Find vector orthogonal to two lines

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SUMMARY

The discussion focuses on finding a vector orthogonal to two skew lines L1 and L2 defined by specific points in 3D space. The equations for L1 and L2 are derived from the points provided, with L1 represented as x(t) = -2 - 6t, y(t) = 36 + 8t, z(t) = 9 + 3t, and L2 as x(s) = 55 - 14s, y(s) = -31 + 15s, z(s) = 7 + 6s. The solution involves determining the equations of the normal planes to both lines and finding a mutual perpendicular line that intersects both L1 and L2 by solving the resulting equations for the parameters C and D.

PREREQUISITES
  • Understanding of vector equations in 3D space
  • Knowledge of dot product and cross product operations
  • Familiarity with parametric equations of lines
  • Concept of skew lines and their properties
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  • Study the properties of skew lines in 3D geometry
  • Learn how to derive equations of normal planes to lines
  • Explore methods for solving systems of equations in multiple variables
  • Investigate applications of mutual perpendiculars in vector calculus
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Homework Statement


L1 pass through the points (-2,36,9) and (-8,44,12)
L2 pass through the points (55,-31,7) and (41,-16,13)
Find a point P on L1 and a point Q on L2 so that the vector \vec{}PQ is orthgonal to both lines.

Homework Equations


Dot product/ Cross product


The Attempt at a Solution


Equation for L1: x(t) = -2 -6t; y(t) = 36+8t; z(t) = 9 + 3t
Equation for L2: x(s) = 55 -14s; y(s) = -31 +15s; z(s) = 7 +6s
I stuck from here. I can find the intersect pt if the 2 lines are intersect, but this look intersect but have a space between them.
 
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Yes, two lines in space have a "mutual perpendicular" only if they are skew lines.

Every normal plane to L1 has equation -6x+ 8y+ 3z= C for some C and every normal plane to L2 has equation -14x+ 15y+ 6z= D for some D. A line lying in both normal planes (as a mutual perpendicular would) must satisfy both equations. Solve both equations for, say, z and set them equal. That gives you a single equation in x and y so that you can solve for, say, y as a function of x. That will give parametric equations for the mutual perpendicular line, still with unknowns C and D. Determine C and D so that this new line does intersect both L1 and L2.
 

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