Consider the point. (3; 5, 6) What is the projection of the point on the xy-plane

Consider the point. (3; 5, 6) What is the projection of the point on the xy-plane? (x,y,z)=( boxed 3,5,0 ) What is the projection of the point on the yz-plane? (x,y,z)=( boxed 0,5,6 ) What is the projection of the point on the -plane ? (x,y,z)=( boxed 3,0,6 ) Draw a rectangular box with the origin and (3.5, 6) as opposite vertices and with its faces parallel to the coordinate planes. Labe all vertices of the box. Find the length of the diagonal of the box chegg

The Correct Answer and Explanation is:

Analysis of the Point in 3D Space

We are given the point P(3, 5, 6) in a three-dimensional Cartesian coordinate system. Let’s analyze its properties, including its projections onto the coordinate planes, the rectangular box it defines with the origin, and the length of the box’s main diagonal.

Projections of the Point

A projection of a point onto a plane is the point on the plane that is closest to the original point. This is found by dropping a perpendicular from the point to the plane. In a Cartesian system, this is simplified by setting the coordinate corresponding to the axis perpendicular to the plane to zero.

• Projection onto the xy-plane: The xy-plane is defined by the equation z = 0. To project the point (3, 5, 6) onto this plane, we keep the x and y coordinates and set the z-coordinate to 0.
  What is the projection of the point on the xy-plane? (x,y,z)=( 3, 5, 0 )
• Projection onto the yz-plane: The yz-plane is defined by the equation x = 0. To project the point (3, 5, 6) onto this plane, we keep the y and z coordinates and set the x-coordinate to 0.
  What is the projection of the point on the yz-plane? (x,y,z)=( 0, 5, 6 )
• Projection onto the xz-plane: The xz-plane is defined by the equation y = 0. To project the point (3, 5, 6) onto this plane, we keep the x and z coordinates and set the y-coordinate to 0.
  What is the projection of the point on the xz-plane? (x,y,z)=( 3, 0, 6 )

Rectangular Box and its Vertices

A rectangular box (or rectangular prism) with its faces parallel to the coordinate planes and with the origin O(0, 0, 0) and the point P(3, 5, 6) as opposite vertices will have dimensions defined by the coordinates of P. The length along the x-axis is 3 units, the width along the y-axis is 5 units, and the height along the z-axis is 6 units. The 8 vertices of this box are all the possible combinations of the coordinate values {0, 3} for x, {0, 5} for y, and {0, 6} for z.

The vertices are:

1. O(0, 0, 0) – The origin.
2. (3, 0, 0) – On the x-axis.
3. (0, 5, 0) – On the y-axis.
4. (0, 0, 6) – On the z-axis.
5. (3, 5, 0) – The projection on the xy-plane.
6. (0, 5, 6) – The projection on the yz-plane.
7. (3, 0, 6) – The projection on the xz-plane.
8. P(3, 5, 6) – The given point, opposite the origin.

Length of the Diagonal of the Box

The main diagonal of the box is the line segment connecting the opposite vertices, O(0, 0, 0) and P(3, 5, 6). The length of this diagonal can be found using the three-dimensional distance formula, which is an extension of the Pythagorean theorem.

The distance formula between two points 

(x1,y1,z1)(x1​,y1​,z1​)

 and 

(x2,y2,z2)(x2​,y2​,z2​)

 is:

d=(x2−x1)2+(y2−y1)2+(z2−z1)2d=(x2​−x1​)2+(y2​−y1​)2+(z2​−z1​)2​

Applying this formula to our vertices O(0, 0, 0) and P(3, 5, 6):

d=(3−0)2+(5−0)2+(6−0)2d=(3−0)2+(5−0)2+(6−0)2​

d=32+52+62d=32+52+62​

d=9+25+36d=9+25+36​

d=70d=70​

The length of the diagonal of the box is √70. This value is approximately 8.37 units.thumb_upthumb_down