If the shortest distance between the lines \( \vec{r}=(\hat{\imath}+2\hat{\jmath}+3\hat{k})+\lambda(2\hat{\imath}+3\hat{\jmath}+4\hat{k}) \) and \( \vec{r}=(a\hat{\imath}+4\hat{\jmath}+5\hat{k})+\mu(4\hat{\imath}+6\hat{\jmath}+8\hat{k}) \), \( a\in N \), where λ and μ are parameters, is \( \sqrt{\displaystyle\frac{5}{29}} \), then \( a \) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
The direction \( 4\hat{\imath}+6\hat{\jmath}+8\hat{k}=2(2\hat{\imath}+3\hat{\jmath}+4\hat{k}) \), so the lines are parallel. For parallel lines \( \vec{r}=\vec{a}_1+\lambda\vec{b} \) and \( \vec{r}=\vec{a}_2+\mu\vec{b} \):
\( \displaystyle d=\frac{|\vec{b}\times(\vec{a}_2-\vec{a}_1)|}{|\vec{b}|} \)
Here \( \vec{b}=2\hat{\imath}+3\hat{\jmath}+4\hat{k} \), \( |\vec{b}|=\sqrt{29} \), and \( \vec{a}_2-\vec{a}_1=(a-1)\hat{\imath}+2\hat{\jmath}+2\hat{k} \).
\( \displaystyle \vec{b}\times(\vec{a}_2-\vec{a}_1)=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\2 & 3 & 4\\a-1 & 2 & 2\end{vmatrix} \)
\( \displaystyle =(6-8)\hat{\imath}-(4-4(a-1))\hat{\jmath}+(4-3(a-1))\hat{k} \)
\( \displaystyle =-2\hat{\imath}+(4a-8)\hat{\jmath}+(7-3a)\hat{k} \)
\( \displaystyle d=\frac{\sqrt{4+(4a-8)^2+(7-3a)^2}}{\sqrt{29}}=\sqrt{\frac{5}{29}} \)
\( \displaystyle 4+(4a-8)^2+(7-3a)^2=5 \)
\( \displaystyle 25a^2-106a+112=0 \)
\( \displaystyle (a-2)(25a-56)=0\ \Rightarrow\ a=2 \text{ or } a=\frac{56}{25} \)
Since \( a\in N \), a = 2.
Shortcut: Put the options into \( (4a-8)^2+(7-3a)^2=1 \): a = 2 gives 0 + 1 = 1. a = 0 gives 64 + 49, a = 1 gives 16 + 16, a = 3 gives 16 + 4, none equal to 1.
Tip: Check first whether the direction vectors are proportional: for parallel lines use the cross-product formula with one direction vector, not the skew-lines formula, whose denominator would be zero.