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

Question

If the vectors $a \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+b \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+\hat{\mathbf{j}}+c \hat{\mathbf{k}}$ are coplanar, where $(a, b, c \neq 1$ ), then the value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=$, 2, 0, -1, 1

MARKS App · Accepted Answer

For coplanar vectors, $\Rightarrow \quad\left|\begin{array}{ccc} a_x & a_y & a_z \ b_x & b_y & b_z \ c_x & c_y & c_z \end{array}\right|=0$ Substituting values we have, $\left|\begin{array}{lll}a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c\end{array}\right|=0$ $R_2 \rightarrow R_2-R_1 \text { and } R_3 \rightarrow R_3 \rightarrow R_1$ $\Rightarrow\left|\begin{array}{ccc} a & 1 & 1 \ 1-a & b-1 & 0 \ 1-a & 0 & c-1 \end{array}\right|=0$ $\Rightarrow a(b-1)(c-1)-(1-a)(c-1)-(1-a)(b-1)=0$ $\Rightarrow$ Dividing by $(1-a)(1-b)(1-c)$, we get $\begin{aligned} \frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} & =0 \ \Rightarrow \quad \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} & =\frac{1}{1-a}-\frac{a}{1-a}=1 \end{aligned}$

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