Question
let $$\vec a = \hat i - \hat j,\,\vec b = \hat j - \hat k,\,\vec c = \hat k - \hat i.$$ if $${\vec d}$$ is a unit vector such that $$\vec a.\vec d = 0 = \left[ {\vec b\,\vec c\,\vec d} \right],$$ 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 :
$$ \pm \frac{{\hat i + \hat j - 2\hat k}}{{\sqrt 6 }}$$
Solution :
$$\eqalign{
& {\text{Let}}\,\,\vec d = x\,\hat i\, + y\,\hat j + z\,\hat k \cr
& {\text{where}}\,\,{x^2} + {y^2} + {z^2}.....(1) \cr
& \left( {\vec d\,\,{\text{being}}\,{\text{unit}}\,{\text{vector}}} \right)\,\,\therefore \vec a.\vec d = 0 \cr
& \Rightarrow x - y = 0\,\,\,\, \Rightarrow x = y.....(2) \cr} $$
\[\left[ {\vec b\,\vec c\,\vec d} \right] = 0 \Rightarrow \left| \begin{array}{l}
\,\,\,\,\,\,0\,\,\,\,1\,\,\,\, - 1\\
- 1\,\,\,\,0\,\,\,\,1 = 0\\
\,\,\,\,\,\,x\,\,\,\,y\,\,\,\,z
\end{array} \right|\]
$$\eqalign{
& \Rightarrow x + y + z = 0 \cr
& \Rightarrow 2x + z = 0\,\,\,\,\,\,\,\,\left( {{\text{using }}\,(2)} \right) \cr
& \Rightarrow z = - 2x.....(3) \cr
& {\text{From (1),}}\,{\text{(2) and (3)}} \cr
& {x^2} + {x^2} + 4{x^2} = 1 \Rightarrow x = \pm \frac{1}{{\sqrt 6 }} \cr
& \therefore d = \pm \left( {\frac{1}{{\sqrt 6 }}\hat i + \frac{1}{{\sqrt 6 }}\hat j - \frac{2}{{\sqrt 6 }}\hat k} \right) = \pm \left( {\frac{{\hat i + \hat j - 2\hat k}}{{\sqrt 6 }}} \right) \cr} $$