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Q.
If $a =\hat{ i }-\hat{ j }, b =\hat{ j }-\hat{ k }, c =\hat{ k }-\hat{ i }$ and $d$ is a unit vector such that $a \cdot d =0=[ b\,\, c\,\, d ]$, then $d$ is/are
Vector Algebra
Solution:
Let $ d=d_1 \hat{i}+d_2 \hat{j}+d_3 \hat{k} $
$ a \cdot d=(\hat{i}-\hat{j}) \cdot\left(d_1 \hat{i}+d_2 \hat{j}+d_3 \hat{k}\right)$
$\Rightarrow d_1-d_2=0 (\because a \cdot d=0)$
$\Rightarrow d_1=d_2...$(i)
Also, $d$ is a unit vector.
$ \Rightarrow d_1^2+d_2^2+d_3^2=1 ...$(ii)
Also, $[b\,\, c\,\, d]=0 $
$\Rightarrow \begin{vmatrix} 0 & 1 & -1 \\-1 & 0 & 1 \\
d_1 & d_2 & d_3\end{vmatrix}=0$
$ \Rightarrow -1\left(-d_3-d_1\right)-1\left(-d_2\right)=0$
$\Rightarrow d_1+d_2+d_3=0$
$\Rightarrow 2 d_1+d_3=0 \text { [from Eq. (i)] } $
$\Rightarrow d_3=-2 d_1 ...$(iii)
Using Eqs. (iii) and (i) in Eq. (ii), we get
$ d_1^2+d_1^2+4 d_1^2=1 \Rightarrow d_1=\pm \frac{1}{\sqrt{6}} $
$ \therefore d_2=\pm \frac{1}{\sqrt{6}} \text { and } d_3=\mp \frac{2}{\sqrt{6}} $
Hence, required vector is $ \pm \frac{1}{\sqrt{6}}(\hat{i}+\hat{j}-2 \hat{k})$.