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 $ $\begin{vmatrix} 1 & 1 & -2 \\[0.3em] k & 2 & 3 \\[0.3em] 3 &k & 2 \end{vmatrix} $ = 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