vec{b}| = 2$, $|\vec{c}| = 3$ and $\vec{a} \cdot \vec{b} = 0$, $\vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a}$ (as the three vectors are mutually perpendicular)
Step-by-Step Solution
Key Concept: For mutually perpendicular vectors, the dot products between different vectors are zero; use vector product properties and orthogonality to simplify expressions.
Since the three vectors are mutually perpendicular, $\vec{a} \cdot \vec{b} = 0$, $\vec{b} \cdot \vec{c} = 0$, and $\vec{c} \cdot \vec{a} = 0$. Expanding $(\vec{a} + \vec{b} + \vec{c}) \times (\vec{b} - \vec{a}) \cdot \vec{c}$: $(\vec{a} + \vec{b} + \vec{c}) \times (\vec{b} - \vec{a}) = \vec{a} \times \vec{b} - \vec{a} \times \vec{a} + \vec{b} \times \vec{b} - \vec{b} \times \vec{a} + \vec{c} \times \vec{b} - \vec{c} \times \vec{a} = \vec{a} \times \vec{b} + \vec{b} \times \vec{a} + \vec{c} \times \vec{b} - \vec{c} \times \vec{a} = 2(\vec{a} \times \vec{b}) + (\vec{c} \times \vec{b}) - (\vec{c} \times \vec{a})$. Taking the dot product with $\vec{c}$ and using orthogonality simplifies to evaluate the scalar triple product using the perpendicularity conditions.
Correct Answer: 3