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Solutions of Assignment (Set-2) Introduction to Three Dimensional Geometry (Solutions)
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Section-A
Q.No. Solution
1. Answer (2)
z -coordinate is the distance of a point P (6, 8, 9) from XY -plane.
2. Answer (3)
x -coordinate is the distance of a point P (–2, 3, 5) from YZ -plane.
3. Answer (3)
Equation of z -axis is x = 0, y = 0, because x and y -coordinates on the z -axis is zero.
4. Answer (4)
Length of perpendicular from the point P (3, 6, 7) on y -axis is
2 2( -coordinate) ( -coordinate) x z = 2 23 7 58 units
5. Answer (3)
(–2, 3, 4) lies in IInd
octant.
6. Answer (3)
y -coordinate on XZ -plane is zero.
Equation of XZ -plane is y = 0.
7. Answer (4)
Any point on y -axis can be taken as (0, y , 0)
Now, 2 2 2(0 2) ( 2) (0 5) 20y
24 ( 2) 25 20y
(y – 2)2 = –9
Which is not true.
12Chapter Introduction to Three Dimensional Geometry
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Introduction to Three Dimensional Geometry (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
8. Answer (2)
Length of diagonal
= 2 2 22 1 2 1 2 1( ) ( ) ( ) x x y y z z
= 2 2 2(6 2) (7 5) (9 3)
= 16 4 36
= 2 14 units
9. Answer (3)
Let A(0, 7, 10), B(–1, 6, 6) and C (–4, 9, 6) be given points
Now, 2 2 2( 1 0) (6 7) (6 10) 3 2 units AB
2 2 2( 4 1) (9 6) (6 6) 3 2 unitsBC
2 2 2( 4 0) (9 7) (6 10) 6 units AC
Since, AB = BC and AB2 + BC
2 = AC
2
10. Answer (3)
Let the required ratio be k : 1
3 2 4 4 5 5
( , , 0), ,1 1 1
k k k x y
k k k
5 5
01
k
k
k = 1
Hence, XY -plane divides the join of given points in 1 : 1.
11. Answer (2)
Let x -axis divides the join of line segment in k : 1. Coordinates of the point are
3 3 2 4 5 10
, ,1 1 1
k k k
k k k
Since, y and z -coordinates on x -axis is zero
2 4
01
k
k
,
5 100
1
k
k
On solving the equation, we get
k = –2
Hence, the required ratio is 2 : 1 externally.
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Solutions of Assignment (Set-2) Introduction to Three Dimensional Geometry (Solutions)
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Q.No. Solution
12. Answer (4)
Let YZ -plane divide the join of line segment (3, 0, 5) and (–2, 3, 2) in k : 1.
Therefore, coordinates of the point are
2 3 3 2 5, ,
1 1 1
k k k
k k k
Since, x -coordinate on YZ -plane is zero.
2 3
01
k
k
3
2k
Hence, the required ratio is 3 : 2.
13. Answer (1)
Length of perpendicular from the point (3, 4, 5) on z -axis is
2 2
( -coordinate) ( -coordinate) x y =2 2
3 4 = 5 units
14. Answer (2)
x -coordinate is the distance of a point from YZ -plane.
15. Answer (2)
Since, z -coordinate in XY -plane is zero.
Therefore, coordinates of the foot of perpendiculars of the point L is (–3, 6).
16. Answer (1)
Since, x -coordinate in YZ -plane is zero.Therefore, coordinates of the foot of perpendiculars of the point is (0, –4, 5).
17. Answer (2)
Image of the point in XY -plane is (–2, 3, –5)
18. Answer (4)
Coordinates of Q is (–4, –3, 5).
19. Answer (2)
Length of diagonal
= 2 2 22 1 2 1 2 1( ) ( ) ( ) x x y y z z
= 2 2 2(3 1) ( 4 2) (3 3)
= 2 2 units
Now, diagonal of square = 2 side
Side = 2 units
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Q.No. Solution
20. Answer (2)
2 2 2(4 0) ( 2 1) (1 2) 26 unitsPQ
2 2 2(0 4) (0 2) (0 1) 21 unitsQR
2 2 2(0 0) (0 1) (0 2) 5 unitsPR
Now, QR 2 + PR
2 = PQ
2
It means PQR is a right angled triangle
2
PRQ
21. Answer (2)
The coordinates of centroid are
1 2 3 1 2 3 1 2 3, ,3 3 3
x x x y y y z z z
=3 4 8 3 2 12 6 8 2
, ,3 3 3
a b c
=3 4 2 15 2 2
, ,3 3 3
a b c
But it is given that, origin is centroid
3 4
0
3
a ,
2 150
3
b and
2 20
3
c
On solving these equations, we get
4
3a ,
15
2b and c = 1
22. Answer (1)
Let the fourth vertex be ( x , y , z ), then mid-point of AC = mid-point of BD
0 3 1 3 5 0 1 2 4
, , , ,2 2 2 2 2 2
x y z
3 5 1 2 4
, 1, , ,2 2 2 2 2
x y z
3 1
2 2
x ,
21
2
y ,
5 4
2 2
z
x = 2, y = 4, z = 9
Hence, the fourth coordinates are (2, 4, 9).
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Solutions of Assignment (Set-2) Introduction to Three Dimensional Geometry (Solutions)
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Q.No. Solution
23. Answer (3)
Let R ( x , 6, z ) divides the join of the given points (–2, 3, 4) and (6, 10, 18) in k : 1.
The coordinates of point are
6 2 10 3 18 4, ,1 1 1
k k k k k k
...(i)
But y -coordinate is given as 6
10 3
61
k
k
10k + 3 = 6k + 6
4k = 3
3
4k
On putting3
4k in (i), we get the coordinates of the point as
10, 6, 10
7
24. Answer (2)
Since points are collinear, it means A lies on BC . Let A divides BC in k : 1.
C
(9, 8, –10)
B
(5, 4, –6)
k 1
A
(3, 2, –4)
Therefore,9 5 8 4 10 6
(3, 2, 4), ,1 1 1
k k k
k k k
9 5
31
k
k
,
8 42
1
k
k
and
10 64
1
k
k
On solving these equations, we get
1
3k
Hence, the required ratio is 1 : 3 externally.
25. Answer (3)
Let the join of A(2, 1, 5) and B(3, 4, 3) is divided by the plane x + 2y – z = 0 in k : 1.
Therefore, the coordinates of the point are
3 2 4 1 3 5, ,
1 1 1
k k k
k k k
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Introduction to Three Dimensional Geometry (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
Since, the point lies on x + 2y – z = 0
3 2 2(4 1) 3 5
01 1 1
k k k
k k k
3k + 2 + 8k + 2 – 3k – 5 = 0
8k – 1 = 0
1
8k
Hence, the required ratio is 1 : 8.
26. Answer (2)
Let the join of A(1, 2, 3) and B(3, 4, 6) is divided by XY -plane in k : 1.
Therefore, the coordinates are
3 1 4 2 6 3, ,1 1 1
k k k k k k
Now, z -coordinate on XY -plane is zero
6 3
01
k
k
1
2k
Hence, the required ratio is 1 : 2.
27. Answer (1)
Let A(5, –4, 2), B(4, –3, 1), C (7, –6, 4) and D(8, –7, 5) be given points
Mid-point of AC =7 5 6 4 4 2
, ,2 2 2
= (6, –5, 3)
Mid-point of BD =8 4 7 3 5 1
, ,2 2 2
= (6, –5, 3)
Since, mid-points are same.
Therefore, ABCD may be a parallelogram.
Now, 2 2 2(7 5) ( 6 4) (4 2) 3 2 units AC
2 2 2(8 4) ( 7 3) (5 1) 4 3 unitsBD
AC BD
ABCD is a parallelogram.
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Solutions of Assignment (Set-2) Introduction to Three Dimensional Geometry (Solutions)
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Q.No. Solution
28. Answer (1)
Any line parallel to YZ -plane is x = C
x -coordinate is same.
29. Answer (1)
x = C is a line parallel to YZ -plane.
30. Answer (1)
XY -plane is uniquely determined by x and y -coordinates.
31. Answer (2)
Three mutually perpendicular plane divide the space into 8 parts. Each known as octant.
32. Answer (2)
Planes intersect in a line, lines of intersection of XY -plane and YZ -plane is y -axis.
33. Answer (3)
Centroid of the triangle formed by the mid-points of the sides of a triangle is equal to centroid of thegiven triangle
Centroid is given by1 3 2 2 0 2 3 1 5
, ,3 3 3
= (2, 0, 1)
34. Answer (3)Shortest distance of the point (a, b, c ) from y -axis is length of foot of perpendicular the given point to y -axis.
Length of perpendicular =2 2( -coordinate) ( -coordinate) x z
= 2 2a c
35. Answer (1)
Let A(3, –5, 4) and B(a, –8, 4) be given two points, then
AB = 5 ( AB)2 = 25
(a – 3)2 + (–8 + 5)
2 + (4 – 4)
2 = 25
a2 + 9 – 6a + 9 = 25
a2 – 6a – 7 = 0
(a – 7) (a + 1) = 0
a = –1 and 7
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Introduction to Three Dimensional Geometry (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
36. Answer (3)
Any point on z -axis is x = 0, y = 0
x 2 = 0 and y
2 = 0
x 2 + y 2 = 0
37. Answer (4)
Let the point P ( x , y , z ) be equal from the given points A(0, 0, 0), B(4, 0, 0), C (0, 6, 0) and D(0, 0, 8),then
PA = PB = PC = PD
Now, PA = PB (PA)2 = (PB)
2
x 2 + y
2 + z
2 = ( x – 4)
2 + y
2 + z
2
( x – 4)2
= x 2
x = 2
Also, PA = PC (PA)2 = (PC )
2
x 2 + y
2 + z
2 = x
2 + (y – 6)
2 + z
2
y 2 = (y – 6)
2
y = 3
Again, PA = PD (PA)2 = (PD)
2
x
2
+ y
2
+ z
2
= x
2
+ y
2
+ (z – 8)
2
z 2 = (z – 8)
2
z = 4
Hence, the required coordinates are (2, 3, 4).
38. Answer (2)
XY -plane divides the join of (1, –1, 5) and (2, 3, 4) in the ratio k : 1, then the coordinates of the pointare
2 1 3 1 4 5
, ,1 1 1
k k k
k k k
Since, z -coordinate on XY -plane is zero
4 5
01
k
k
5
4k
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Solutions of Assignment (Set-2) Introduction to Three Dimensional Geometry (Solutions)
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Q.No. Solution
39. Answer (2)
Length of the edges are given by
( x 2 – x 1), (y 2 – y 1) and (z 2 – z 1)
x 2 – x 1 = 6 – 1 = 5
y 2 – y 1 = 8 – 2 = 6
z 2 – z 1 = 18 – 3 = 15
40. Answer (1)
Let the point on XY -plane divides the join of (3, 4, 1) and (5, 1, 6) in k : 1.
Therefore, the coordinates of the point are
5 3 4 6 1, ,
1 1 1
k k k
k k k
...(i)
z -coordinate on XY -plane is zero
6 1
01
k
k
1
6k
On putting the value of1
6
k in (i), we get the coordinates of the point as
13 23, , 0
5 5
41. Answer (4)
The point of intersection of median is known as centroid. Let the third vertex is ( x , y , z ), then
3 1 4 3 2 3(0, 3, 1), ,
3 3 3
x y z
4 7 5 (0, 3, 1), ,3 3 3
x y z
4
03
x ,
73
3
y and
51
3
z
x = –4, y = 2 and z = –8
Hence, the required point is (–4, 2, –8).
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Q.No. Solution
42. Answer (2)
Let A(a, b, 3), B(2, 0, –1) and C (1, –1, –3) be given points. Also, let A divides the BC in k : 1.
Therefore,2 3 1
( , , 3), ,
1 1 1
k k k a b
k k k
3 1
31
k
k
–3k – 1 = 3k + 3
6k = –4
2
3k
Now,2
1
k a
k
and
1
k b
k
a = 4 and b = –2
43. Answer (4)
Let the join of the given points (1, 0, 0) and (1, 3, –5) is divided by the plane 2 x + 3y + 5z = 1 in k : 1.
Therefore, the coordinates of the given point are
1 3 5, ,
1 1 1
k k k
k k k
Since, the point lies on 2 x + 3y + 5z = 1
9 25
2 11 1
k k
k k
16
11
k
k
–16k = –k – 1
1
15k
Hence, the required ratio is 1 : 15 internally.
44. Answer (1)
Let P ( x , y , z ) divides the join of (3, 3, 7) and (8, 3, 1) in 2 : 1
( x , y , z ) =16 3 6 3 2 7
, ,3 3 3
( x , y , z ) =19
, 3, 33
Hence, the required coordinates are19
, 3, 33
.
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Q.No. Solution
45. Answer (2)
2 2 2( 0) (0 1) (1 2) 27a
a2 + 2 = 27
5a
46. Answer (2)
x 2 + y
2 = 0 is the equation of z -axis.
Shortest distance from (1, 3, 5) on z -axis is
2 2( -coordinate) ( -coordinate) x y
= 2 21 3 10 units
47. Answer (4)
Let P ( x , y , z ), A(a, 0, 0) and B(–a, 0, 0) be given points such that
(PA)2 + (PB)
2 = 2c
2
( x – a)2 + y
2 + z
2 + ( x + a)
2 + y
2 + z
2 = 2c
2
2[ x 2 + a
2 + y
2 + z
2] = 2c
2
x 2 + a
2 = c
2 – y
2 – z
2
48. Answer (3)
22 2 units AB
2 2(1 2) ( 3 0) 2 unitsBC
221 2 2
3 2 units03 3
CD
AD =
222 1 2 2
0(1 0) 03 3
=1 8
13 3
= 2 units
and diagonal,
2 2(1 0) ( 3 0) 2 units AC
222 1 2 2
0(2 1) 2 units03 3
BD
Since, the all four sides and diagonals are equal. Therefore given points are vertices of a rectangulartetrahedron.
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Q.No. Solution
49. Answer (2)
Let the XY -plane divides the join of (2, 4, 5) and (–4, 3, –2) in k : 1.
Therefore, the coordinates of the point are
4 2 3 4 2 5, ,1 1 1
k k k
k k k
z -coordinate on XY -plane is zero
2 5
01
k
k
5
2k
Hence, the required is 5 : 2
50. Answer (4)
Let A(–1, 3, 2), B(–4, 2, –2) and C (5, 5, y ) be the given points and C divides the join of AB in k : 1.
4 1 2 3 2 2
(5, 5, ), ,1 1 1
k k k y
k k k
4 1
51
k
k
,
2 35
1
k
k
and
2 2
1
k y
k
2
3k
and y = 10
51. Answer (1)
Centroid is given by
1 1 ( 2) 1 2 2 0 1 ( 1), ,
3 3 3
=5
0, , 03
52. Answer (3)
Let A(1, 4, 5) and B(2, 2, 3) are given points
AB = 2 2 2(2 1) (2 4) (3 5)
= 1 4 4
= 3 units
53. Answer (3)
The minimum distance of the point (1, 2, 3) from x -axis is
2 2( -coordinate) ( -coordinate)y z
= 2 22 3 13 units
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Q.No. Solution
54. Answer (1)
Circumcentre, centroid and orthocentre are collinear. Also centroid is a point of trisection and is closerto the circumcentre. Let the coordinates of circumcentre are ( x , y , z ). Point G divides line segment PO in the ratio 1 : 2.
O(–3, 5, 1)
P x y z ( , , )
1 2
G(3, 3, –1)
So,3 2
33
x ,
5 23
3
y and
1 21
3
z
x = 6, y = 2 and z = –2
Hence, the required coordinates are (6, 2, –2).
Section-B
Q.No. Solution
1. Answer (1, 4)
54 –458 –3222 a (a + 3)
2 + 9 = 25
(a + 3)2 = 16 a + 3 = 4, –4
a = 1, –7
2. Answer (2, 4)
Coordinates of P are
23
725 –3,
23
3 –243,
23
221 –3 =
5
1 –,
5
6,
5
1 for internal division, and
2 –3
72 –5 –3,
2 –3
3 –2 –43,
2 –3
22 –1 –3 = (–7, 18, –29) for external division.
3. Answer (2, 4)
a1 = 1 + , b1 = 1 – , c 1 = 2
a2 = 0, b2 = 1, c 2 = 0
2
22
1
212121cos
aa
c c bbaa
222222
0102 –11
20 –1110
2
1
62
–1
2
1
2
22 + 6 = 4(1 – )
2 2 2 – 8 – 2 = 0 2
– 4 – 1 = 0 52
4. Answer (1, 2)
a = 3, b = 4, c = 12
131243 222222 c ba l , m, n are13
12,
13
4,
13
3
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Q.No. Solution
5. Answer (3, 4)
Direction Ratio’s are 3, –2, 6
Direction cosine’s are7
6,
7
2 –,
7
3
Components are
63
7
6,63
7
2 –,63
7
3 = ± (27, –18, 54)
6. Answer (1, 3)
Direction ratio’s of AP are 1, 2, –2
Equation of AP is
say r z y x
2 –
1 –
2
1 –
1
1 –
To put r as the distance between points P and Q, direction ratios must be converted to directioncosines.
r z y x
3
2 –
1 –
3
2
1 –
3
1
1 –
Put r = 3, and –3, the coordinates of Q are (2, 3, 1) and (0, –1, 3).
7. Answer (1, 2, 3, 4)
Eliminating ‘n’ between the given equations
(6l + 5m) (3m + 5l ) + 2lm = 0
30l 2 + 15m
2 + 45 lm = 0 0132
2
m
l
m
l
1 –1
1 m
l ,
2
1 –
2
2 m
l
Similarly, eliminating ’m’ between the given equations 02 –53
5 –6
l n
l nnl
10l 2 – 5nl – 5n
2 = 0 01 – –2
2
n
l
n
l
1,2
1 –
2
2
1
1 n
l
n
l
16
1
211 –211 – 222
21
21
21111
nml nml
6
1
1 –21 –1 –21 – 222
22
22
22222
nml nml
The direction-cosines of the lines are
;6
2,
6
1,
6
1 –;
6
2 –,
6
1 –,
6
1
6
1,
6
2 –,
6
1;
6
1 –,
6
2,
6
1 –
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Solutions of Assignment (Set-2) Introduction to Three Dimensional Geometry (Solutions)
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Q.No. Solution
8. Answer (1, 2)
Let a point on sphere divides joining of two points in the ratio : 1
Its co-ordianates are27 12 9 4 18 8
, ,1 1 1
It lies on the sphere, therefore
(27 + 12)2 + (–9 – 4)
2 + (18 + 8)
2 = 504( + 1)
2
7292 + 144 + 648 + 812
+ 16 + 72 + 3242 + 64 + 288
= 5042 + 1008 + 504
6302 = 280
2 4 2
9 3
(1) & (2) are correct
Section-C
Q.No. Solution
Comprehension I
1. Answer (1)
2 2 21 1 1 1 l m n
21
1 11
3 3 n
1
1
3 n
2. Answer (2)
1 2 1 2 1 2cos60 l l m m n n
1 2 1 2 1 2
1
2 l l m m n n
1 1 2 1 1 2 1 1 2( ) ( ) ( ) l l + l m m m n n n
= 2 2 21 1 1 1 2 1 2 1 2( ) ( ) l + l l m n m m n n
=1 3
12 2
3. Answer (3)
cos =1 1 1 1 1 1
2 2 2 2 2 2
=1 1 1
04 4 2
= 90°
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Solutions of Assignment (Set-2) Introduction to Three Dimensional Geometry (Solutions)
Aakash Educational Services Pvt. Ltd. Redg. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.011-47623456
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Section-D
Q.No. Solution
1. Answer (1)
Let A, B, C be the points (1, 3, 5), (2, 4, 6) and (0, 5, 7) respectively.
Direction Ratio’s of AB, BC , and CA are 1, 1, 1 ; –2, 1, 1 ; 1, –2, –2
respectively.
Clearly, –2(1) + 1(1) + 1(1) = 0
AB BC .
In vector from OAOB AB –
= k j i ˆˆˆ 0 AB
OBOC BC – = k j i ˆˆˆ
2 – 0BC
BC AB . = –2(1) + 1(1) + 1(1) = 0
2. Answer (4)
Let points (2, 3, 5), (7, 5, 7) and (–3, 1, 3) be represented by A, B and C respectively.
335 –73 –52 –7222 AB
337 –35 –17 –3 –222 BC
ABC is not equilateral
3. Answer (3)
The angle given as = cos –1
(l 1 l 2 + m1 m2 + n1 n2) is the angle between the lines whose direction-cosines are l 1, m1, n1 and l 2, m2, n2 (not direction ratio’s)
4. Answer (2)
Centroid of a tetrahedron with vertices ( x 1, y 1, Z 1), ( x 2, y 2, z 2), ( x 3, y 3, z 3), ( x 4, y 4, z 4) is
4,
4,
4432143214321 z z z z y y y y x x x x
The centroid of the tetrahedron with given vertices is
3,2 –,14
12000,
4
08 –00,
4
0040
Statement 2 is true, but does not explain statement 1.
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Introduction to Three Dimensional Geometry (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
5. Answer (4)
The given equations are
x – y + z = 1
x + y – z = –1
x – 3y + 3z = 2
The system of equations can be put in matrix form as Ax = B
i.e.
1 1 1 1
1 1 1 1
1 3 3 2
x
y
z
1 1 1 1
~ 0 2 2 2
0 2 2 1
x
y
z
2 2 1
3 3 1
R R R
R R R
1 1 1 1
~ 0 2 2 2
0 0 0 1
x
y
z
which is inconsistent as
The three planes do not have a common point.
Statement–2 is true.
Since planes P 1, P 2, P 3 are pairwise intersecting, their lines of intersection are parallel.
Statement–1, is false.
Section-E
Q.No. Solution
1. Answer : A(q), B(p), C(s), D(r)
(A) 5,4,33
357,
3
273,
3
162
(B) 5,6,12
1 –11,
2
39,
2
5 –7
(C) Any point on532
z y x can be taken as (2k , 3k , 5k ).
20 –50 –30 –2222 k k k
38
2
k
Required point is
38
10,
38
6,
38
4
(D)
32
3052,
32
5302,
32
5302
= (3, 3, 2)
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Solutions of Assignment (Set-2) Introduction to Three Dimensional Geometry (Solutions)
Aakash Educational Services Pvt. Ltd. Redg. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.011-47623456
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Q.No. Solution
2. Answer : A(s), B(q), C(r), D(p)
Since cos2 + cos
2 + cos2 = 1
(1 – sin2) + (1 – sin
2) + (1 – sin2) = 1
sin
2
+ sin
2
+ sin
2
= 2
cos2 + cos
2 + cos2 = 2(cos
2 + cos2 + cos
2) – 3
= 2(1) – 3 = – 1
Since, 2cos( + ) cos ( – ) = cos2 + cos2
2 –coscoscos2 –coscos2 222
3. Answer A(p), B(p, q, s, t), C(q, r, s), D(q, r, s, t)
(A) 1 × 1 + 2 × 1 + 3 × (–1) = 0
hence = 90°
(B)1 2 3
2 4 6 , hence angle between the lines is zero.
(C) 1 2 1 1 1 1 2 2 2
cos31 1 1 4 1 1 3 6 3 2
1.411 1cos 0.47 60
3 2
o
(D) 1 × 0 + 0 × 1 + 0 × 0 = 0 = 90°
Section-F
Q.No. Solution
1. Answer (4)
(1 × ) + (( + 1)) + 2 = 0
+ 2 + + 2 = 0
2 + 4 = 0
= 0, –4
sum = 0 – 4 = –4Modulus of sum = 4
2. Answer (1)
Projection = 1 1 1
3 2 4 3 5 4 33 3 3
Integral value = 1
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Introduction to Three Dimensional Geometry (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
3. Answer (1)
Direction cosine =2 3 6
, ,7 7 7
= (cos, cos, cos)
But cos > 0
Hence directions cosines are
2 3 6, ,
7 7 7
sum =2 3 6
17
Modulus of sum = 1
4. Answer (5)
1 2 3
23
2 1 1 4
3 3
3 0 4 7
3 3
[] = 2 , [] = 1, [] = 2
[] + [] + [] = 5
Section-G
Q.No. Solution
1. Answer (1)
STATEMENT-1 distance = 2 23 4 5
STATEMENT-2 distance = 131
STATEMENT-3 length = 1 4 9 14
2. Answer (1)
STATEMENT-1
cos = 1 × (–12) + (2 × 3) + (3 ×2) = –12 + 6 + 6 = 0 , = 90°
STATEMENT-2
Direction cosine =3 4 12
, ,13 13 13
STATEMENT-3
cos2 + cos
2 + cos2 = 1
1 – sin2 + 1 – sin
2 + 1 – sin2 = 1
sin2 + sin
2 + sin
2 = 2
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Solutions of Assignment (Set-2) Introduction to Three Dimensional Geometry (Solutions)
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Section-H
Q.No. Solution
1. Let l 1, m1, n1 and l 2, m2, n2 be the direction cosines of the two lines satisfying the relations
pl + qm + rn = 0 and al 2 + bm2 + cn2 = 0
Eliminating n, we get
al 2 + bm2 + c
2
0 pl qm
r
2 2 2 2 2 2 2 2 2 0al r bm r cp l cq m cplqm
2
2 2 2 2( ) 2 ( ) 0l l
ar cp cpq br cqm m
2 2
1 2
2 2
1 2
l l br cq
m m ar cp
1 2 1 2
2 2 2 2
l l m m
br cq ar cp
from symmetry of the result we can write
1 2 1 2 1 2
2 2 2 2 2 2
l l m m n n
br cq ar cp aq bp
The two straight lines are perpendicular if
1 2 1 2 1 20l l m m n n
2 2 2 2 2 2 0br cq ar cp aq bp
2 2 2( ) ( ) ( ) 0 p b c q c a r a b
Also, for the two lines to be parallel, their direction cosines are equal and hence the discriminant ofthe above equation is equal to zero.
2 2 2 2 2 2 2
4 4( )( ) 0c p q ar cp br cq
2 2 2 4 2 2 2 2 2 2 2 0c p q abr acq r bcp r c p q
2 2 2 0abr acq bcp
2 2 2
0 p q r
a b c .
2. Let required point be P ( x , y , z )
d.r. of AB (11 + 9, 0 – 4, –1–5)
(20, –4, –6)
Or (10, –2, –3)
Also d.r. of line AB can be taken as
( x – 11, y – 0, z + 1)
11 1
10 2 3
x y z k
(say)
x = 10k + 11
y = – 2k
z = –3k – 1
A(–9, 4, 5) P x y z ( , , ) (11, 0, –1)
O(0, , )0 0
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Introduction to Three Dimensional Geometry (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
D.R. of OP are (10k + 11, –2k , –3k – 1)
AB is perpendicular to OP
(10k + 11) 10 + (–2k )(–2) + (–3k – 1) (–3) = 0
113k + 113 = 0 k = – 1
Co-ordinate of P are (10 –1 + 11, –2 –1, –3 (–1 –1))
= (1, 2, 2)
3. A (5, –1, 1), C = (1, –6, 10)
B (7, –4, 7), D = (–1, –3, 4)
2 2 2(2) (3) (6) 4 9 36 7 AB
2 2 2(6) ( 2) (3) 36 4 9 7BC
2 2 2(2) ( 3) (6) 4 9 36 7CD
2 2 2(5 1) ( 1 3) (1 4) 36 4 9 7DA
Also d.r. of AC = (4, 5, –9)
d.r. of BD = (8, –1, 3)
If ‘’ be angle between AC and BD then
2 2 2 2 2 2
(4.8) (5 1) ( 9 3)cos
4 5 ( 9) 8 ( 1) (3)
= 0
= 90°, diagonals are perpendicular
ABCD represents rhombus
4. Direction ratio’s of RS are 6, 2, 3
Direction consines of RS are7
3,
7
2,
7
6
Projection of PQ on RS = 7
32 –4
7
25 –6
7
62 –3 = 2
5.The Direction-consines of the four diagonals of a cube are given by
3
1,
3
1,
3
1;
3
1 –,
3
1,
3
1;
3
1 –,
3
1 –,
3
1;
3
1,
3
1 –,
3
1
If the direction-cosines of the line are l , m, n, then
nml nml
3
1
333cos ,
nml –3
1cos , nm – –1
3
1cos
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Solutions of Assignment (Set-2) Introduction to Three Dimensional Geometry (Solutions)
Aakash Educational Services Pvt Ltd Redg Office : Aakash Tower Plot No -4 Sector-11 Dwarka New Delhi-110075 Ph 011-47623456
Q.No. Solution
cos = nml – –3
1
cos2 + cos
2 + cos2 + cos
2
= 2222 –
3
1 – –
3
1 –
3
1
3
1nml nml nml nml
= 2222 – – – –
3
1nml nml nml nml
= 2222 –223
1nml nml ( (a + b)
2 + (a – b)
2 = 2(a
2 + b
2))
= 222 –23
2nmnml = 222 22
3
2nml = 222
3
4nml =
3
4
9(cos2 + cos2 + cos2 + cos2)2
= 22 1 –cos29 = 22222 4 –coscoscoscos29
=2
4 –3
429
=
2
3
4 –9
= 16
6. Clearly (1, 2, 5) and (–1, –1, –1) will be the end points of the diagonal
Length of diagonal = 222151211 = 7