If the vectors \(a\hat i + \hat j + \hat k,\;\hat i + b\hat j + \hat k\) and \(\hat i + \hat j + c\hat k\;\left( {a,\;b,\;c \ne 1} \right)\) are coplanar, then the value of \(\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}}\) is equal to

Concept:

Scaler triple product of the vectors:

The scaler triple product of the vectors \({\rm{\bar A}} = {{\rm{x}}_1}\hat i + \;{y_1}\hat j + {\rm{\;}}{{\rm{z}}_1}\hat k,\;\bar B = \;{x_2}\hat i + \;{y_2}\hat j + {z_2}\hat k\) and \({\rm{\bar C}} = {\rm{\;}}{{\rm{x}}_3}\hat i + \;{y_3}\hat j + {\rm{\;}}{{\rm{z}}_3}\hat k\) is given by:

Coplaner vectors:

Three vectors \({\rm{\bar A}} = {{\rm{x}}_1}\hat i + \;{y_1}\hat j + {\rm{\;}}{{\rm{z}}_1}\hat k,\;\bar B = \;{x_2}\hat i + \;{y_2}\hat j + {z_2}\hat k\) and \({\rm{\bar C}} = {\rm{\;}}{{\rm{x}}_3}\hat i + \;{y_3}\hat j + {\rm{\;}}{{\rm{z}}_3}\hat k\) are said to be coplaner if the scaler triple product \({\bf{\bar A}} \cdot \left[ {{\bf{\bar B}} \times {\bf{\bar C}}} \right] = 0.\)

Solution:

Let the given vectors be \({\rm{\bar A}} = {\rm{a}}\hat i + \;\hat j + {\rm{\;}}\hat k,\;\bar B = \;\hat i + \;b\hat j + \hat k\) and \({\rm{\bar C}} = {\rm{\;}}\hat i + \;\hat j + {\rm{c}}\hat k\).

It is given that the vectors are coplaner therefore the scaler triple product \({\rm{\bar A}} \cdot \left[ {{\rm{\bar B}} \times {\rm{\bar C}}} \right] = 0.\)

Therefore,

\(\left| {\begin{array}{*{20}{c}} {\rm{a}}&1&1\\ 1&{\rm{b}}&1\\ 1&1&{\rm{c}} \end{array}} \right| = 0\)

Perform the coloumn operation C₁ – C₂ as follows:

\(\left| {\begin{array}{*{20}{c}} {{\rm{a}} - 1}&1&1\\ {1 - {\rm{b}}}&{\rm{b}}&1\\ 0&1&{\rm{c}} \end{array}} \right| = 0\)

Now perform another coloum operation C₂ – C₃ as follows:

\(\left| {\begin{array}{*{20}{c}} {{\rm{a}} - 1}&0&1\\ {1 - {\rm{b}}}&{{\rm{b}} - 1}&1\\ 0&{1 - {\rm{c}}}&{\rm{c}} \end{array}} \right| = 0\)

Take (1 - a)(1 - b)(1 - c) common. Note that it is given that a,b,c ≠ 0 therefore this action is jstified.

\(\left( {1 - {\rm{a}}} \right)\left( {1 - {\rm{b}}} \right)\left( {1 - {\rm{c}}} \right)\left| {\begin{array}{*{20}{c}} { - 1}&0&{\frac{1}{{1 - {\rm{a}}}}}\\ 1&{ - 1}&{\frac{1}{{1 - {\rm{b}}}}}\\ 0&1&{\frac{{\rm{c}}}{{1 - {\rm{c}}}}} \end{array}} \right| = 0\)

Since a, b, c ≠ 0 therefore (1 - a)(1 - b)(1 - c) ≠ 0.Therefore the determinant has to be zero.

\( - 1\left( {\frac{{\rm-{c}}}{{1 - {\rm{c}}}} - {\rm{}}\frac{1}{{1 - {\rm{b}}}}} \right) + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 0\)

\(\frac{1}{{1 - {\rm{b}}}} + {\rm{\;}}\frac{{\rm{c}}}{{1 - {\rm{c}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 0\)

Simplify the above expression as follows:

\(\frac{1}{{1 - {\rm{b}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{c}}}} - {\rm{\;}}1 + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 0\)

\(\frac{1}{{1 - {\rm{b}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{c}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 1\)