Question

Find the distance between the two given lines.

The x axis and=j-k+(2i-3j+k)t.

Short answer

The distance between both the lines are$\frac{2}{\sqrt{10}}$ .

Step-by-step solution

Given Condition

The given lines are x axis and $r=j-k+\left(2i-3j+k\right)t$.

Condition Used

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

To calculate the distance d between two skew lines

Then $\left|PQ\times n\right|$where n a unit vector is perpendicular to both lines, is the distance d. Now if A and A are vectors along the two lines, then $A\times B$is perpendicular to both, and n is just $A\times B$divided by $\left|A\times B\right|$. In this problem, consider the two points by putting =0.

This gives,P=(1,0,0) and Q=(0,1,-1) which means.

PQ=(0,1,-1)-(1,0,0)=(-1,1,-1)

The two vectors along the lines are as follows:

$$\begin{gathered} \mathrm{A}=\mathrm{i} \\ \mathrm{B}=2\mathrm{i}-3\mathrm{j}+\mathrm{k} \\ \mathrm{A}\times \mathrm{B}=\left(\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 1& 0& 0\\ 2& -3& 1\end{array}\right) \\ =-\mathrm{j}-3\mathrm{k} \\ \mathrm{n}=\frac{\mathrm{A}\times \mathrm{B}}{\left|\mathrm{A}\times \mathrm{B}\right|} \\ =\frac{-\mathrm{j}-3\mathrm{k}}{\sqrt{{\left(-1\right)}^2+{\left(-3\right)}^2}} \\ =\frac{-\mathrm{j}-3\mathrm{k}}{\sqrt{10}} \end{gathered}$$

Calculation of distance

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

$\mathrm{d}=\left|\mathrm{PQ}\times \mathrm{n}=\frac{1}{\sqrt{10}}\left|\left[\left(-1\right)\left(0\right)+\left(1\right)\left(-1\right)+\left(-1\right)\left(-3\right)\right]\right|=\frac{1}{\sqrt{10}}\times 2\right|=\frac{2}{\sqrt{10}}$

The distance between the two lines is $\frac{2}{\sqrt{10}}$.