SATHEE: Vectors 3 Question 22

Vectors 3 Question 22

22. If the vectors $a \hat{i} + \hat{j} + \hat{k}, \hat{i} + b \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + c \hat{k} (a \neq b \neq c \neq 1)$ are coplanar, then the value of $\frac{1}{(1 - a)} + \frac{1}{(1 - b)} + \frac{1}{(1 - c)} = \dots \dots .$ .

(1987, 2M)

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Answer:

Correct Answer: 22. (1)

Solution:

Since, vectors are coplanar.

$∴ | \begin{array}{lll} a & 1 & 1 1 & b & 1 1 & 1 & c \end{array} | = 0$

Applying $R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$ ,

$\begin{array}{cc} | \begin{array}{ccc} a & 1 & 1 \\ 1 - a & b - 1 & 0 \\ 1 - a & 0 & c - 1 \end{array} | = 0 \\ \Rightarrow & | \begin{array}{ccc} a / (1 - a) & 1 / (1 - b) & 1 / (1 - c) \\ 1 & - 1 & 0 \\ 1 & 0 & - 1 \end{array} | = 0 \\ \Rightarrow \frac{a}{1 - a} (1) - \frac{1}{1 - b} (- 1) + \frac{1}{1 - c} (1) = 0 \\ \Rightarrow \frac{a}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = 0 \\ \Rightarrow \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - b} + \frac{1}{1 - c} = 0 \\ \Rightarrow & = 1 \end{array}$

Vectors 3 Question 22

22. If the vectors $a \hat{i} + \hat{j} + \hat{k}, \hat{i} + b \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + c \hat{k} (a \neq b \neq c \neq 1)$ are coplanar, then the value of $\frac{1}{(1 - a)} + \frac{1}{(1 - b)} + \frac{1}{(1 - c)} = \dots \dots .$ .

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Vectors 3 Question 22

22. If the vectors ai^+j^+k^,i^+bj^+k^ and i^+j^+ck^(a≠b≠c≠1) are coplanar, then the value of 1(1−a)+1(1−b)+1(1−c)=……..

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22. If the vectors $a \hat{i} + \hat{j} + \hat{k}, \hat{i} + b \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + c \hat{k} (a \neq b \neq c \neq 1)$ are coplanar, then the value of $\frac{1}{(1 - a)} + \frac{1}{(1 - b)} + \frac{1}{(1 - c)} = \dots \dots .$ .