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Question: Let $\vec{a} = \hat{i}-\hat{j}, \vec{b} = \hat{j}-\hat{k}, \vec{c} = \hat{k}-\hat{i}$. If $\hat{d}$ ...

Let a⃗=i^−j^,b⃗=j^−k^,c⃗=k^−i^\vec{a} = \hat{i}-\hat{j}, \vec{b} = \hat{j}-\hat{k}, \vec{c} = \hat{k}-\hat{i}. If d^\hat{d} is a unit vector such that a⃗⋅d^=0=[b⃗c⃗d^]\vec{a} \cdot \hat{d} = 0 = [\vec{b}\vec{c}\hat{d}], then d^\hat{d} equals :

A

i^+j^−2k^6\frac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}}

B

i^+j^−k^3\frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}

C

i^+j^+k^3\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}

D

k^\hat{k}

Answer

±i^+j^−2k^6\pm \frac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}}

Explanation

Solution

The given conditions are:

  1. a⃗⋅d^=0\vec{a} \cdot \hat{d} = 0: This implies that the unit vector d^\hat{d} is orthogonal to the vector a⃗\vec{a}.
  2. [b⃗c⃗d^]=0[\vec{b}\vec{c}\hat{d}] = 0: This is the scalar triple product of b⃗\vec{b}, c⃗\vec{c}, and d^\hat{d}. A zero scalar triple product means the three vectors are coplanar. This implies that d^\hat{d} is orthogonal to the vector formed by the cross product of b⃗\vec{b} and c⃗\vec{c}, i.e., d^⊥(b⃗×c⃗)\hat{d} \perp (\vec{b} \times \vec{c}).

Given vectors: a⃗=i^−j^\vec{a} = \hat{i}-\hat{j} b⃗=j^−k^\vec{b} = \hat{j}-\hat{k} c⃗=k^−i^\vec{c} = \hat{k}-\hat{i}

First, calculate the cross product b⃗×c⃗\vec{b} \times \vec{c}: b⃗×c⃗=(j^−k^)×(k^−i^)\vec{b} \times \vec{c} = (\hat{j}-\hat{k}) \times (\hat{k}-\hat{i}) b⃗×c⃗=∣i^j^k^01−1−101∣=i^(1−0)−j^(0−1)+k^(0−(−1))=i^+j^+k^\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -1 \\ -1 & 0 & 1 \end{vmatrix} = \hat{i}(1-0) - \hat{j}(0-1) + \hat{k}(0-(-1)) = \hat{i} + \hat{j} + \hat{k} Let v⃗=b⃗×c⃗=i^+j^+k^\vec{v} = \vec{b} \times \vec{c} = \hat{i} + \hat{j} + \hat{k}.

Now, d^\hat{d} must be orthogonal to both a⃗\vec{a} and v⃗\vec{v}. Therefore, d^\hat{d} must be parallel to the cross product of a⃗\vec{a} and v⃗\vec{v}. Calculate a⃗×v⃗\vec{a} \times \vec{v}: a⃗×v⃗=(i^−j^)×(i^+j^+k^)\vec{a} \times \vec{v} = (\hat{i}-\hat{j}) \times (\hat{i}+\hat{j}+\hat{k}) a⃗×v⃗=∣i^j^k^1−10111∣=i^(−1−0)−j^(1−0)+k^(1−(−1))=−i^−j^+2k^\vec{a} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 1 & 1 \end{vmatrix} = \hat{i}(-1-0) - \hat{j}(1-0) + \hat{k}(1-(-1)) = -\hat{i} - \hat{j} + 2\hat{k}

Since d^\hat{d} is a unit vector, it must be in the direction of a⃗×v⃗\vec{a} \times \vec{v} or its opposite direction, with a magnitude of 1. d^=±−i^−j^+2k^∣−i^−j^+2k^∣\hat{d} = \pm \frac{-\hat{i} - \hat{j} + 2\hat{k}}{|-\hat{i} - \hat{j} + 2\hat{k}|} The magnitude is: ∣−i^−j^+2k^∣=(−1)2+(−1)2+(2)2=1+1+4=6|-\hat{i} - \hat{j} + 2\hat{k}| = \sqrt{(-1)^2 + (-1)^2 + (2)^2} = \sqrt{1+1+4} = \sqrt{6} So, d^=±−i^−j^+2k^6=±i^+j^−2k^6\hat{d} = \pm \frac{-\hat{i} - \hat{j} + 2\hat{k}}{\sqrt{6}} = \pm \frac{\hat{i} + \hat{j} - 2\hat{k}}{\sqrt{6}} This matches option (A).