Question

Show that the lines $\dfrac{x+1}{3}=\dfrac{y+3}{5}=\dfrac{z+5}{7}$ and $\dfrac{x-2}{1}=\dfrac{y-4}{3}=\dfrac{z-6}{5}$ intersect. Also find the point of intersection.

Answer 1

Hint: Suppose that $\dfrac{x+1}{3}=\dfrac{y+3}{5}=\dfrac{z+5}{7}=m$ and $\dfrac{x-2}{1}=\dfrac{y-4}{3}=\dfrac{z-6}{5}=n$. Now, take two general points in the form of $m$ and $n$ i.e. $\left( 3m-1,5m-3,7m-5 \right)$ for first line and $\left( n+2,3n+4,5n+6 \right)$for the second line. Now in order to find whether these two lines intersect or not, equate these two points, form three equations in $m\And n$ and after solving them get the value of $m$ and $n$. So, if $m$ and $n$ exists then these two lines intersect and the point of intersection can be found by putting the value of $m$ or $n$ in any of the general points.

Complete step by step answer:
We have given two lines $\dfrac{x+1}{3}=\dfrac{y+3}{5}=\dfrac{z+5}{7}$ and $\dfrac{x-2}{1}=\dfrac{y-4}{3}=\dfrac{z-6}{5}$ . Now we have to find whether these two intersect or not.
Let’s take a general point on these two lines.
Suppose that $\dfrac{x+1}{3}=\dfrac{y+3}{5}=\dfrac{z+5}{7}=m$ and $\dfrac{x-2}{1}=\dfrac{y-4}{3}=\dfrac{z-6}{5}=n$.
Now for the first line, we have
$\Rightarrow x=3m-1,y=5m-3$ and $z=7m-5$
Similarly, for the second line, we have
$\Rightarrow x=n+2,y=3n+4$ and $z=5n+6$
So, the general points on these lines are $\left( 3m-1,5m-3,7m-5 \right)$ and $\left( n+2,3n+4,5n+6 \right)$.
Hence, lines intersect if $3m-1=n+2$, $5m-3=3n+4$ and $7m-5=5n+6$ for some $m$ and $n$
The above three equations can also be written as,
$3m-n=3..............(1)$
$5m-3n=7............(2)$
$7m-5n=11...........(3)$
From equation (1), we have $3m-n=3\Rightarrow n=3m-3$
Now, substituting the value of $n$ in equation (2), we get
$\Rightarrow 5m-3(3m-3)=7$
$\Rightarrow 5m-9m+9=7$
$\Rightarrow -4m=7-9=-2$
$\Rightarrow m=\dfrac{-2}{-4}=\dfrac{1}{2}$
Now as $m=\dfrac{1}{2}$, so $n=3\times \dfrac{1}{2}-3=-\dfrac{3}{2}$
Putting the value of $m$ and $n$ in equation (3), we get
$\Rightarrow 7\times \dfrac{1}{2}-5\times -\dfrac{3}{2}=\dfrac{7}{2}+\dfrac{15}{2}=11$
Hence, $\left( m=\dfrac{1}{2},n=-\dfrac{3}{2} \right)$ satisfies all the three equations which means that these two lines intersect with each other.
Now, in order to find the point of intersection, putting the value of $m$ in the general point of first line
$\left( 3m-1,5m-3,7m-5 \right)$.
$\Rightarrow \left( 3m-1,5m-3,7m-5 \right)=\left( 3\times \dfrac{1}{2}-1,5\times \dfrac{1}{2}-3,7\times \dfrac{1}{2}-5 \right)=\left( \dfrac{1}{2},\dfrac{-1}{2},\dfrac{-3}{2} \right)$
Hence, the point of intersection of these two lines is $\left( \dfrac{1}{2},\dfrac{-1}{2},\dfrac{-3}{2} \right)$.

Note:
This question involves the basic knowledge of 3D lines and calculating their point of intersection. The process to check whether two lines are parallel or not , if at least a common point exists on these two lines then we can say that those two lines intersect. While solving for $m$ and $n$ we have three equations but only two variables, so we can find the value of these two variables by solving any two from these three equations. Now students often make one mistake in this step that they forget to check whether the value that they got from solving any two equations satisfies the third equation or not. If it satisfies the third equation then only we can say that the lines will intersect and hence get the point of intersection.