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.
$$-5$$
B.
5
C.
2
D.
$$-2$$
Answer :
$$-5$$
Solution :
The two lines intersect if shortest distance between them is zero i.e.
$$\frac{{\left( {{{\vec a}_2} - {{\vec a}_1}} \right).{{\vec b}_1} \times {{\vec b}_2}}}{{\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|}} = 0\,\,\,\,\,\, \Rightarrow \left( {{{\vec a}_2} - {{\vec a}_1}} \right).{{\vec b}_1} \times {{\vec b}_2} = 0$$
Where,
$$\eqalign{
& {{\vec a}_1} = \hat i + 2\hat j + 3\hat k,\,\,\,\,{{\vec b}_1} = k\hat i + 2\hat j + 3\hat k \cr
& {{\vec a}_2} = 2\hat i + 3\hat j + \hat k,\,\,\,\,{{\vec b}_2} = 3\hat i + k\hat j + 2\hat k \cr} $$
\[ \Rightarrow \left| \begin{array}{l}
1\,\,\,\,\,1\,\,\,\, - 2\\
k\,\,\,\,\,2\,\,\,\,\,\,\,\,\,3\\
3\,\,\,\,\,k\,\,\,\,\,\,\,\,\,2
\end{array} \right| = 0\]
$$\eqalign{
& \Rightarrow 1\left( {4 - 3k} \right) - 1\left( {2k - 9} \right) - 2\left( {{k^2} - 6} \right) = 0 \cr
& \Rightarrow - 2{k^2} - 5k + 25 = 0 \cr
& \Rightarrow k = - 5{\text{ or }}\frac{5}{2} \cr} $$
$$\because \,\,k$$ is an integer, therefore $$k=-5$$