Question

A point p is 20mm below h.P. And lies in the third quadrant. Its shortest distance from xy is 40mm. Draw its projections

Answer 1

Solution: A point p which is, AB= 20 mm below horizontal plane.

  Shortest distance from xy i.e origin=OB = 40 mm

→ ΔOAB forms a right angled triangle.

→OB² = OA² + AB² [ By pythagoras theorem]

→ 40² = OA²  + 20²

→1600 - 400 = OA²

→ OA² = 1200

→ OA =  √1200       = $\sqrt{10\times10\times2\times2\times3}$

         = 10 ×2× √3= 20√3 mm

The vector is → x  = 20 √3 i and y =  - 20 j

The two vectors x and y are perpendicular.

Projection of AB on OA = $\frac{x.y}{\text{Magnitude of x}}$= 0

As i .j= 0, therefore there is no projection of AB on OA.

But if you consider the whole vector OB = OA +AB=x i -yj=20√3 i -20 j, it's Horizontal component= 40 cos 30°=40 ×√3/2=20√3

and vertical component is 40 sin30°= 40×1/2=20 .along negative y axis.

Answer 2

Answer:

A point p is 20mm below h.P. And lies in the third quadrant. Its shortest distance from xy is 40mm. Draw its projections