) | , where | ≠ Proof The two skew lines r sb and r td are denoted by L and L respectively. Let A and C be the points on L and L with position vectors a and c respectively. Fig. .
Fig. . L L D L L ( ) B c ( ) A a Vector - - From the given equations of skew lines, we observe that L is parallel to the vector b L is parallel to the vector d . So, b is perpendicular to the lines L and L .
Let SD be the line segment perpendicular to both the lines L L . Then the vector SD is perpendicular to the vectors b and d and therefore it is parallel to the vector b So, is a unit vector in the direction of SD . Then, the shortest distance | SD is the absolute value of the projection of AC on SD . That is, δ = | | | SD AC .
(Unit vector in the direction of SD )| δ = | ( ) ( ) | , where | ≠ . Remark (i) It follows from theorem ( . ) that two straight lines r sb and r td intersect each other (that is, coplanar) if ( ) ( ( ) If two lines intersect each other (that is, coplanar), then we have Example . Find the parametric form of vector equation of a straight line passing through the point of intersection of the straight lines ( ) , and perpendicular to both straight lines.
The Cartesian equations of the straight line ( ) is s (say) Then any point on this line is of the form ( , , ) s s s ... ( ) The Cartesian equation of the second line is = (say) Then any point on this line is of the form ( , , ) ... ( ) Fig. .