Question

If the straight lines $\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x-2}{3}=\frac{y-3}{k}=\frac{z-1}{2}$ intersect at a point, then the integer k is equal to

• A. -2
• B. 5
• C. 2
• D. -5

Answer 1

Answer: (D)

The two lines intersect if shortest distance between them is zero i.e.,
$\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot \overrightarrow{b_1}\times \overrightarrow{b_2}}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}$ = 0 $\Rightarrow (\overrightarrow{a_2}-\overrightarrow{a_1})\cdot \overrightarrow{b_1}\times \overrightarrow{b_2}$ = 0 where
$\overrightarrow{a_1}=\hat{i}+2\hat{j}+3\hat{k},\overrightarrow{b_1}=k\hat{i}+2\hat{j}+3\hat{k}$
$\overrightarrow{a_2}=2\hat{i}+3\hat{j}+\hat{k},\overrightarrow{b_2}=3\hat{i}+k\hat{j}+2\hat{k}$
$\therefore$ $\left|\begin{array}{ccc}1& 1& -2\\ k& 2& 3\\ 3& k& 2\end{array}\right|$ = 0
$\Rightarrow$ $1(4-3k)-(2k-9)-2(k^2-6)$ = 0
$\Rightarrow$ $-2k^2-5k+25$ = 0 $\Rightarrow$ = -3 or $\frac{5}{2}$ But $k$ is an integer. $\therefore$ $k$ =.- 5