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SB
Numerade Educator

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Problem 79 Hard Difficulty

Let $ L_1 $ be the line through the origin and the point $ (2, 0 , -1) $. Let $ L_2 $ be the line through the points $ (1, -1 , 1) $ and $ (4, 1 , 3) $. Find the distance between $ L_1 $ and $ L_2 $.

Answer

$\frac{13}{\sqrt{69}}$

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Video Transcript

the question there asking to find The distance between the lines. L one Line passes through original and the point A 20-1. Underline L two. Mhm Passes through the Point B 1 -1 one and C41 three. So these two lines are having these coordinates and we have to find the distance between these. So in order to find first we have to find the equation of these two lines. So the line one has origin, that is the coordinates Or the original 000. And and the another point that passes through this line is 20 minus one. So therefore the direction vector poor line L one B one vector is equal to a vector that can be found using these two points Uh to zero is 20 minus 00 and minus one minus zero minus ones. Therefore the direction vector Poland Elvan is 20 minus one. And the point We can consider any of the points that passes through this line. So let us consider the origin 000. So using the direction vector and the point we can find the equation of L one. So uh the direction vector into x minus the corresponding coordinate value. Um All of the of values of each of the coordinates together Some equal to zero in order to find the equation of a line. So uh after equating this value, the equation of line L one is equal to X- that is equal to zero. So therefore we have to find also the direction vectors and the point of flying L two. So the two points that passes through the line delta is given by be one minus 11 and c 41 and three. So therefore the direction vector is equal to B. C vector, that is equal to 3 to 2. The direction numbers for this victim. Uh Line L two. And therefore um we can consider any of the points that passes to this line. Let us consider B. And so therefore the equation for the line will be three expressed to a place to the recorded three for line L two. So now, in order to find the distance between the line L one and L2, we have to find the value of the formula. That is uh used for calculating the distance between the lines. That is uh the any two points uh that exists in these two lines. Um So uh we have to find the yeah direction vector of uh these two uh using these two points along the line that connects these two lines. So that will be P one, p two victor. Uh with the dot product of the normal vector uh of both of these lines divided by the scalar value of the normal vector of these two lines. So therefore we first we have to calculate the cross product of both the direction victims of these two lines. That is the normal vector of the mhm of line. That will be the cross product of these two victors of like L. One and L two. So after calculating the cross product, the value of uh the normal vector is equal to two minus seven and four. So therefore in order to find P one P two we have To find now the any of the coordinates we take for the line L2 that is um the point B. Um for this and um and any of the point for from one we can take that to the origin. So in order to find the um direction vector of the line that passes through each of these lines is p one, P two vector That is equal to 1 -1 and zero. So therefore in order to find the distance between these two lines, stuff according to the formula we put all the values that is the um P one p two vector dot with the normal vector divided by the scalar value of the normal victor. So the scalable normal victories route under two squared plus seven square plus four square. So therefore the value of D is equal to two plus seven plus four. After completing the dot product of uh the value in the numerous. Don't you want me to vector dot n victor? And so this divided by the value uh in skill er uh for the normal vector of these two lines so therefore the value of the is equal to Um 13 divided by Route 69. So hence this is the required answer of the given question.