**RD Sharma Solutions Class 11 Maths Chapter 28 Exercise 28.1: **We will cover concepts related to coordinates of a point in space, as well as signs of coordinates of a point, in this exercise. Students can utilise **RD Sharma Class 11 Solutions** to help them get good grades on their board exams. The pdf of **RD Sharma Solutions Class 11 Maths Chapter 28** Exercise 28.1 is available in the links below, which can be readily downloaded and saved for future reference.

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## Download RD Sharma Class 11 Solutions Chapter 28 Exercise 28.1 Free PDF

RD Sharma Solutions Class 11 Maths Chapter 28 Exercise 28.1

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### Access answers to RD Sharma Solutions Class 11 Maths Chapter 28 Exercise 28.1- Important Question with Answers

**1. 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)**

**Solution:**

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

**2. 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**

**Solution:**

**(i)** (-2, 3, 4)

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

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

**3. A cube of side 5 has one vertex at the point (1, 0, 1), and the three edges from this vertex are, respectively, parallel to the negative x and y-axes and positive z-axis. Find the coordinates of the other vertices of the cube.**

**Solution:**

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 the 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)

**4. Planes are drawn parallel to the coordinates planes through the points (3, 0, -1) and (-2, 5, 4). Find the lengths of the edges of the parallelepiped so formed.**

**Solution:**

Given:

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

We need to find the lengths of the edges of the parallelepiped formed.

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

A plane parallel to coordinate planes of x_{1} and x_{2} is yz-plane

A plane parallel to coordinate planes of y_{1} and y_{2} is xz-plane

A plane parallel to coordinate planes of z_{1} and z_{2} is xy-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

∴The edges of parallelepiped is 5, 5, 5

**5. Planes are drawn through the points (5, 0, 2) and (3, -2, 5) parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelepiped so formed.**

**Solution:**

Given:

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

We need to find the lengths of the edges of the parallelepiped formed

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

A plane parallel to coordinate planes of x_{1} and x_{2} is yz-plane

A plane parallel to coordinate planes of y_{1} and y_{2} is xz-plane

A plane parallel to coordinate planes of z_{1} and z_{2} is xy-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 edges of parallelepiped is 2, 2, 3

**6. Find the distances of the point P (-4, 3, 5) from the coordinate axes.**

**Solution:**

Given:

The point P (-4, 3, 5)

The distance of the point from the 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:

**7. The coordinates of a point are (3, -2, 5). Write down the coordinates of seven points such that the absolute values of their coordinates are the same as those of the coordinates of the given point.**

**Solution:**

Given:

Point (3, -2, 5)

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

√(x^{2} + y^{2} + z^{2})

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

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

The 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)

We have provided complete details of RD Sharma Solutions Class 11 Maths Chapter 28 Exercise 28.1. If you have any queries related to **CBSE** Class 11, feel free to ask us in the comment section below.

## FAQs on RD Sharma Class 11 Solutions Chapter 28 Exercise 28.1

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