If the shortest distance between the straight lines 3 ( x - 1 ) = 6 ( y - 2 ) = 2 ( z - 1 ) and 4 ( x - 2 )…
If the shortest distance between the straight lines and is then the integral value of is equal to:
3
2
5
-1
Solution
The shortest distance \(d\) between two skew lines is given by \(d=\frac{|(\vec{a_{2}}-\vec{a_{1}})\cdot (\vec{b_{1}}\times \vec{b_{2}})|}{|\vec{b_{1}}\times \vec{b_{2}}|}\). Given \(d=\frac{1}{\sqrt{38}}\), we have \(\frac{|14-5\lambda |}{\sqrt{38}}=\frac{1}{\sqrt{38}}\). This implies \(|14-5\lambda |=1\). Solve for \(\lambda \): Two cases arise from the absolute value: Case 1: \(14-5\lambda =1\implies 5\lambda =13\implies \lambda =\frac{13}{5}\). Case 2: \(14-5\lambda =-1\implies 5\lambda =15\implies \lambda =3\). Determine the integral value of \(\lambda \): The integral value of \(\lambda \) from the possible solutions is \(3\).