Le t a→=i^−j^,b→=j^−k^ and c→=k^−i^. If d→ is a unit vector such that a→⋅d→=0=[b→c→d→], then d→ equals

Let d→=xi^+yj^+zk^ where x2+y2+z2=1-----1 (d→ being a unit vector) ∴ a→⋅d→=0⇒ x−y=0 or x=y----2[b→c→d→]=0⇒01−1−101xyz=0 or x+y+z=0or 2x+z=0 [using 2]or z=−2x-----3From (1), (2) and (3), we havex2+x2+4x2=1x=±16∴ d=±16i^+16j^−26k→ =±i^+j^−2k^6

Le t a→=i^−j^,b→=j^−k^ and c→=k^−i^. If d→ is a unit vector such that a→⋅d→=0=[b→c→d→], then d→ equals

Let d→=xi^+yj^+zk^ where x2+y2+z2=1-----1 (d→ being a unit vector) ∴ a→⋅d→=0⇒ x−y=0 or x=y----2[b→c→d→]=0⇒01−1−101xyz=0 or x+y+z=0or 2x+z=0 [using 2]or z=−2x-----3From (1), (2) and (3), we havex2+x2+4x2=1x=±16∴ d=±16i^+16j^−26k→ =±i^+j^−2k^6

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$Le t \vec{a} = \hat{i} - \hat{j}, \vec{b} = \hat{j} - \hat{k} and \vec{c} = \hat{k} - \hat{i} . If \vec{d} is a unit vector such that \vec{a} \cdot \vec{d} = 0 = [\vec{b} \vec{c} \vec{d}], then \vec{d} equals$

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$\pm \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

$\pm \frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}}$

$\pm \frac{\hat{i} + \hat{j} - 2 \hat{k}}{\sqrt{6}}$

$\pm \hat{k}$

answer is A.

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Detailed Solution

$Let \vec{d} = x \hat{i} + y \hat{j} + z \hat{k}$

$where x^{2} + y^{2} + z^{2} = 1 - - - - - (1)$ $(\vec{d} being a unit vector)$

$\begin{array}{l} ∴ \vec{a} \cdot \vec{d} = 0 \\ \Rightarrow x - y = 0 or x = y - - - - (2) \\ [\vec{b} \vec{c} \vec{d}] = 0 \\ \Rightarrow |\begin{array}{rrr} 0 & 1 & - 1 \\ - 1 & 0 & 1 \\ x & y & z \end{array}| = 0 \end{array}$

$or x + y + z = 0$

$\begin{array}{l} or 2 x + z = 0 [using 2] \\ or z = - 2 x - - - - - (3) \end{array}$

From (1), (2) and (3), we have

$\begin{array}{l} x^{2} + x^{2} + 4 x^{2} = 1 \\ x = \pm \frac{1}{\sqrt{6}} \\ ∴ \begin{aligned} d & = \pm (\frac{1}{\sqrt{6}} \hat{i} + \frac{1}{\sqrt{6}} \hat{j} - \frac{2}{\sqrt{6}} \vec{k}) \\ = \pm (\frac{\hat{i} + \hat{j} - 2 \hat{k}}{\sqrt{6}}) \end{aligned} \end{array}$