RD Sharma Class 11 Solutions Chapter - Introduction to 3D Coordinate Geometry | Mathematics (Maths) Class 11 - Commerce PDF Download

 Page 1


28. Introduction to 3-D Coordinate Geometry
Exercise 28.1
1. Question
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
Given: Points are given
To find: name of the octant
Formula used:
Notation of octants:
If x, y and z all three are positive, then octant will be XOYZ
If x is negative and y and z are positive, then the octant will be X’OYZ
If y is negative and x and z are positive, then the octant will be XOY’Z
If z is negative and x and y are positive, then the octant will be XOYZ’
If x and y are negative and z is positive, then the octant will be X’OY’Z
If z and y are negative and x is positive, then the octant will be XOY’Z’
If x and z are negative and x is positive, then the octant will be X’OYZ’
If x, y and z all three are negative, then octant will be X’OY’Z’
(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’
Page 2


28. Introduction to 3-D Coordinate Geometry
Exercise 28.1
1. Question
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
Given: Points are given
To find: name of the octant
Formula used:
Notation of octants:
If x, y and z all three are positive, then octant will be XOYZ
If x is negative and y and z are positive, then the octant will be X’OYZ
If y is negative and x and z are positive, then the octant will be XOY’Z
If z is negative and x and y are positive, then the octant will be XOYZ’
If x and y are negative and z is positive, then the octant will be X’OY’Z
If z and y are negative and x is positive, then the octant will be XOY’Z’
If x and z are negative and x is positive, then the octant will be X’OYZ’
If x, y and z all three are negative, then octant will be X’OY’Z’
(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. Question
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
(i) Given: Point is (-2, 3, 4)
To find: the image of the point in yz-plane
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) Given: Point is (-5, 4, -3)
To find: image of the point in xz-plane
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) Given: Point is (5, 2, -7)
To find: the image of the point in xy-plane
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) Given: Point is (-5, 0, 3)
To find: image of the point in xz-plane
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) Given: Point is (-4, 0, 0)
To find: image of the point in xy-plane
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)
3. Question
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.
Answer
Given: A cube has side 4 having one vertex at (1, 0, 1)
Page 3


28. Introduction to 3-D Coordinate Geometry
Exercise 28.1
1. Question
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
Given: Points are given
To find: name of the octant
Formula used:
Notation of octants:
If x, y and z all three are positive, then octant will be XOYZ
If x is negative and y and z are positive, then the octant will be X’OYZ
If y is negative and x and z are positive, then the octant will be XOY’Z
If z is negative and x and y are positive, then the octant will be XOYZ’
If x and y are negative and z is positive, then the octant will be X’OY’Z
If z and y are negative and x is positive, then the octant will be XOY’Z’
If x and z are negative and x is positive, then the octant will be X’OYZ’
If x, y and z all three are negative, then octant will be X’OY’Z’
(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. Question
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
(i) Given: Point is (-2, 3, 4)
To find: the image of the point in yz-plane
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) Given: Point is (-5, 4, -3)
To find: image of the point in xz-plane
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) Given: Point is (5, 2, -7)
To find: the image of the point in xy-plane
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) Given: Point is (-5, 0, 3)
To find: image of the point in xz-plane
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) Given: Point is (-4, 0, 0)
To find: image of the point in xy-plane
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)
3. Question
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.
Answer
Given: A cube has side 4 having one vertex at (1, 0, 1)
To find: coordinates of the other vertices of the cube.
Let 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)
4. Question
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.
Answer
Given: Points are (3, 0, -1) and (-2, 5, 4)
To find: 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
Plane parallel to coordinate planes of x
1
 and x
2
 is yz-plane
Plane parallel to coordinate planes of y
1
 and y
2
 is xz-plane
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
Page 4


28. Introduction to 3-D Coordinate Geometry
Exercise 28.1
1. Question
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
Given: Points are given
To find: name of the octant
Formula used:
Notation of octants:
If x, y and z all three are positive, then octant will be XOYZ
If x is negative and y and z are positive, then the octant will be X’OYZ
If y is negative and x and z are positive, then the octant will be XOY’Z
If z is negative and x and y are positive, then the octant will be XOYZ’
If x and y are negative and z is positive, then the octant will be X’OY’Z
If z and y are negative and x is positive, then the octant will be XOY’Z’
If x and z are negative and x is positive, then the octant will be X’OYZ’
If x, y and z all three are negative, then octant will be X’OY’Z’
(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. Question
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
(i) Given: Point is (-2, 3, 4)
To find: the image of the point in yz-plane
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) Given: Point is (-5, 4, -3)
To find: image of the point in xz-plane
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) Given: Point is (5, 2, -7)
To find: the image of the point in xy-plane
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) Given: Point is (-5, 0, 3)
To find: image of the point in xz-plane
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) Given: Point is (-4, 0, 0)
To find: image of the point in xy-plane
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)
3. Question
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.
Answer
Given: A cube has side 4 having one vertex at (1, 0, 1)
To find: coordinates of the other vertices of the cube.
Let 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)
4. Question
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.
Answer
Given: Points are (3, 0, -1) and (-2, 5, 4)
To find: 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
Plane parallel to coordinate planes of x
1
 and x
2
 is yz-plane
Plane parallel to coordinate planes of y
1
 and y
2
 is xz-plane
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
Hence, edges of parallelepiped is 5, 5, 5
5. Question
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.
Answer
Given: Points are (5, 0, 2) and (3, -2, 5)
To find: 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
Plane parallel to coordinate planes of x
1
 and x
2
 is yz-plane
Plane parallel to coordinate planes of y
1
 and y
2
 is xz-plane
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
Hence, edges of parallelepiped is 2, 2, 3
6. Question
Find the distances of the point P (-4, 3, 5) from the coordinate axes.
Answer
Given: Point P(-4, 3, 5)
To find: distances of the point P from coordinate axes
The distance of the point from x-axis will be given by,
The distance of the point from y-axis will be given by,
Page 5


28. Introduction to 3-D Coordinate Geometry
Exercise 28.1
1. Question
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
Given: Points are given
To find: name of the octant
Formula used:
Notation of octants:
If x, y and z all three are positive, then octant will be XOYZ
If x is negative and y and z are positive, then the octant will be X’OYZ
If y is negative and x and z are positive, then the octant will be XOY’Z
If z is negative and x and y are positive, then the octant will be XOYZ’
If x and y are negative and z is positive, then the octant will be X’OY’Z
If z and y are negative and x is positive, then the octant will be XOY’Z’
If x and z are negative and x is positive, then the octant will be X’OYZ’
If x, y and z all three are negative, then octant will be X’OY’Z’
(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. Question
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
(i) Given: Point is (-2, 3, 4)
To find: the image of the point in yz-plane
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) Given: Point is (-5, 4, -3)
To find: image of the point in xz-plane
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) Given: Point is (5, 2, -7)
To find: the image of the point in xy-plane
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) Given: Point is (-5, 0, 3)
To find: image of the point in xz-plane
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) Given: Point is (-4, 0, 0)
To find: image of the point in xy-plane
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)
3. Question
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.
Answer
Given: A cube has side 4 having one vertex at (1, 0, 1)
To find: coordinates of the other vertices of the cube.
Let 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)
4. Question
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.
Answer
Given: Points are (3, 0, -1) and (-2, 5, 4)
To find: 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
Plane parallel to coordinate planes of x
1
 and x
2
 is yz-plane
Plane parallel to coordinate planes of y
1
 and y
2
 is xz-plane
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
Hence, edges of parallelepiped is 5, 5, 5
5. Question
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.
Answer
Given: Points are (5, 0, 2) and (3, -2, 5)
To find: 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
Plane parallel to coordinate planes of x
1
 and x
2
 is yz-plane
Plane parallel to coordinate planes of y
1
 and y
2
 is xz-plane
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
Hence, edges of parallelepiped is 2, 2, 3
6. Question
Find the distances of the point P (-4, 3, 5) from the coordinate axes.
Answer
Given: Point P(-4, 3, 5)
To find: distances of the point P from coordinate axes
The distance of the point from x-axis will be given by,
The distance of the point from y-axis will be given by,
The distance of the point from z-axis will be given by,
= 5
7. Question
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.
Answer
Given: Point (3, -2, 5)
To find: the coordinates of 7 more points such that the absolute values of all 8 coordinates are the same
Formula used:
Absolute value of any point(x, y, z) is given by,
We need to make sure that absolute value to be the same for all points
In the formula of absolute value, there is square of the coordinates. So when we change the sign of any of
the coordinates, it will not affect the absolute value.
Let 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)
Exercise 28.2
1 A. Question
Find the distance between the following pairs of points :
P(1, -1, 0) and Q (2, 1, 2)
Answer
Given: P(1, -1, 0) and Q(2, 1, 2)
To find: Distance between given two points
Formula used:
The distance between any two points (a, b, c) and (m, n, o) is given by,