Finding Shortest Distances

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Presentation transcript:

Finding Shortest Distances

Shortest Distance from a point to a line

Draw the given projections of line AB and the point C

Draw the reference line x1y1 parallel to say the front view of the line

Project the auxiliary front views of the line a1’b1’ and the point c1’ Project the auxiliary front views of the line a1’b1’ and the point c1’.the length of the line a1’b1’represents the true length of the line.

Through c1’ draw a perpendicular c1’d1’ to a1’b1’

Draw the reference line x2y2 parallel to c1’d1’.

Project the normal views C1D1 of the line CD

C1D1 is the shortest Distance of the point C from line AB

Shortest distance between two skew lines

Draw the given projection of line AB and CD

Draw the reference line X1Y1 parallel to AB at any convenient Distance

Project the auxiliary front views of the line a1b1 and c1d1 Project the auxiliary front views of the line a1b1 and c1d1.a1’b1’ is the normal view of the line AB represents the true length of it.

Draw the reference line x2y2 perpendicular to a1’b1’ at any convenient distance from it

Obtain secondary auxiliary view a1b1 and c1d1 Obtain secondary auxiliary view a1b1 and c1d1. the point a1b1 represents the edge view of the line AB

The perpendicular distance between a1b1 to the line c1d1 is the required shortest distance

Shortest Distance Between two parallel lines

Draw the given projections of the parallel lines AB and CD

Draw the reference line X1Y1 parallel to say the front view of the lines

Project the normal views of the lines a1’b1’ and c1’d1’

Draw the reference line X2Y2 perpendicular to the above normal views

Project edge view of lines a1b1 & c1d1

The distance between the edge views is the required shortest distance between the lines

Shortest Distance from a point to a plane

Draw the given projections of the plane ABC and the point P

Obtain the edge view of the plane a1b1c1 and locate p1 with respect to it.

Draw perpendicular p1 to a1b1c1 intersecting at O1 Draw perpendicular p1 to a1b1c1 intersecting at O1. the length of p1o1 is the required shortest distance

To show the line PO with respect to the given projections of the plane Draw a line through p’ , parallel to X1Y1

Through O1,draw a line perpendicular to X1Y1, meeting the above line at O’

Join p’o’

> Draw projectors through o’, perpendicular to XY > Draw projectors through o’, perpendicular to XY. > mark point O on it such that its distance from XY is the same as the distance O1 from X1Y1. > Join PO

P’O’ and PO are the projectors of the line PO