Show that the lines \frac{x-2}{1}= \frac{y-2}{3}= \frac{ z-3}{1}  and \frac{x-2}{1}= \frac{y-3}{4}= \frac{ z-4}{2} intersect.

Also, find the coordinates of the point of intersection. Find the equation of the plane containing the two lines

 

 

Answers (1)
S safeer

Given lines:

 \frac{x-2}{1}= \frac{y-2}{3}= \frac{ z-3}{1}=k 

Point on line: (2+k,2+3k,k+3) 

\frac{x-2}{1}= \frac{y-3}{4}= \frac{ z-4}{2}=l

Point on line: (2+l,3+4l,4+2l)

At intersect point 

 (2+l,3+4l,4+2l)\equiv (2+k,2+3k,k+3)

2+l=2+k \Rightarrow l=k

\\3+4l= 2+3k \\ \text{put k=l} \Rightarrow 3+4l= 2+3l \\

k=l=-1

Point of intersection (1,-1,2).

Equation of lien passing through two lines:

\left|\begin{array}{ccc} x-2 & y-2 & z-3 \\ 1 & 3 & 1 \\ 1 & 4 & 2 \end{array}\right|=0

\\ (x-2)(6-4) - (y-2)(2-1) +(z-3)(4-3)=0\\ 2x -4 -y+2+z-3 = 0 \\ 2x-y+z = 5

 

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