Question

Let \(\vec{a}, \vec{b}, \vec{c}\) be three vectors mutually perpendicular to each other and have same magnitude. If a vector \(\vec{r}\) satisfies \(\vec{a} \times \{(\vec{r} - \vec{b}) \times \vec{a}\} + \vec{b} \times \{(\vec{r} - \vec{c}) \times \vec{b}\} + \vec{c} \times \{(\vec{r} - \vec{a}) \times \vec{c}\} = \vec{0}\), then \(\vec{r}\) is equal to :

• A. \(\frac{1}{2}(\vec{a} + \vec{b} + 2\vec{c})\)
• B. \(\frac{1}{2}(\vec{a} + \vec{b} + \vec{c})\)
• C. \(\frac{1}{3}(\vec{a} + \vec{b} + \vec{c})\)
• D. \(\frac{1}{3}(2\vec{a} + \vec{b} - \vec{c})\)

Hint:

For any vector \(\vec{v}\) in an orthogonal basis \(\{\vec{e}_1, \vec{e}_2, \vec{e}_3\}\) with equal magnitudes \(\lambda\), the identity \(\sum (\vec{v} \cdot \vec{e}_i)\vec{e}_i = \lambda^2 \vec{v}\) is a powerful shortcut for vector algebra problems.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves vector triple products and the property of orthogonal bases. We expand the vector triple product \(\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}\) and utilize the fact that \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular.
Step 2: Key Formula or Approach:
Given \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{a} \cdot \vec{c} = 0\) and \(|\vec{a}| = |\vec{b}| = |\vec{c}| = \lambda\) (say).
For any vector \(\vec{r}\), in this basis: \(\vec{r} = \frac{(\vec{r} \cdot \vec{a})}{\lambda^2}\vec{a} + \frac{(\vec{r} \cdot \vec{b})}{\lambda^2}\vec{b} + \frac{(\vec{r} \cdot \vec{c})}{\lambda^2}\vec{c}\).
Step 3: Detailed Explanation:
The first term is \(\vec{a} \times \{(\vec{r} - \vec{b}) \times \vec{a}\}\). Expanding using the triple product formula: \[ (\vec{a} \cdot \vec{a})(\vec{r} - \vec{b}) - (\vec{a} \cdot (\vec{r} - \vec{b}))\vec{a} \] \[ = \lambda^2 \vec{r} - \lambda^2 \vec{b} - (\vec{a} \cdot \vec{r} - \vec{a} \cdot \vec{b})\vec{a} \] Since \(\vec{a} \cdot \vec{b} = 0\), it becomes: \(\lambda^2 \vec{r} - \lambda^2 \vec{b} - (\vec{a} \cdot \vec{r})\vec{a}\).
Similarly, the other two terms are: Second term: \(\lambda^2 \vec{r} - \lambda^2 \vec{c} - (\vec{b} \cdot \vec{r})\vec{b}\).
Third term: \(\lambda^2 \vec{r} - \lambda^2 \vec{a} - (\vec{c} \cdot \vec{r})\vec{c}\).
Summing all terms and setting to \(\vec{0}\): \[ 3\lambda^2 \vec{r} - \lambda^2(\vec{a} + \vec{b} + \vec{c}) - [(\vec{a} \cdot \vec{r})\vec{a} + (\vec{b} \cdot \vec{r})\vec{b} + (\vec{c} \cdot \vec{r})\vec{c}] = \vec{0} \] From the orthogonal basis property, we know \((\vec{a} \cdot \vec{r})\vec{a} + (\vec{b} \cdot \vec{r})\vec{b} + (\vec{c} \cdot \vec{r})\vec{c} = \lambda^2 \vec{r}\).
Substituting this back into the equation: \[ 3\lambda^2 \vec{r} - \lambda^2(\vec{a} + \vec{b} + \vec{c}) - \lambda^2 \vec{r} = \vec{0} \] \[ 2\lambda^2 \vec{r} = \lambda^2(\vec{a} + \vec{b} + \vec{c}) \] Dividing by \(\lambda^2\): \[ 2\vec{r} = \vec{a} + \vec{b} + \vec{c} \implies \vec{r} = \frac{1}{2}(\vec{a} + \vec{b} + \vec{c}) \]
Step 4: Final Answer:
The vector \(\vec{r}\) is \(\frac{1}{2}(\vec{a} + \vec{b} + \vec{c})\).