Question

Find the shortest distance between the lines

X / 4 = Y + 1 / 3 = Z- 2 / 2 and


5X - 2Y - 3Z + 6 = 0 = X - 3Y + 2Z - 3
​

Answer 1

Answer

Step-by-step explanation:

45

Answer 2

Step-by-step explanation:

Step 1. Expressing the first straight line in terms of direction cosines

The given straight line is

$\quad\dfrac{x}{4}=\dfrac{y+1}{3}=\dfrac{z-2}{2}$

The equations in terms of direction cosines be

$\quad\dfrac{x}{\frac{4}{\sqrt{29}}}=\dfrac{y+1}{\frac{3}{\sqrt{29}}}=\dfrac{z-2}{\frac{2}{\sqrt{29}}}$

Step 2. Expressing the second straight line in the symmetrical form

The given straight line is

$\quad 5x-2y-3z+6=0=x-3y+2z-3$

Let the direction ratios of the above straight line be $l,m,n$.

Since the straight line is perpendicular to the normals of both the planes, we have

$\quad\quad 5l-2m-3n=0$

$\quad\quad l-3m+2n=0$

By cross-multiplication, we get

$\quad\frac{l}{-4-9}=\frac{m}{-3-10}=\frac{n}{-15+2}$

$\Rightarrow \frac{l}{-13}=\frac{m}{-13}=\frac{n}{-13}$

$\Rightarrow \frac{l}{1}=\frac{m}{1}=\frac{n}{1}$

Hence the direction ratios of the straight line are $(1,1,1)$.

Let us take $z=0$. Then,

$\quad\quad 5x-2y+6=0$

$\quad\quad x-3y-3=0$

Solving, we get $x=-\frac{24}{13},y=-\frac{21}{13}$.

Therefore the point on the straight line is $(-\frac{24}{13},-\frac{21}{13},0)$.

Hence the equations of the straight line are

$\quad \dfrac{x+\frac{24}{13}}{1}=\dfrac{y+\frac{21}{13}}{1}=\dfrac{z}{1}$

Here the direction cosines of the above line are

$\quad\quad \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$.

So the line can be rewritten as,

$\quad\quad \dfrac{x+\frac{24}{13}}{\frac{1}{\sqrt{3}}}=\dfrac{y+\frac{21}{13}}{\frac{1}{\sqrt{3}}}=\dfrac{z}{\frac{1}{\sqrt{3}}}$

Step 3. Applying distance formula to find the shortest distance between the two given lines

We have found two straight lines

$\quad\quad\dfrac{x}{\frac{4}{\sqrt{29}}}=\dfrac{y+1}{\frac{3}{\sqrt{29}}}=\dfrac{z-2}{\frac{2}{\sqrt{29}}}$

$\quad\quad \dfrac{x+\frac{24}{13}}{\frac{1}{\sqrt{3}}}=\dfrac{y+\frac{21}{13}}{\frac{1}{\sqrt{3}}}=\dfrac{z}{\frac{1}{\sqrt{3}}}$

Applying the shortest distance formula between two straight lines, we get the required distance as,

$\dfrac{\left|\begin{array}{ccc}-\frac{24}{13}&1-\frac{21}{13}&2\\ \frac{4}{\sqrt{29}}&\frac{3}{\sqrt{29}}&\frac{2}{\sqrt{29}}\\ \frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}}\end{array}\right|}{\sqrt{(\frac{3}{\sqrt{87}}-\frac{2}{\sqrt{87}})^{2}+(\frac{4}{\sqrt{87}}-\frac{2}{\sqrt{87}})^{2}+(\frac{4}{\sqrt{87}}-\frac{3}{\sqrt{87}})^{2}}}$ units

$=\dfrac{\frac{1}{13\sqrt{29}\sqrt{3}}\left|\begin{array}{ccc}-24&-8&26\\ 4&3 &2\\ 1&1&1\end{array}\right|}{\sqrt{\frac{1}{87}+\frac{4}{87}+\frac{1}{87}}}$ units

$=\dfrac{\frac{1}{13\sqrt{87}}[-24(3-2)+8(4-2)+26(4-3)]}{\sqrt{\frac{6}{87}}}$ units [ Expanding along the first row ]

$=\dfrac{\frac{1}{13\sqrt{87}}[-24+16+26]}{\sqrt{\frac{6}{87}}}$ units

$=\dfrac{\frac{18}{13\sqrt{87}}}{\frac{\sqrt{6}}{\sqrt{87}}}$ units

$=\dfrac{18}{13\sqrt{6}}$ units

$=\dfrac{3\times\sqrt{6}\times\sqrt{6}}{13\times\sqrt{6}}$ units

$=\dfrac{3}{13}\sqrt{6}$ units

Answer:

The shortest distance is $\frac{3}{13}\sqrt{6}$ units.

Formula to find the shortest distance between to skew lines:

Let two skew lines be

$\quad\frac{x-x_{1}}{l_{1}}=\frac{y-y_{1}}{m_{1}}=\frac{z-z_{1}}{n_{1}}$

$\quad\frac{x-x_{2}}{l_{2}}=\frac{y-y_{2}}{m_{2}}=\frac{z-z_{2}}{n_{2}}$

where $(l_{1},m_{1},n_{1})$ and $(l_{2},m_{2},n_{2})$ are direction cosines.

The shortest distance formula is

$\dfrac{\left|\begin{array}{ccc}x_{2}-x_{1}&y_{2}-y_{1}&z_{2}-z_{1}\\l_{1}&m_{1}&n_{1}\\l_{2}&m_{2}&n_{2}\end{array}\right|}{\sqrt{\sum(m_{1}n_{2}-m_{2}n_{1})^{2}}}$