Question

If the vectors e₁ = (1, 0, 2), e₂ = (0, 1, 0) and e₃ = (−2, 0, 1) form an orthogonal basis of the three dimensional real space ℝ³, then the vector u = (4, 3,−3) ∈ ℝ³ can be expressed as
1. \({\rm{u}} = - \frac{2}{5}{{\rm{e}}_1} - 3{{\rm{e}}_2} - \frac{{11}}{5}{{\rm{e}}_3}\)
2. \({\rm{u}} = - \frac{2}{5}{{\rm{e}}_1} - 3{{\rm{e}}_2} + \frac{{11}}{5}{{\rm{e}}_3}\)
3. \({\rm{u}} = - \frac{2}{5}{{\rm{e}}_1} + 3{{\rm{e}}_2} + \frac{{11}}{5}{{\rm{e}}_3}\)
4. \({\rm{u}} = - \frac{2}{5}{{\rm{e}}_1} + 3{{\rm{e}}_2} - \frac{{11}}{5}{{\rm{e}}_3}\)

Answer 1

Correct Answer - Option 4 : \({\rm{u}} = - \frac{2}{5}{{\rm{e}}_1} + 3{{\rm{e}}_2} - \frac{{11}}{5}{{\rm{e}}_3}\)

Given \({{\rm{e}}_1} = \left( {1,0,2} \right) = {\rm{\vec i}} + 2{\rm{\vec k}}\)

\(\begin{array}{l} {{\rm{e}}_2} = \left( {0,1,0} \right) = {\rm{\vec j}}\\ {{\rm{e}}_3} = \left( { - 2,0,1} \right) = - 2{\rm{\vec i}} + {\rm{\vec k}} \end{array}\)

\({\rm{u}} = \left( {4,3, - 3} \right) = 4{\rm{\vec i}} + 3{\rm{\vec j}} - 3{\rm{\vec k}}\)    ---(1)

Consider \({\rm{u}} = \left( {\frac{{ - 2}}{5}{{\rm{e}}_1} + 3{{\rm{e}}_2} - \frac{{11}}{5}{{\rm{e}}_3}} \right)\)

\(= \frac{{ - 2}}{5}\left( {{\rm{\hat i}} + 2{\rm{\hat k}}} \right) + 3\left( {{\rm{\hat j}}} \right) - \frac{{11}}{5}\left( { - 2{\rm{\hat i}} + {\rm{\hat k}}} \right)\)

\(\\ = \left( {\frac{{ - 2}}{5}{\rm{\hat i}} + \frac{{22}}{5}{\rm{\hat i}}} \right) + 3{\rm{\hat j}} + {\rm{\hat k}}\left( {\frac{{ - 4}}{5}{\rm{\;\;}}\frac{{ - 11}}{5}} \right) \)

\(= 4{\rm{\hat i}} + 3{\rm{\hat j}} - 3{\rm{\hat k}}\)    ---(2)

∴ (1) = (2) ⇒ so option (D) correct