Question

A point p is 20 mm below the hp lies in 3rd quadrant ,the shortest distance from x-y line is 40mm draw its projection

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}

10×10×2×2×3

= 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}}

Magnitude of x

x.y

= 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.