RD Chapter 28- Introduction to 3D coordinate geometry Ex-28.2 |
RD Chapter 28- Introduction to 3D coordinate geometry Ex-28.3 |

Name the octants in which the following points lie:

(i) (5, 2, 3)

(ii) (-5, 4, 3)

(iii) (4, -3, 5)

(iv) (7, 4, -3)

(v) (-5, -4, 7)

(vi) (-5, -3, -2)

(vii) (2, -5, -7)

(viii) (-7, 2, -5)

**Answer
1** :

**Solution:**

**(i)** (5,2, 3)

In this case, since x, y and z all three are positive thenoctant will be XOYZ

**(ii)** (-5,4, 3)

In this case, since x is negative and y and z are positive thenthe octant will be X′OYZ

**(iii)** (4,-3, 5)

In this case, since y is negative and x and z are positive thenthe octant will be XOY′Z

**(iv)** (7,4, -3)

In this case, since z is negative and x and y are positive thenthe octant will be XOYZ′

**(v)** (-5,-4, 7)

In this case, since x and y are negative and z is positive thenthe octant will be X′OY′Z

**(vi)** (-5,-3, -2)

In this case, since x, y and z all three are negative thenoctant will be X′OY′Z′

**(vii)** (2,-5, -7)

In this case, since z and y are negative and x is positive thenthe octant will be XOY′Z′

**(viii)** (-7,2, -5)

In this case, since x and z are negative and x is positive thenthe octant will be X′OYZ′

Find the image of:

(i) (-2, 3, 4) in the yz-plane

(ii) (-5, 4, -3) in the xz-plane

(iii) (5, 2, -7) in the xy-plane

(iv) (-5, 0, 3) in the xz-plane

(v) (-4, 0, 0) in the xy-plane

**Answer
2** :

(i) (-2, 3, 4)

Since we need to find its image in yz-plane, a sign of its x-coordinate will change

So, Image of point (-2, 3, 4) is (2, 3, 4)

(ii)(-5, 4, -3)

Since we need to find its image in xz-plane, sign of its y-coordinate will change

So, Image of point (-5, 4, -3) is (-5, -4, -3)

(iii) (5, 2, -7)

Since we need to find its image in xy-plane, a sign of its z-coordinate will change

So, Image of point (5, 2, -7) is (5, 2, 7)

(iv) (-5, 0, 3)

Since we need to find its image in xz-plane, sign of its y-coordinate will change

So, Image of point (-5, 0, 3) is (-5, 0, 3)

(v) (-4, 0, 0)

Since we need to find its image in xy-plane, sign of its z-coordinate will change

So, Image of point (-4, 0, 0) is (-4, 0, 0)

**Answer
3** :

Given: A cube has side 4 having one vertex at (1, 0, 1)

Side of cube = 5

We need to find the coordinates of the other vertices of the cube.

So let the Point A(1, 0, 1) and AB, AD and AE is parallel to –ve x-axis, -ve y-axis and +ve z-axis respectively.

Since side of cube = 5

Point B is (-4, 0, 1)

Point D is (1, -5, 1)

Point E is (1, 0, 6)

Now, EH is parallel to –ve y-axis

Point H is (1, -5, 6)

HG is parallel to –ve x-axis

Point G is (-4, -5, 6)

Now, again GC and GF is parallel to –ve z-axis and +ve y-axis respectively

Point C is (-4, -5, 1)

Point F is (-4, 0, 6)

**Answer
4** :

Given:

Points are (3, 0, -1) and (-2, 5, 4)

We need to find the lengths of the edges of the parallelepipedformed.

For point (3, 0, -1)

x_{1} = 3, y_{1} = 0 and z_{1} =-1

For point (-2, 5, 4)

x_{2} = -2, y_{2} = 5 and z_{2} =4

Plane parallel to coordinate planes of x_{1} and x_{2} isyz-plane

Plane parallel to coordinate planes of y_{1} and y_{2} isxz-plane

Plane parallel to coordinate planes of z_{1} and z_{2} isxy-plane

Distance between planes x_{1} = 3 and x_{2} =-2 is 3 – (-2) = 3 + 2 = 5

Distance between planes x_{1} = 0 and y_{2} =5 is 5 – 0 = 5

Distance between planes z_{1} = -1 and z_{2} =4 is 4 – (-1) = 4 + 1 = 5

∴Theedgesof parallelepiped is 5, 5, 5

**Answer
5** :

Given:

Points are (5, 0, 2) and (3, -2, 5)

We need to find the lengths of the edges of the parallelepipedformed

For point (5, 0, 2)

x_{1} = 5, y_{1} = 0 and z_{1} =2

For point (3, -2, 5)

x_{2} = 3, y_{2} = -2 and z_{2} =5

Plane parallel to coordinate planes of x_{1} and x_{2} isyz-plane

Plane parallel to coordinate planes of y_{1} and y_{2} isxz-plane

Plane parallel to coordinate planes of z_{1} and z_{2} isxy-plane

Distance between planes x_{1} = 5 and x_{2} =3 is 5 – 3 = 2

Distance between planes x_{1} = 0 and y_{2} =-2 is 0 – (-2) = 0 + 2 = 2

Distance between planes z_{1} = 2 and z_{2} =5 is 5 – 2 = 3

∴The edgesof parallelepiped is 2, 2, 3

**Answer
6** :

Given:

The point P (-4, 3, 5)

The distance of the point from x-axis is given as:

The distance of the point from y-axis is given as:

The distance of the point from z-axis is given as:

**Answer
7** :

Given:

Point (3, -2, 5)

The Absolute value of any point(x, y, z) is given by,

√(x2 + y2 + z2)

We need to make sure that absolute value to be the same for all points.

So let the point A(3, -2, 5)

Remaining 7 points are:

Point B(3, 2, 5) (By changing the sign of y coordinate)

Point C(-3, -2, 5) (By changing the sign of x coordinate)

Point D(3, -2, -5) (By changing the sign of z coordinate)

Point E(-3, 2, 5) (By changing the sign of x and y coordinate)

Point F(3, 2, -5) (By changing the sign of y and z coordinate)

Point G(-3, -2, -5) (By changing the sign of x and z coordinate)

Point H(-3, 2, -5) (By changing the sign of x, y and z coordinate)

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