Question

Find the shortest distance between the lines
$\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}and\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$

Answer 1

The given lines are
$\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}and\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$
For line 1st dr's = (7, -6, 1) and passes through (-1, -1, -1) then equation of given lines (in vector form) are
$\Rightarrow r_1=-\hat{i}-\hat{j}-\hat{k}+\lambda (7\hat{i}-6\hat{j}+\hat{k})$
Similarly, $r_2=3\hat{i}+5\hat{j}+7\hat{k}+\mu (\hat{i}-2\hat{j}+\hat{k})$
Which are of the form $r_1=a_1+\lambda b_{1}and{r}_2=a_2+\mu b_2$
where, $a_1=-\hat{i}-\hat{j}-\hat{k},b_1=7\hat{i}-6\hat{j}+\hat{k}$
and $a_2=3\hat{i}+5\hat{j}+7\hat{k},b_2=\hat{i}-2\hat{j}+\hat{k}$
Now, $a_2-a_1=(3\hat{i}+5\hat{j}+7\hat{k})-(-\hat{i}-\hat{j}-\hat{k})=4\hat{i}+6\hat{j}+8\hat{k}$
and $b_1\times b_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 7& -6& 1\\ 1& -2& 1\end{array}\right|=\hat{i}(-6+2)-\hat{j}(7-1)+\hat{k}(-14+6)=-4\hat{i}-6\hat{j}-8\hat{k}$
$|b_1\times b_2|=\sqrt{(-4{)}^2+(-6{)}^2+(-8{)}^2}=\sqrt{16+36+64}=\sqrt{116}=2\sqrt{29}$
$\therefore$ SD between the given lines
$d=\left|\frac{(b_1\times b_2).(a_2-a_1)}{|b_1\times b_2|}\right|=\frac{|(-4\hat{i}-6\hat{j}-8\hat{k}).(4\hat{i}+6\hat{j}+8\hat{k})|}{2\sqrt{29}}=\frac{|(-4)\times 4+(-6)\times 6+(-8)\times 8|}{2\sqrt{29}}=\frac{|-16-36-64|}{2\sqrt{29}}=\frac{116}{2\sqrt{29}}=\frac{58}{\sqrt{29}}=2\sqrt{29}units$