Let the square of the shortest distance between the lines \( \mathrm{L_1}:\vec{r}=\left(\hat{\imath}+2\hat{\jmath}-4\hat{k}\right)+\lambda\left(2\hat{\imath}+3\hat{\jmath}+6\hat{k}\right) \) and \( \mathrm{L_2}:\vec{r}=\left(3\hat{\imath}+3\hat{\jmath}-5\hat{k}\right)+\mu\left(2\hat{\imath}+3\hat{\jmath}+6\hat{k}\right) \) is \( \displaystyle\frac{\mathrm{m}}{n} \), where \( \gcd(m,n)=1 \). Then \( m+n \) is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 342
Both lines have the same direction \( \vec{b}=2\hat{\imath}+3\hat{\jmath}+6\hat{k} \), so they are parallel. For parallel lines:
\( \displaystyle d=\frac{\left|\vec{b}\times(\vec{a_2}-\vec{a_1})\right|}{|\vec{b}|} \)
\( \displaystyle \vec{a_2}-\vec{a_1}=2\hat{\imath}+\hat{\jmath}-\hat{k} \)
\( \displaystyle |\vec{b}|=\sqrt{4+9+36}=7 \)
\( \displaystyle \vec{b}\times(\vec{a_2}-\vec{a_1})=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\2 & 3 & 6\\2 & 1 & -1\end{vmatrix} \)
\( \displaystyle =(-3-6)\hat{\imath}-(-2-12)\hat{\jmath}+(2-6)\hat{k} \)
\( \displaystyle =-9\hat{\imath}+14\hat{\jmath}-4\hat{k} \)
\( \displaystyle \left|\vec{b}\times(\vec{a_2}-\vec{a_1})\right|^{2}=81+196+16=293 \)
\( \displaystyle d^{2}=\frac{293}{49} \)
293 is a prime number, so gcd(293, 49) = 1. Hence m = 293, n = 49 and
\( \displaystyle m+n=293+49=342 \)
Tip: Check the direction vectors first: equal (or proportional) directions mean parallel lines, and the skew-lines formula with \( \vec{b_1}\times\vec{b_2} \) cannot be used because that cross product is zero.