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

To determine the value of \(\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}\), given that the vectors \(a\^i+\^j+\^k\), \(\^i+b\^j+\^k\), and \(\^i+\^j+c\^k\) are coplanar, follow these steps:

Vectors are said to be coplanar if their scalar triple product is zero. The scalar triple product of the given vectors is calculated as:

1. Represent the vectors as: 

• \(\vec{v}_1 = a\hat{i} + \hat{j} + \hat{k}\)
• \(\vec{v}_2 = \hat{i} + b\hat{j} + \hat{k}\)
• \(\vec{v}_3 = \hat{i} + \hat{j} + c\hat{k}\)

2. The scalar triple product is given by the determinant of a matrix formed by these vectors:

\(\left|\begin{array}{ccc} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \\ \end{array}\right|\)

3. Calculate the determinant:

\(\begin{array}{l} = a(bc - 1) - 1(c - 1) + 1(1 - b) \\ = abc - a - c + 1 + 1 - b \\ = abc - a - b - c + 2 \end{array}\)

4. Set the determinant to zero (condition for coplanarity):

\(abc - a - b - c + 2 = 0\)

5. Rearranging the equation, we get:

\(abc = a + b + c - 2\)

6. Now, find the value of \(\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}\). This can be transformed using the relation found above:

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

7. The numerator simplifies to:

\((1-b)(1-c) + (1-a)(1-c) + (1-a)(1-b) = 3 - 2(a + b + c) + ab + bc + ca\)

8. Using \(abc = a + b + c - 2\)\) equation; substitute and simplify:

Numerator becomes: \\( 3 - 2(a + b + c) + (a+b+c-2) + 2 \\(a+b+c) - 6 + abc \\)

Combine and simplify terms:

\\( ab + bc + ca - a - b - c + 1 = 1\\)

Since the scalar triple product is zero (due to coplanarity), we conclude that:

The value of \(\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}\) is 1.