MHT CET · Maths · Vector Algebra
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
- A \(+\frac{1}{\sqrt{3}}(1,1,1)\)
- B \(+\frac{1}{\sqrt{3}}(-1,-1,1)\)
- C \(+\frac{1}{\sqrt{6}}(1,1,-2)\)
- D \(+\frac{1}{\sqrt{2}}(1,1,0)\)
Answer & Solution
Correct Answer
(C) \(+\frac{1}{\sqrt{6}}(1,1,-2)\)
Step-by-step Solution
Detailed explanation
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)
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)
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