Mathematics › Vectors › Scalar Triple Product
If $[(\overline{\mathrm{a}}+2 \overline{\mathrm{b}}+3 \overline{\mathrm{c}}) \times(\overline{\mathrm{b}}+2…
If $[(\overline{\mathrm{a}}+2 \overline{\mathrm{b}}+3 \overline{\mathrm{c}}) \times(\overline{\mathrm{b}}+2 \overline{\mathrm{c}}+3 \overline{\mathrm{a}})] \cdot(\overline{\mathrm{c}}+2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}})=54$, then the value of $\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$ is
0 1 3 2
Solution
R.H.S. of the given equality can be written as
$\begin{aligned}
&(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}+\overline{\mathrm{c}}) \cdot[(\overline{\mathrm{a}}+2 \overline{\mathrm{b}}+3 \overline{\mathrm{c}}) \times(3 \overline{\mathrm{a}}+\overline{\mathrm{b}}+2 \overline{\mathrm{c}})] \\
&=(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}+\overline{\mathrm{c}}) \cdot[3(\overline{\mathrm{a}} \times \overline{\mathrm{a}})+(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+2(\overline{\mathrm{a}} \times \overline{\mathrm{c}}) \\
&+6(\overline{\mathrm{b}} \times \overline{\mathrm{a}})+2(\overline{\mathrm{b}} \times \overline{\mathrm{b}})+4(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \\
&+9(\overline{\mathrm{c}} \times \overline{\mathrm{a}})+3(\overline{\mathrm{c}} \times \overline{\mathrm{b}})+6(\overline{\mathrm{c}} \times \overline{\mathrm{c}})] \\
&=(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}+\overline{\mathrm{c}})[0+(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+2(\overline{\mathrm{a}} \times \overline{\mathrm{c}}) \\
&-6(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+0+4(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \\
&=(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}+\overline{\mathrm{c}})[-5(\overline{\mathrm{a}} \times \overline{\mathrm{c}})-3(\overline{\mathrm{b}} \times \overline{\mathrm{c}})+(\overline{\mathrm{b}} \times \overline{\mathrm{c}})-7(\overline{\mathrm{a}} \times \overline{\mathrm{c}})] \\
&=-10[\overline{\mathrm{a}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{b}})]+2[\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]-14[\overline{\mathrm{a}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})] \\
&-15[\overline{\mathrm{b}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{b}})]+3[\overline{\mathrm{b}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]-21[\overline{\mathrm{b}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})] \\
&-5[\overline{\mathrm{c}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{b}})]+[\overline{\mathrm{c}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]-7[\overline{\mathrm{c}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})] \\
&= 0+2[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]+0 \\
&+0+0 \\
&=-5[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]+0+21[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}] \\
&= 18[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}} \overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=54
\end{aligned}$
$\Rightarrow[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=3$
Asked in: MHT CET 2023 (09 May Shift 1)
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