MHT CET · Maths · Vector Algebra
If \(\overline{\mathrm{a}}, \overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) are unit coplanar vectors, then the scalar triple product \(\left[\begin{array}{lll}2 \bar{a}-\bar{b} & 2 \bar{b}-\bar{c} & 2 \bar{c}-\bar{a}\end{array}\right]\) has the value
- A 0
- B \(-\sqrt{3}\)
- C 1
- D \(\sqrt{3}\)
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
\(\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}\) are coplanar vectors.
\(\Rightarrow\left[\begin{array}{lll}
\overline{\mathrm{a}} & \overline{\mathrm{~b}} & \overline{\mathrm{c}}
\end{array}\right]=0\)
Let \(\bar{\alpha}=2 \overline{\mathrm{a}}-\overline{\mathrm{b}}, \bar{\beta}=2 \overline{\mathrm{~b}}-\overline{\mathrm{c}}\) and \(\gamma=2 \overline{\mathrm{c}}-\overline{\mathrm{a}}\). Then,
\(\begin{aligned}
& {\left[\begin{array}{ccc}
\bar{\alpha} & \bar{\beta} & \bar{\gamma}
\end{array}\right]=\left|\begin{array}{ccc}
2 & -1 & 0 \\
0 & 2 & -1 \\
-1 & 0 & 2
\end{array}\right|\left[\begin{array}{lll}
\overline{\mathrm{a}} & \overline{\mathrm{~b}} & \overline{\mathrm{c}}
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{lll}
\bar{\alpha} & \bar{\beta} & \bar{\gamma}
\end{array}\right]=7\left[\begin{array}{lll}
\overline{\mathrm{a}} & \overline{\mathrm{~b}} & \overline{\mathrm{c}}
\end{array}\right]=7(0)=0
\end{aligned}\)
\(\Rightarrow\left[\begin{array}{lll}
\overline{\mathrm{a}} & \overline{\mathrm{~b}} & \overline{\mathrm{c}}
\end{array}\right]=0\)
Let \(\bar{\alpha}=2 \overline{\mathrm{a}}-\overline{\mathrm{b}}, \bar{\beta}=2 \overline{\mathrm{~b}}-\overline{\mathrm{c}}\) and \(\gamma=2 \overline{\mathrm{c}}-\overline{\mathrm{a}}\). Then,
\(\begin{aligned}
& {\left[\begin{array}{ccc}
\bar{\alpha} & \bar{\beta} & \bar{\gamma}
\end{array}\right]=\left|\begin{array}{ccc}
2 & -1 & 0 \\
0 & 2 & -1 \\
-1 & 0 & 2
\end{array}\right|\left[\begin{array}{lll}
\overline{\mathrm{a}} & \overline{\mathrm{~b}} & \overline{\mathrm{c}}
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{lll}
\bar{\alpha} & \bar{\beta} & \bar{\gamma}
\end{array}\right]=7\left[\begin{array}{lll}
\overline{\mathrm{a}} & \overline{\mathrm{~b}} & \overline{\mathrm{c}}
\end{array}\right]=7(0)=0
\end{aligned}\)
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