91Ó°ÊÓ

The given lines are $r=(4,3,-1)+(1,1,1)tandr=(4,-1,1)+(1,-2,-1)t$.

Introduction of a unit vector perpendicular to both lines and then solving for the value of the distance between the given vector and the unit vector gives the distance between the given lines.

Then $\left|PQ.n\right|$where n a unit vector is perpendicular to both lines, is the distance . Now if A and A are vectors along the two lines, then $A×B$is perpendicular to both, and n is just $A×B$divided by $\left|A×B\right|$. In this problem, consider the two points by putting t=0. This gives,$P=\left(4,3,-1\right)$and localid="1658818007480" $Q=\left(4,-1,1\right)$which means the following.

$PQ=(4,-1,-1)-(4,3,-1)=(0,-4,-2)$

The two vectors along the lines are as follows.

$$\begin{gathered} A=i+j+k \\ B=i-2j-k \\ A×B=\left(\begin{array}{ccc}i& j& k\\ 1& 1& 1\\ 1& -2& 1\end{array}\right) \\ =i+2j-3k \end{gathered}$$

$$\begin{gathered} n=\frac{A×B}{\left|A×B\right|} \\ =\frac{i+2j-3k}{\sqrt{{\left(1\right)}^2+{\left(2\right)}^2+{\left(-3\right)}^2}} \\ =\frac{i+2j-3k}{\sqrt{14}} \end{gathered}$$

The formula for distance between two lines is given as follows.

$d=\left|PQâ‹\dots n=\frac{1}{\sqrt{14}}\left|\right[\left(0\right)\left(1\right)+(-4)\left(2\right)+\left(2\right)(-3)\left]\right|=\frac{1}{\sqrt{14}}×14)\right|=\sqrt{14}$

The distance between the two lines is $\sqrt{14}$.