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Show that the lines \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}  and  \frac{x-4}{2}=\frac{y}{0}=\frac{z+1}{3}  intersect. Find their point of intersection.

 

 

 

 
 
 
 
 

Answers (1)

L_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}=\lambda   and  L_2: \frac{x-4}{2}=\frac{y}{0}=\frac{z+1}{3}=\mu

Therefore, coordinates of random points on the given lines are P(3\lambda +1,-\lambda +1,-1)\; \; \&\; \; Q(2\mu +4,0,3\mu -1)

If the lines intersect, then they have a common point for some value of \lambda and  \mu.

3\lambda -2\mu=3\; \; \; -(i),\; \; \; -\lambda +1=0\; \; \; -(ii),\; \; \; 3\mu -1=-1\; \; \;- (iii)

Solving (i) and (ii) we get : \lambda =1  and  \mu = 0

As LHS of (i) : 3\lambda -2\mu =3\times 1-2\times 0=3 =RHS of (i)

Since \lambda =1  and  \mu =0  satisfy the equation (i).

\therefore The given lines intersect. The point of intersection is P or Q (4,0,-1)

 

Posted by

Ravindra Pindel

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