Question

A point is 2 units away from the vertical plane and 3 units away from profile plane and 7 units away from horizontal plane in 1st quadrant then the projections are drawn on paper the shortest distance from top view and side view of point is _____________

(a) 10.29

(b) 5.14

(c) 9

(d) 7

This question was addressed to me during an interview for a job.

My question is taken from Projection of Points in First Quadrant topic in chapter Projection of Points of Engineering Drawing

Answer 1

Correct option is (c) 9

The best explanation: Since here distance from side view and top view is asked for that we need  the distance between the front view and side view (3+2); front view and top view (7+2)and these lines which form perpendicular to each other gives needed distance, answer is square root of squares of both the distances √(5^2+9^2 ) =10.29 units.