Question

\(\bar{a} = \hat{i} - \hat{j}, \bar{b} = \hat{j} - \hat{k}, \bar{c} = \hat{k} - \hat{i}\) then a unit vector \(\bar{d}\) such that \(\bar{a} \cdot \bar{d} = 0 = [\bar{b} \bar{c} \bar{d}]\) is

• A. \(\pm \left( \frac{\hat{i} + \hat{j} + 3\hat{k}}{\sqrt{11}} \right)\)
• B. \(\pm \left( \frac{-\hat{j} + \hat{k}}{\sqrt{2}} \right)\)
• C. \(\pm \left( \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} \right)\)
• D. \(\pm \left( \frac{\hat{i} + \hat{j} - 2\hat{k}}{\sqrt{6}} \right)\)

Hint:

If scalar triple product is zero, the vectors are coplanar. That is often the fastest clue in vector algebra problems.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
\(\bar{a} \cdot \bar{d} = 0\) means \(\bar{d}\) is perpendicular to \(\bar{a}\).
\([\bar{b} \bar{c} \bar{d}] = 0\) means \(\bar{d}, \bar{b}, \bar{c}\) are coplanar, so \(\bar{d}\) is perpendicular to \(\bar{b} \times \bar{c}\).
Step 2: Key Formula or Approach:
\(\bar{d}\) is parallel to \(\bar{a} \times (\bar{b} \times \bar{c})\).
However, it's easier to find \(\bar{n} = \bar{b} \times \bar{c}\) and then \(\bar{v} = \bar{n} \times \bar{a}\).
Step 3: Detailed Explanation:
\(\bar{b} \times \bar{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
0 & 1 & -1
-1 & 0 & 1 \end{vmatrix} = \hat{i}(1) - \hat{j}(-1) + \hat{k}(1) = \hat{i} + \hat{j} + \hat{k}\).
Now, \(\bar{d}\) is perpendicular to \((1, -1, 0)\) and \((1, 1, 1)\).
Direction \(\bar{d} \parallel \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -1 & 0
1 & 1 & 1 \end{vmatrix} = \hat{i}(-1) - \hat{j}(1) + \hat{k}(2) \implies -\hat{i} - \hat{j} + 2\hat{k}\).
Magnitude is \(\sqrt{1+1+4} = \sqrt{6}\).
Unit vector \(\bar{d} = \pm \frac{\hat{i} + \hat{j} - 2\hat{k}}{\sqrt{6}}\).
Step 4: Final Answer:
The unit vector is \(\pm \left( \frac{\hat{i} + \hat{j} - 2\hat{k}}{\sqrt{6}} \right)\).