Question

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

• A. $\frac{13}{6}$
• B. $\frac{5}{17}$
• C. $\frac{17}{5}$
• D. $\frac{6}{13}$

Answer 1

Answer: (A)

$\frac{x-1}{1}=\frac{2y+1}{2}=\frac{z+1}{-1}$
$\frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-1}$
$\text{ Points : }A(-1,k,0),B(2,k,-1),C(1,1,2)$
$\overrightarrow{CA}=-2\hat{i}+(k-1)\hat{j}-2\hat{k}$
$\overrightarrow{CB}=\hat{i}+(k-1)\hat{j}-3\hat{k}$
$\overrightarrow{CA}\times \overrightarrow{CB}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& k-1& -2\\ 1& k-1& -3\end{array}\right|$
$=\hat{i}(-3k+3+2k-2)-\hat{j}(6+2)+\hat{k}(-2k+2-k+1)$
$=(1-k)\hat{i}-8\hat{j}+(3-3k)\hat{k}$
The line $\frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-1}$ is perpendicular to normal vector.
$\therefore 1\cdot (1-k)+1(-8)+(-1)(3-3k)=0$
$\Rightarrow 1-k-8-3+3k=0$
$\Rightarrow 2k=10\Rightarrow k=5$
$\therefore \frac{k^2+1}{(k-1)(k-2)}=\frac{26}{4\cdot 3}=\frac{13}{6}$