Let a, b, c be three distinct real numbers, none equal to one. If the vectors \(a\hat{i}+\hat{j}+\hat{k},\hat{i}+b\hat{j}+\hat{k}\) and \(\hat{i}+\hat{j}+c\hat{k}\) are coplanar, then \(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\) is equal to
- -1
- 1
- -2
- 2
The Correct Option is B
Solution and Explanation
Top Questions on Vectors
Match the LIST-I with LIST-II
LIST-I (Expressions) LIST-II (Values) A. \( i^{49} \) I. 1 B. \( i^{38} \) II. \(-i\) C. \( i^{103} \) III. \(i\) D. \( i^{92} \) IV. \(-1\)
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A. \(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = 0\)
B. \(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = 0\)
C. \(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = 0\)
D. \(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \neq 0\)
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