Question

Draw the projection of a point P lying 30 mm above HP and in first quadrant, if its shortest distance from the line of intersection of planes is 50 mm. Also find the distance of the point from V.P

Explain with diagram

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Answer 1

Answer:

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Explanation:

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Answer 2

The distance of the point from the Vertical plane is = 40 mm.

Given: i) A point P is lying in the first quadrant.

          ii) It is 30 mm above the horizontal plane  

            iii) Its shortest distance from the intersection of the Horizontal and Vertical planes is 50   mm.

To find:  The distance of the point from the Vertical plane.

Solution:

According to the question,

The point is 30 mm above the horizontal plane.

⇒ P = 30 mm.

Its shortest distance from the intersection of the Horizontal and Vertical planes is 50   mm.

⇒ d = 50 mm.

We need to find its distance from the vertical plane i.e., we have to find B.

Using the formula,

d² = B² + P²

50² = 30² + B²

2500 = 900 + B²

B² = 2500 – 900

B² = 1600

Or

B = 40 mm

B = 40 mm is the required distance.

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