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MHT CET · Maths · Vector Algebra

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

  1. A 0
  2. B 1
  3. C 3
  4. D 2
Verified Solution

Answer & Solution

Correct Answer

(C) 3

Step-by-step Solution

Detailed explanation

R.H.S. of the given equality can be written as
\((2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}+\overline{\mathrm{c}}) \cdot[(\overline{\mathrm{a}}+2 \overline{\mathrm{b}}+3 \overline{\mathrm{c}}) \times(3 \overline{\mathrm{a}}+\overline{\mathrm{b}}+2 \overline{\mathrm{c}})] \)
\( =(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}+\overline{\mathrm{c}}) \cdot[3(\overline{\mathrm{a}} \times \overline{\mathrm{a}})+(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+2(\overline{\mathrm{a}} \times \overline{\mathrm{c}}) \)
\( +6(\overline{\mathrm{b}} \times \overline{\mathrm{a}})+2(\overline{\mathrm{b}} \times \overline{\mathrm{b}})+4(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \)
\( +9(\overline{\mathrm{c}} \times \overline{\mathrm{a}})+3(\overline{\mathrm{c}} \times \overline{\mathrm{b}})+6(\overline{\mathrm{c}} \times \overline{\mathrm{c}})] \)
\( =(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}+\overline{\mathrm{c}})[0+(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+2(\overline{\mathrm{a}} \times \overline{\mathrm{c}}) \)
\( -6(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+0+4(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \)
\( =(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}+\overline{\mathrm{c}})[-5(\overline{\mathrm{a}} \times \overline{\mathrm{c}})-3(\overline{\mathrm{b}} \times \overline{\mathrm{c}})+(\overline{\mathrm{b}} \times \overline{\mathrm{c}})-7(\overline{\mathrm{a}} \times \overline{\mathrm{c}})] \)
\( =-10[\overline{\mathrm{a}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{b}})]+2[\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]-14[\overline{\mathrm{a}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})] \)
\( -15[\overline{\mathrm{b}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{b}})]+3[\overline{\mathrm{b}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]-21[\overline{\mathrm{b}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})] \)
\( -5[\overline{\mathrm{c}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{b}})]+[\overline{\mathrm{c}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})]-7[\overline{\mathrm{c}} \cdot(\overline{\mathrm{a}} \times \overline{\mathrm{c}})] \)
\( = 0+2[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]+0 \)
\( +0+0 \)
\( =-5[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]+0+21[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}] \)
\( = 18[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}} \overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=54\)
\(\Rightarrow[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=3\)