Question

If \( \overline{a}=\overline{i}+\overline{j}+\overline{k} \), \( \overline{a}.\overline{b}=1 \) and \( \overline{a}\times\overline{b}=\overline{j}-\overline{k} \), then \( \overline{b}= \)

• A. \( \overline{i}-\overline{j}+\overline{k} \)
• B. \( 2\overline{j}-\overline{k} \)
• C. \( \overline{i} \)
• D. \( 2\overline{i} \)

Hint:

Alternatively, you can test the options directly. Take Option C (\( \overline{b} = \overline{i} \)): \[ \overline{a} \cdot \overline{b} = (\overline{i}+\overline{j}+\overline{k}) \cdot \overline{i} = 1 \] \[ \overline{a} \times \overline{b} = \begin{vmatrix}\overline{i}&\overline{j}&\overline{k}\\1&1&1\\1&0&0\end{vmatrix} = \overline{j} - \overline{k} \] Both given conditions are satisfied perfectly within 10 seconds!

Answer 1

The Correct Option is C

Concept: We can use the vector triple product identity to directly isolate vector \( \overline{b} \): \[ \overline{a} \times (\overline{a} \times \overline{b}) = (\overline{a} \cdot \overline{b})\overline{a} - (\overline{a} \cdot \overline{a})\overline{b} \]

Step 1: Evaluating basic component scalar operations.
Given \( \overline{a} = \overline{i} + \overline{j} + \overline{k} \): \[ \overline{a} \cdot \overline{a} = 1^2 + 1^2 + 1^2 = 3 \] We are also given that \( \overline{a} \cdot \overline{b} = 1 \).

Step 2: Computing the left-hand cross product.
We are given \( \overline{a}\times\overline{b} = \overline{j} - \overline{k} \). Let us cross this with \( \overline{a} \): \[ \overline{a} \times (\overline{a} \times \overline{b}) = \begin{vmatrix} \overline{i} & \overline{j} & \overline{k} \\ 1 & 1 & 1 \\ 0 & 1 & -1 \end{vmatrix} \] \[ = \overline{i}(-1 - 1) - \overline{j}(-1 - 0) + \overline{k}(1 - 0) = -2\overline{i} + \overline{j} + \overline{k} \]

Step 3: Substituting values into the triple product equation to solve for \( \overline{b} \).
\[ -2\overline{i} + \overline{j} + \overline{k} = (1)(\overline{i} + \overline{j} + \overline{k}) - 3\overline{b} \] Rearranging terms to isolate \( 3\overline{b} \): \[ 3\overline{b} = (\overline{i} + \overline{j} + \overline{k}) - (-2\overline{i} + \overline{j} + \overline{k}) \] \[ 3\overline{b} = 3\overline{i} \implies \overline{b} = \overline{i} \]