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

Concept:
The condition \[ [\bar{b}\ \bar{c}\ \bar{d}]=0 \] means the scalar triple product is zero. So \(\bar{d}\) lies in the plane of \(\bar{b}\) and \(\bar{c}\), or equivalently is perpendicular to \(\bar{b}\times\bar{c}\). ip

Step 1: Write the vectors in component form.
\[ \bar{a}=(1,-1,0) \] \[ \bar{b}=(0,1,-1) \] \[ \bar{c}=(-1,0,1) \] ip

Step 2: Use the scalar triple product condition.
\[ [\bar{b}\ \bar{c}\ \bar{d}]=0 \] means \(\bar{d}\) must be coplanar with \(\bar{b}\) and \(\bar{c}\). Also note: \[ \bar{a}+\bar{b}+\bar{c}=0 \] So \(\bar{a},\bar{b},\bar{c}\) are coplanar in the same plane. Therefore, \(\bar{d}\) also lies in this plane. ip

Step 3: Use the condition \(\bar{a}\cdot \bar{d}=0\).
We need a unit vector in the plane of \(\bar{a},\bar{b},\bar{c}\) and perpendicular to \(\bar{a}\). The vector normal to the plane of \(\bar{a},\bar{b},\bar{c}\) is proportional to: \[ \hat{i}+\hat{j}+\hat{k} \] Checking option (C): \[ \bar{d}=\pm\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} \] Then \[ \bar{a}\cdot \bar{d} = (1,-1,0)\cdot \frac{(1,1,1)}{\sqrt{3}} = \frac{1-1+0}{\sqrt{3}}=0 \] So the condition is satisfied. ip Hence, the correct answer is:
\[ \boxed{(C)\ \pm \left( \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} \right)} \]