Question

Let a, b, c be three distinct real numbers, none equal to one. If the vectors \(a\^i+\^j+\^k , \^i+b\^j+\^k\) and \( \^i+\^j+c\^k \) are coplanar, then  \(\frac{1}{ 1 − a }+ \frac{1}{ 1 − b} + \frac{1 }{1 − c }\) is equal to 

• A. -1
• B. 1
• C. -2
• D. 2

Hint:

Coplanar vectors have a scalar triple product of zero. Be careful when manipulating equations and look for algebraic techniques to simplify expressions

Answer 1

The Correct Option is B

We start with the determinant:

\[ \begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0 \]

Perform row operations to simplify the determinant:

\( C_2 \rightarrow C_2 - C_1, \quad C_3 \rightarrow C_3 - C_1 \)

This results in:

\[ \begin{vmatrix} a & 1-a & 1-a \\ 1 & b-1 & 0 \\ 1 & 0 & c-1 \end{vmatrix} \] 

Expand the determinant:

\( a(b - 1)(c - 1) - (1 - a)(c - 1) + (1 - a)(1 - b) = 0 \)

Simplify the equation:

\( a(1 - b)(1 - c) + (1 - a)(1 - c) + (1 - a)(1 - b) = 0 \)

Divide through by \( (1 - a)(1 - b)(1 - c) \) (assuming none of the terms are zero):

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

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

Rearrange the terms to isolate the result:

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

Therefore:

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