Question

A point is 6 units away from the vertical plane and profile plane and 10 units away from horizontal plane in 3rd quadrant then the projections are drawn on paper the distance between the side view and top view of point is ________________

(a) 15

(b) 16

(c) 12

(d) 20

I have been asked this question by my college professor while I was bunking the class.

My query is from Projection of Points in Third Quadrant in section Projection of Points of Engineering Drawing

Answer 1

Right option is (d) 20

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 (6+6); front view and top view (10+6)and these lines which form perpendicular to each other gives needed distance, answer is square root of squares of both the distances √(12^2+16^2 ) = 20 units.