A spacecraft located at î + 2ĵ + 3k̂ is subjected to a force λ k̂ by firing a rocket. The spacecraft is subjected to a moment of magnitude

Concept:

1. Moment: It will be equal to the cross product of a position vector r from the point to anywhere on the line of action of the force and the force vector itself.⇔ \(\vec M = \vec r\; \times \vec F\)
2. Cross product of two vectors: The cross product can be written as a determinant.
   Let \(\vec a\) and \(\vec b\) be the two vectors. ⇒ \(\vec a = \;{a_1}\vec i + {b_1}\vec j + \;{c_1}\vec k\;\)and \(\vec b = \;{a_2}\vec i + {b_2}\vec j + \;{c_2}\vec k\)
   \(\therefore \;\vec a\; \times \;\vec b = \;\left| {\begin{array}{*{20}{c}} {\vec i}&{\vec j}&{\vec k\;}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right|\)
3. If  \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\) then determinant of A is given by:
   |A| = a₁₁ × {(a₂₂ × a₃₃) – (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) – (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) – (a₂₂ × a₃₁)}
    
4. Magnitude of vector: Let \(\vec a = x\;\vec i + y\;\vec j + z\;\vec k\) 
   Magnitude of vector of a = \(\left| {\vec a} \right| = \;\sqrt {{x^2} + \;{y^2} + {z^2}} \)

Calculation:

Given: \({\rm{\vec r}} = {\rm{\;\;\hat i}} + {\rm{\;}}2{\rm{\hat j}} + {\rm{\;}}3{\rm{\hat k\;and\;\vec F}} = {\rm{\;\lambda \;\hat k}}\)

We have to find magnitude of moment,

We know that, \({\rm{\vec M}} = {\rm{\vec r\;}} \times {\rm{\vec F}}\)

\(\therefore \;{\rm{\vec M}} = \;\left| {\begin{array}{*{20}{c}} {\vec i}&{\vec j}&{\vec k\;}\\ 1&2&3\\ 0&0&{\rm{\lambda }} \end{array}} \right|\) 
\( \Rightarrow \;{\rm{\vec M}} = \;\vec i\left( {2{\rm{\lambda }} - 0} \right) - \;\vec j\;\left( {{\rm{\lambda }} - 0} \right) + \;\vec k\;\left( {0 - 0} \right) = \;2{\rm{\lambda }}\vec i - \;{\rm{\lambda }}\vec j\;\)

Now,