Let \[\vec{a}=\hat{i}-\hat{j},\vec{b}=\hat{j}-\hat{k}\] 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

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

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

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

Let \[\vec{a}=\hat{i}-\hat{j},\vec{b}=\hat{j}-\hat{k}\] and \[\vec{c}=

1.	Let \[\vec{a}=\hat{i}-\hat{j},\vec{b}=\hat{j}-\hat{k}\] 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
A.	\[\pm \frac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}}\]
B.	\[\pm \frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}\]
C.	\[\pm \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}\]
D.	\[\pm \,\hat{k}\]
Answer» B. \[\pm \frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}\]

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Let \[\vec{a}=\hat{i}-\hat{j},\vec{b}=\hat{j}-\hat{k}\] 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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