Three Dimensional Geometry Question 23
Question 23 - 30 January - Shift 2
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
(1) $\frac{17}{5}$
(2) $\frac{5}{17}$
(3) $\frac{6}{13}$
(4) $\frac{13}{6}$
Show Answer
Answer: (4)
Solution:
Formula: Equation of a plane, Vector product of two vectors
$ \begin{aligned} & \frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1} \\ & \frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-1} \end{aligned} $
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}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & k-1 & -2 \\ 1 & k-1 & -3\end{vmatrix} $
$=\hat{i}(-3 k+3+2 k 2)-\hat{j}(6+2)+\hat{k}(-2 k+2-k+1 )$
$=(1-k) \hat{i}-8 \hat{j}+(3-3 k) \hat{k}$
The line $\frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-1}$ is perpendicular to normal vector.
$\therefore 1(1-k)+1(-8)+(-1)(3-3 k)=0$
$\Rightarrow 1-k-8-3+3 k=0$
$\Rightarrow 2 k=10 \Rightarrow k=5$
$\therefore \frac{k^{2}+1}{(k-1)(k-2)}=\frac{26}{4 \cdot 3}=\frac{13}{6}$
