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

Step 1: Pick the right identity.
The triple product identity is $\overline{a}\times(\overline{a}\times\overline{b}) = (\overline{a}\cdot\overline{b})\overline{a} - (\overline{a}\cdot\overline{a})\overline{b}$.
Step 2: Find the simple dot products.
With $\overline{a} = \overline{i}+\overline{j}+\overline{k}$, we get $\overline{a}\cdot\overline{a} = 3$, and we are given $\overline{a}\cdot\overline{b} = 1$.
Step 3: Compute the left cross product.
Cross $\overline{a}$ with $\overline{a}\times\overline{b} = \overline{j}-\overline{k}$ using the determinant: result is $-2\overline{i} + \overline{j} + \overline{k}$.
Step 4: Put both sides together.
\[ -2\overline{i}+\overline{j}+\overline{k} = (1)(\overline{i}+\overline{j}+\overline{k}) - 3\overline{b} \]
Step 5: Isolate the b term.
$3\overline{b} = (\overline{i}+\overline{j}+\overline{k}) - (-2\overline{i}+\overline{j}+\overline{k}) = 3\overline{i}$.
Step 6: Solve for b.
Divide by 3 to get $\overline{b} = \overline{i}$. \[ \boxed{\overline{b} = \overline{i}} \]