JEE Main 2016 — Three Dimensional Geometry Question with Solution

Let the direction vectors of the two lines be given by-$a_1\hat{i}+b_1\hat{j}+c_1\hat{k}=2\hat{i}+2\hat{j}+\hat{k}$ and $a_2\hat{i}+b_2\hat{j}+c_2\hat{k}=-\hat{i}+8\hat{j}+4\hat{k}$

The shortest distance between the two lines is given by-

$\frac{\left(\left(x_2-x_1\right)\hat{i}+\left(y_2-y_1\right)\hat{j}+\left(z_2-z_1\right)\hat{k}\right).\left[\left(a_1\hat{i}+b_1\hat{j}+c_1\hat{k}\right)\times \left(a_2\hat{i}+b_2\hat{j}+c_2\hat{k}\right)\right]}{\left|\left(a_1\hat{i}+b_1\hat{j}+c_1\hat{k}\right)\times \left(a_2\hat{i}+b_2\hat{j}+c_2\hat{k}\right)\right|}$

$=\left|\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-\hspace{0.17em}{y}_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\left|\begin{array}{ccc}i& j& k\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}\right|$

$=\left|\frac{\left|\begin{array}{ccc}-2& 4& 5\\ 2& 2& 1\\ -1& 8& 4\end{array}\right|}{\left|\begin{array}{ccc}i& j& k\\ 2& 2& 1\\ -1& 8& 4\end{array}\right|}\right|=\frac{54}{\sqrt{9^2+{18}^2}}=\frac{6}{\sqrt{5}}$

Now $\frac{6}{\sqrt{9}}<\frac{6}{\sqrt{5}}<\frac{6}{\sqrt{4}}\Rightarrow 2<\frac{6}{\sqrt{5}}<3$