If a plane passes through the points (-1, k, 0),(2, k,-1),(1,1,2) and is parallel to the l…

Question

If a plane passes through the points $(-1, k, 0),(2, k,-1),(1,1,2)$ and is parallel to the line $\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}$, then the value of $\frac{k^2+1}{(k-1)(k-2)}$ is :

Accepted Answer

Let $$P \equiv ( - 1,k,0),Q \equiv (2,k, - 1)$$ \& $R(1,1,2)$

$$\overrightarrow P R = 2\widehat i + (1 - k)\widehat j + 2\widehat k$$

\& $$\overrightarrow Q R =  - \widehat i + (1 - k)\widehat j + 3\widehat k$$

$$\therefore$$ Normal to plane will be

$$\left| {\begin{pmatrix} {\widehat i} & {\widehat j} & {\widehat k}  \
   2 & {(1 - k)} & 2  \
   { - 1} & {(1 - k)} & 3  \

\end{pmatrix} } \right| = \widehat i(1 - k) - \widehat j(8) + 3\widehat k(1 - k)$$

If line is parallel to this we have

$1(1 - k) + 1( - 8) + ( - 3)(1 - k) = 0$

$$ \Rightarrow 2(1 - k) =  - 8$$

$$ \Rightarrow 1 - k =  - 4 \Rightarrow k = 5$$

$$\therefore$$ $${{{k^2} + 1} \over {(k - 1)(k - 2)}} = {{26} \over {4.3}} = {{13} \over 6}$$

Solutions · 1