Let a,b,c be three distinct reals, none equal to 1.

Question

Let $a,b,c$ be three distinct reals, none equal to 1. If $a\hat i+\hat j+\hat k$, $\hat i+b\hat j+\hat k$, $\hat i+\hat j+c\hat k$ are coplanar, then $\dfrac1{1-a}+\dfrac1{1-b}+\dfrac1{1-c}$ is:

Accepted Answer

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

Using column operations $C_2\to C_2-C_1$, $C_3\to C_3-C_1$:
$$\begin{vmatrix}a&1-a&1-a\1&b-1&0\1&0&c-1\end{vmatrix}=0$$

Expanding: $a(b-1)(c-1) - (1-a)(c-1)+(1-a)(1-b)=0$, which rearranges to
$a(1-b)(1-c)+(1-a)(1-c)+(1-a)(1-b)=0$

Dividing through by $(1-a)(1-b)(1-c)$:
$$\frac a{1-a} + \frac1{1-b} + \frac1{1-c} = 0 \implies -1+\frac1{1-a}+\frac1{1-b}+\frac1{1-c}=0$$
$$\frac1{1-a}+\frac1{1-b}+\frac1{1-c} = 1$$

Solutions · 1