My Math Forum Exercise on two straight lines

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 July 9th, 2016, 05:17 AM #1 Newbie   Joined: Jan 2016 From: Naples Posts: 4 Thanks: 0 Exercise on two straight lines I have this exercise that I really do not understand how to do. In a Euclidean space $E^3$, we have an orthonormal system and considering two straight lines $r$ and $s$ of the following equations: $r:\{(2y + 5z - 17 = 0),(x + 1 = 0)\}$ $s:\{(4x + (a - 4)z + 16 - 3a = 0),(4y + (a + 6)z - 7a - 18 = 0)\}$ with $a$ real parameter. 1) Determine the reciprocal position of $r$ and $s$ when $a$ varies. 2) For what values of $a$ are the two straight lines orthogonal? 3) In case $r$ and $s$ are coplanar, determine the plane that contains both. I know that in order to find the reciprocal position I need the two directional vectors of the straight lines, find a point that for each straight line, and then calculate the vector product of the two directional vectors. But in this case I have a parameter that can vary. How should I proceed in this case? Last edited by skipjack; July 23rd, 2016 at 08:30 AM.
 July 13th, 2016, 04:15 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Just leave the parameter in the calculation. The first line has x = -1 and, taking parameter t= 10z, 2y + 5z - 17 = 0 becomes 2y + 2t - 17 = 0 so y = 17 - t. Parametric equations are x = -1, y = 17- t, z = t/10. The "direction vector" of the first line is <0, -1, 1/10>. For the second line we have 4x = (4 - a)z + 3a - 16 so x = (1 - a/4)z + (3/4)a - 4 and 4y = -(a + 6)z + 7a + 18 so y = -(a/4 + 3/2)z + 7a/4 + 9/2 so parametric equations are x = (1 - a/4)t - 4, y = -(a/4 + 3/2)t + 7a/4 + 9/2, z = t. Its "direction vector" is <1- a/4, -(a/4+ 3/2), 1>. I frankly don't know what "reciprocal position" of two lines means and your description doesn't make sense. The two lines will be orthogonal when the dot product of their directional vectors is 0. Take the dot product of the two vectors, set it equal to 0 and solve for a. To find the equation of the plane containing the two lines, take the cross product of the two direction vectors. That will give the normal vector to the plane. The equation of a plane containing point $\displaystyle (x_0, y_0, z_0)$ and with normal vector is $\displaystyle A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$. Last edited by skipjack; July 23rd, 2016 at 08:34 AM.

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