If $\mathrm{A} \equiv(1,-1,0), \mathrm{B} \equiv(0,1,-1)$ and $\mathrm{C} \equiv(-1,0,1)$, then the unit…
If $\mathrm{A} \equiv(1,-1,0), \mathrm{B} \equiv(0,1,-1)$ and $\mathrm{C} \equiv(-1,0,1)$, then the unit vector $\overline{\mathrm{d}}$ such that $\overline{\mathrm{a}}$ and $\overline{\mathrm{d}}$ are perpendiculars and $\overline{\mathrm{b}}, \overline{\mathrm{c}}, \overline{\mathrm{d}}$ are coplanar is
$+\frac{1}{\sqrt{3}}(1,1,1)$
$+\frac{1}{\sqrt{3}}(-1,-1,1)$
$+\frac{1}{\sqrt{6}}(1,1,-2)$
$+\frac{1}{\sqrt{2}}(1,1,0)$
Solution
Let $\bar{d}=p \hat{i}+q \hat{j}+r \hat{k}$, where $p, q, r \in R$
As $\overline{\mathrm{b}}, \overline{\mathrm{c}}, \overline{\mathrm{d}}$ are coplanar, we get
$\begin{array}{ll}
& \left|\begin{array}{ccc}
0 & 1 & -1 \\
-1 & 0 & 1 \\
p & q & r
\end{array}\right|=0 \\
\therefore & -1(-r-p)-1(-q)=0 ...(i)\\
\therefore & p+q+r=0
\end{array}$ Also, given that $\overline{\mathrm{a}}$ and $\overline{\mathrm{d}}$ are perpendiculars.
$\begin{array}{ll}
\therefore & \overline{\mathrm{a}} \cdot \overline{\mathrm{~d}}=0 \\
\therefore & \mathrm{p}-\mathrm{q}=0 \\
\therefore & \mathrm{p}=\mathrm{q}...(ii)
\end{array}$
Among the given options only option (C) satisfies equations (i) and (ii)