A point is 5 units away from the vertical plane and 4 units away from profile plane and 3 units away from horizontal plane in 1st quadrant then the projections are drawn on paper the distance between the front view and top view of point is _____________a)7 unitsb)8 unitsc)9 unitsd)5 unitsCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

To find the distance between the front view and top view of a point, we need to determine the coordinates of the point in 3D space and then project those coordinates onto the respective planes.

Let's assume the point is located at (x, y, z) in the 3D space. Given the information in the question, we know the following distances:

- The point is 5 units away from the vertical plane: This means the x-coordinate of the point is 5.
- The point is 4 units away from the profile plane: This means the y-coordinate of the point is 4.
- The point is 3 units away from the horizontal plane: This means the z-coordinate of the point is 3.

So, the coordinates of the point are (5, 4, 3).

Now, let's project this point onto the front view and top view planes.

- Front View Projection:
In the front view projection, the z-coordinate is ignored, so the point is projected onto the xy-plane. The x and y coordinates remain the same, so the front view projection of the point is (5, 4).

- Top View Projection:
In the top view projection, the y-coordinate is ignored, so the point is projected onto the xz-plane. The x and z coordinates remain the same, so the top view projection of the point is (5, 3).

Now, we can find the distance between the front view and top view of the point by calculating the distance between the projected points.

Using the distance formula, we have:

Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
= sqrt((5 - 5)^2 + (4 - 3)^2)
= sqrt(0 + 1)
= sqrt(1)
= 1

Therefore, the distance between the front view and top view of the point is 1 unit.

The correct answer is option B) 1 unit.

Similar JEE Doubts

Particle is thrown in a way that it's maximum hide achieve and horizontal range are same and equal to h the maximum height that can be achieved in particle is thrown vertically upward is as and maximam distance it can be throw away from point of projection is are then?

two projectiles are thrown from the same point simultaneously with same velocity 10m/s .one goes straight vertically while other at 60 degree with the vertical. what will be the distance of separation between the two after 1 second of their throw?

Top Courses for JEEView all

Physics for JEE Main & Advanced

Mock Tests for JEE Main and Advanced 2025

Mathematics (Maths) for JEE Main & Advanced

Chemistry for JEE Main & Advanced

Study Plan for JEE

Top Courses for JEE

Top Courses for JEE

Related Content

JEE Courses

Mock Test Series

Past Year Papers

Additional Courses

Company