3-d geometry (package solutions)

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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

- 303 -

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



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.





- 305 -

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





- 306 -

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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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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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


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


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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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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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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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.


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.





- 314 -

Q.No. Solution

49. Answer (2)

Let the XY -plane divides the join of (2, 4, 5) and (–4, 3, –2) in k : 1.


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





- 315 -

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





- 316 -

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 –





- 317 -

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°





- 319 -

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.





- 320 -

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)





- 321 -

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





- 322 -

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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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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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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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

3-d geometry (package solutions)

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