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JEE Mathematics 2023 Question with Solution

Let a,b,ca, b, c be three distinct real numbers, none equal to one. If the vectors ai^+j^+k^, i^+bj^+k^, ai^+j^+ck^a \hat{i} + \hat{j} + \hat{k}, \, \hat{i} + b\hat{j} + \hat{k}, \, a \hat{i} + \hat{j} + c \hat{k} are coplanar, then 11−a+11−b+11−c\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} is equal to:

  • A

    11

  • B

    −1-1

  • C

    −2-2

  • D

    22

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The vectors ai^+j^+k^a \hat{i} + \hat{j} + \hat{k}, i^+bj^+k^\hat{i} + b\hat{j} + \hat{k} and ai^+j^+ck^a \hat{i} + \hat{j} + c \hat{k} are coplanar.

Find: The value of 11−a+11−b+11−c\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}.

For coplanar vectors, the scalar triple product is zero. Hence,

∣a111b111c∣=0\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0

Apply the column operations C2→C2−C1C_2 \rightarrow C_2 - C_1 and C3→C3−C1C_3 \rightarrow C_3 - C_1. Then the determinant becomes

∣a1−a1−a1b−1010c−1∣=0\begin{vmatrix} a & 1-a & 1-a \\ 1 & b-1 & 0 \\ 1 & 0 & c-1 \end{vmatrix} = 0

Expanding this determinant,

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

which can be written as

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

Now divide throughout by (1−a)(1−b)(1−c)(1-a)(1-b)(1-c), which is valid because none of a,b,ca, b, c equals 11:

a1−a+11−b+11−c=0\frac{a}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0

Use

a1−a=11−a−1\frac{a}{1-a} = \frac{1}{1-a} - 1

So the equation becomes

−1+11−a+11−b+11−c=0-1 + \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0

Therefore,

11−a+11−b+11−c=1\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1

Therefore, the required value is 11. The solution working gives option A. The solution labels the correct option as B, which disagrees with its own derivation.

Why the simplification works

The key algebraic step is rewriting

a1−a\frac{a}{1-a}

as

a1−a=1−(1−a)1−a=11−a−1\frac{a}{1-a} = \frac{1-(1-a)}{1-a} = \frac{1}{1-a} - 1

This converts the expression directly into the required sum. Without this identity, students often stop one step early.

Common mistakes

  • Using the determinant incorrectly for coplanarity. For three vectors, coplanarity means their scalar triple product is zero, so the determinant must be set equal to 00. Do not use a dot-product condition instead.

  • Making a sign error after column operations. When converting terms like (b−1)(c−1)(b-1)(c-1), keep track that this equals (1−b)(1−c)(1-b)(1-c), but isolated factors such as (c−1)(c-1) become −(1−c)-(1-c). Recheck each sign carefully.

  • Dividing by (1−a)(1−b)(1−c)(1-a)(1-b)(1-c) without using the given condition. This division is valid only because none of a,b,ca, b, c equals 11. Always verify denominators are nonzero before dividing.

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