Question

Let \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{b} = \hat{j} - \hat{k}\). If \(\vec{c}\) is a vector such that \(\vec{a} \times \vec{c} = \vec{b}\) and \(\vec{a} \cdot \vec{c} = 3\), then \(\vec{a} \cdot (\vec{b} \times \vec{c})\) is equal to :

• A. \(-2\)
• B. 2
• C. \(-6\)
• D. 6

Hint:

Properties of the triple product like \(\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}\) are essential. When you have an equation like \(\vec{a} \times \vec{c} = \vec{b}\), applying another cross product often leads directly to the solution.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The expression \(\vec{a} \cdot (\vec{b} \times \vec{c})\) is the scalar triple product, which is equal to \((\vec{a} \times \vec{b}) \cdot \vec{c}\). We can use the vector triple product identity on \(\vec{a} \times (\vec{a} \times \vec{c})\) to find information about \(\vec{c}\).
Step 2: Key Formula or Approach:
1. \(\vec{a} \times (\vec{a} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c}\). 2. \(\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}\).
Step 3: Detailed Explanation:
Given \(\vec{a} \times \vec{c} = \vec{b}\). Cross both sides with \(\vec{a}\): \[ \vec{a} \times (\vec{a} \times \vec{c}) = \vec{a} \times \vec{b} \] Using the identity: \[ (\vec{a} \cdot \vec{c})\vec{a} - |\vec{a}|^2 \vec{c} = \vec{a} \times \vec{b} \] Substitute \(\vec{a} \cdot \vec{c} = 3\) and \(|\vec{a}|^2 = 1^2 + 1^2 + 1^2 = 3\): \[ 3\vec{a} - 3\vec{c} = \vec{a} \times \vec{b} \implies 3\vec{c} = 3\vec{a} - (\vec{a} \times \vec{b}) \] Now calculate \(\vec{a} \times \vec{b}\): \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 1 & 1
0 & 1 & -1 \end{vmatrix} = (-2)\hat{i} - (-1)\hat{j} + (1)\hat{k} = -2\hat{i} + \hat{j} + \hat{k} \] We need \(\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}\). From \(3\vec{c} = 3\vec{a} - (\vec{a} \times \vec{b})\), dot with \((\vec{a} \times \vec{b})\): \[ 3(\vec{a} \times \vec{b} \cdot \vec{c}) = 3\vec{a} \cdot (\vec{a} \times \vec{b}) - |\vec{a} \times \vec{b}|^2 \] Since \(\vec{a} \perp (\vec{a} \times \vec{b})\), the first term is zero. \[ 3(\vec{a} \times \vec{b} \cdot \vec{c}) = - ((-2)^2 + 1^2 + 1^2) = -6 \implies (\vec{a} \times \vec{b} \cdot \vec{c}) = -2 \]
Step 4: Final Answer:
The scalar triple product value is \(-2\).