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Numerade Educator



Problem 21 Medium Difficulty

Determine whether the lines $ L_1 $ and $ L_2 $ are parallel, skew, or intersecting. If they intersect, find the point of intersection.

$ L_1 : \frac{x - 2}{1} = \frac{y - 3}{-2} = \frac{z - 1}{-3} $
$ L_2 : \frac{x - 3}{1} = \frac{y + 4}{3} = \frac{z - 2}{-7} $



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Video Transcript

the question they are asking to determine whether lines L one and L two are parallel skew or intersecting and if they intersect then by the point of their intersection, everyone is given us X -2 x one equal to 1 -3. Where -2 is equal to that -1 by minus three. This can be equated to be equal to T. Okay. Mhm. Oh constant value and nine. L 2 is given by x minus three by one is equal to Y plus four by three is equal to zero minus two by minus seven. This can be equated to be equal to S this is another constant value. So From these two questions we can find the value of L one for each of the coordinates X equal to P plus two by equal to minus 20 plus three and that equal to minus 30 plus one and L two has coordinates X equal to x plus three, Y equals three, x minus four and they're equal to minus seven is plus two. So the first condition to take L one and help to our panel or not. So you have to compute the issues of the coefficients of the coordinates of border lines. That is one by one is not equal to minus two way three is not equal to minus three by minus seven. Hence lying L one and L two are not parallel to each other. Mhm. And the second condition is to take L one and L two are intersecting or not to take this, we have to equate the coordinates of the two lines L one and L two. So the equations are for the three coordinates express trees equality plus two, three years minus 40 to minus two. D plus three and minus seven is plus two is equal to minus three T plus one. So from the first equation we can get the value of in terms of S. That is equal to X plus one and putting the value of this T. In another equation we get three years minus four is equal to minus two. Deep mystery where S. Is equal to one. After finding the value of S. We put the value of S. In other equation to find the value of T. That is minus seven is plus two is equal to minus three, T plus one, where T. Is equal to two is the value and putting border values is equal to one and equal to in Mhm. Yeah. Now if we put the value of S. one and T. Two that is one and two values of S and T. In the equations L. One and L two respectively. Then we can find the point of intersections of these lines. So oh if you put the value of equal to in like L one then the value of experience ideas poor minus one and minus five respectively. And If we put the value of 0-1 in line L two equation then we can find the same code in net That is the point of intersection of these two lines, L one and L two, and hence this is the required point of intersection for the given question.