Question

If \[a\hat i + \hat j + \hat k,\hat i + b\hat j + \hat k,\hat i + \hat j + c\hat k\]are coplanar then \[\dfrac{1}{{1 - a}} + \dfrac{1}{{1 - b}} + \dfrac{1}{{1 - c}} = \]
A. 2
B. 1
C. -1
D. 3

Answer 1

Hint: Here we use the concept of coplanar vectors and that their scalar triple product is equal to zero. Write the matrix form of the scalar triple product and perform row transformations to make some values similar to the denominator of terms in the question. Equate the determinant equal to zero.

Complete step-by-step answer:
We have three vectors \[a\hat i + \hat j + \hat k,\hat i + b\hat j + \hat k,\hat i + \hat j + c\hat k\]
Let us name the vectors as
\[\overrightarrow A = a\hat i + \hat j + \hat k,\overrightarrow B = \hat i + b\hat j + \hat k,\overrightarrow C = \hat i + \hat j + c\hat k\]
We know the vectors \[\overrightarrow A ,\overrightarrow B ,\overrightarrow C \] are coplanar.
So we can write
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {\overrightarrow A }&{\overrightarrow B }&{\overrightarrow C }
\end{array}} \right] = 0\]
We write the vectors in matrix form
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  a&1&1 \\
  1&b&1 \\
  1&1&c
\end{array}} \right] = 0\]
Now we apply row transformation\[{R_2} \to {R_2} - {R_1}\]
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  a&1&1 \\
  {1 - a}&{b - 1}&0 \\
  1&1&c
\end{array}} \right] = 0\]
Now we apply row transformation\[{R_3} \to {R_3} - {R_1}\]
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  a&1&1 \\
  {1 - a}&{b - 1}&0 \\
  {1 - a}&0&{c - 1}
\end{array}} \right] = 0\]
Now we find the determinant of the matrix on both sides of the equation
\[ \Rightarrow a\left\{ {(b - 1)(c - 1) - 0} \right\} - 1\left\{ {(1 - a)(c - 1) - 0} \right\} + 1\left\{ {0 - (1 - a)(b - 1)} \right\} = 0\]
\[ \Rightarrow a\left\{ {(b - 1)(c - 1)} \right\} - \left\{ {(1 - a)(c - 1)} \right\} + \left\{ { - (1 - a)(b - 1)} \right\} = 0\]
Multiply the negative sign with the term inside the bracket that makes it look like the term in the denominator of the question.
\[ \Rightarrow a\left\{ {(b - 1)(c - 1)} \right\} + \left\{ {(1 - a)(1 - c)} \right\} + \left\{ {(1 - a)(1 - b)} \right\} = 0\]
Now divide both sides of the equation by \[(1 - a)(1 - b(1 - c)\]
\[ \Rightarrow \dfrac{{a\left\{ {(b - 1)(c - 1)} \right\} + \left\{ {(1 - a)(1 - c)} \right\} + \left\{ {(1 - a)(1 - b)} \right\}}}{{(1 - a)(1 - b)(1 - c)}} = \dfrac{0}{{(1 - a)(1 - b)(1 - c)}}\]
Separate the terms in LHS
\[ \Rightarrow \dfrac{{a(b - 1)(c - 1)}}{{(1 - a)(1 - b)(1 - c)}} + \dfrac{{(1 - a)(1 - c)}}{{(1 - a)(1 - b)(1 - c)}} + \dfrac{{(1 - a)(1 - b)}}{{(1 - a)(1 - b)(1 - c)}} = 0\]
Cancel same terms from numerator and denominator in each fraction in LHS
\[ \Rightarrow \dfrac{a}{{(1 - a)}} + \dfrac{1}{{(1 - b)}} + \dfrac{1}{{(1 - c)}} = 0\]
Add 1 on both sides of the equation
\[ \Rightarrow 1 + \dfrac{a}{{(1 - a)}} + \dfrac{1}{{(1 - b)}} + \dfrac{1}{{(1 - c)}} = 1 + 0\]
Take LCM of 1 with fraction including a
\[ \Rightarrow \dfrac{{1 - a + a}}{{(1 - a)}} + \dfrac{1}{{(1 - b)}} + \dfrac{1}{{(1 - c)}} = 1\]
Cancel terms with opposite signs in the numerator.
\[ \Rightarrow \dfrac{1}{{(1 - a)}} + \dfrac{1}{{(1 - b)}} + \dfrac{1}{{(1 - c)}} = 1\]
So the value of \[\dfrac{1}{{(1 - a)}} + \dfrac{1}{{(1 - b)}} + \dfrac{1}{{(1 - c)}}\]is 1

So, the correct option is B.

Note: Students many times make mistakes while writing the determinant as they don’t take negative signs before counting the second term. Also, while performing row transformations, carefully add or subtract the corresponding value from each element.
* Scalar triple product of three vectors is the dot product of one of the vectors with the cross product of the other two vectors. It is denoted as\[\left[ {\begin{array}{*{20}{c}}
  {\overrightarrow p }&{\overrightarrow q }&{\overrightarrow r }
\end{array}} \right] = \overrightarrow p \bullet (\overrightarrow q \times \overrightarrow r )\]
* Determinant of a matrix \[\left[ {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right] = a(ei - hf) - b(di - fg) + c(dh - eg)\]
* Row transformation is a way of transforming elements of a row using operations like addition, multiplication etc with respect to any other row of the matrix.