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# Solution- Three Dimensional Geometry Test-2 Class 12 Notes | EduRev

## Class 12 : Solution- Three Dimensional Geometry Test-2 Class 12 Notes | EduRev

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CBSE TEST PAPER-07
CLASS - XII MATHEMATICS (Vectors & Three Dimensional Geometry)
Topic: - Three Dimensional Geometry [ANSWERS]
Ans1. x 2 – x 1, y 2 – y, and z 2-z 1 are the direction ratio of the line segment PQ.
Ans2. Comparing the given equation with the standard equation form
? ? ? ?
1 1 1
( 3 5 6 ) (2 4 2 )
x x y y z z
a b c
r i j k i j k ?
- - - = =
= - + + + + +

? ?
Ans3. x 1=-3, y 1 = 1, z 1 = 5
a 1 = -3, b 1=1, c 1= 5
x 2 = -1, y 2=2, z 2 = 5
a 2 = -1, b 2 = 2, c 2 = 5
2 1 2 1 2 1
1 1 1
2 2 2
2 1 0
3 1 5 0
1 2 5
x x y y z z
a b c
z b c
- - - = - =
-
Therefore lines are coplanar.
Ans4.
? ? ?
1 1 , 2 a i j b i j k = + = - +
 
? ?
? ? ? ?
?
? ?
? ?
2 2
2 1
1 2
1 2
1 2 2 1
1 2
2 , 3 5 2
2 1 1
3 5 2
3 7
59
3 0 7
( ).( ) 10
59 59
a i j k b i j k
a a i k
i j k
b b
i j k
b b
b b a a
d
b b
= + - = - +
- = - × = - - = - - × =
- +
× - = = =
×
 
? ?
 
?
?
 
?
 
   
 
Page 2

CBSE TEST PAPER-07
CLASS - XII MATHEMATICS (Vectors & Three Dimensional Geometry)
Topic: - Three Dimensional Geometry [ANSWERS]
Ans1. x 2 – x 1, y 2 – y, and z 2-z 1 are the direction ratio of the line segment PQ.
Ans2. Comparing the given equation with the standard equation form
? ? ? ?
1 1 1
( 3 5 6 ) (2 4 2 )
x x y y z z
a b c
r i j k i j k ?
- - - = =
= - + + + + +

? ?
Ans3. x 1=-3, y 1 = 1, z 1 = 5
a 1 = -3, b 1=1, c 1= 5
x 2 = -1, y 2=2, z 2 = 5
a 2 = -1, b 2 = 2, c 2 = 5
2 1 2 1 2 1
1 1 1
2 2 2
2 1 0
3 1 5 0
1 2 5
x x y y z z
a b c
z b c
- - - = - =
-
Therefore lines are coplanar.
Ans4.
? ? ?
1 1 , 2 a i j b i j k = + = - +
 
? ?
? ? ? ?
?
? ?
? ?
2 2
2 1
1 2
1 2
1 2 2 1
1 2
2 , 3 5 2
2 1 1
3 5 2
3 7
59
3 0 7
( ).( ) 10
59 59
a i j k b i j k
a a i k
i j k
b b
i j k
b b
b b a a
d
b b
= + - = - +
- = - × = - - = - - × =
- +
× - = = =
×
 
? ?
 
?
?
 
?
 
   
 
Ans5.
? ?
1 3 2 6 b i j k = + +

?
? ?
2 2 2 b i j k = + +

?
The angle ? between them is given by
? ? ? ?
? ? ? ?
1 2
1 2
.
cos
(3 2 6 ).( 2 2 )
3 2 6 2 2
3 4 12
49 9
19 19
7 3 21
b b
b b
i j k i j k
i j k i j k
? =
+ + + +
+ + + +
+ +
= =
×
 
 
? ?
? ?
Ans6.
5 ( 2) 0
7 5 1
x y z - - - - = =
- 1 1 1
2 2 2
0 0 0
1 2 3
7, 5, 1
1, 2, 3
x y z
a b c
a b c
- - - = =
= = - =
= = =
For ?
a 1a 2+b 1b 2+c 1c 2=0
L.H. S
1 2
7 1 ( 5 2) 1 3
7 10 3
0
hence l l
= × + - × + ×
= - +
=
?
Ans7. Let
? ?
? ?
? ?
2 5 3
2 3 5
5 3 3
a i j k
b i j k
c i j k
= + - = - - +
= + - 
?

?

?
Vector equation is
? ? ? ? ?
( ).[( ) ( )] 0
[ (2 5 3 )].[( 4 8 8 ) (3 2 )] 0
r a b a c a
r i j k i j k i j
- - × - =
- + - - - + × - =
     

? ? ?
Page 3

CBSE TEST PAPER-07
CLASS - XII MATHEMATICS (Vectors & Three Dimensional Geometry)
Topic: - Three Dimensional Geometry [ANSWERS]
Ans1. x 2 – x 1, y 2 – y, and z 2-z 1 are the direction ratio of the line segment PQ.
Ans2. Comparing the given equation with the standard equation form
? ? ? ?
1 1 1
( 3 5 6 ) (2 4 2 )
x x y y z z
a b c
r i j k i j k ?
- - - = =
= - + + + + +

? ?
Ans3. x 1=-3, y 1 = 1, z 1 = 5
a 1 = -3, b 1=1, c 1= 5
x 2 = -1, y 2=2, z 2 = 5
a 2 = -1, b 2 = 2, c 2 = 5
2 1 2 1 2 1
1 1 1
2 2 2
2 1 0
3 1 5 0
1 2 5
x x y y z z
a b c
z b c
- - - = - =
-
Therefore lines are coplanar.
Ans4.
? ? ?
1 1 , 2 a i j b i j k = + = - +
 
? ?
? ? ? ?
?
? ?
? ?
2 2
2 1
1 2
1 2
1 2 2 1
1 2
2 , 3 5 2
2 1 1
3 5 2
3 7
59
3 0 7
( ).( ) 10
59 59
a i j k b i j k
a a i k
i j k
b b
i j k
b b
b b a a
d
b b
= + - = - +
- = - × = - - = - - × =
- +
× - = = =
×
 
? ?
 
?
?
 
?
 
   
 
Ans5.
? ?
1 3 2 6 b i j k = + +

?
? ?
2 2 2 b i j k = + +

?
The angle ? between them is given by
? ? ? ?
? ? ? ?
1 2
1 2
.
cos
(3 2 6 ).( 2 2 )
3 2 6 2 2
3 4 12
49 9
19 19
7 3 21
b b
b b
i j k i j k
i j k i j k
? =
+ + + +
+ + + +
+ +
= =
×
 
 
? ?
? ?
Ans6.
5 ( 2) 0
7 5 1
x y z - - - - = =
- 1 1 1
2 2 2
0 0 0
1 2 3
7, 5, 1
1, 2, 3
x y z
a b c
a b c
- - - = =
= = - =
= = =
For ?
a 1a 2+b 1b 2+c 1c 2=0
L.H. S
1 2
7 1 ( 5 2) 1 3
7 10 3
0
hence l l
= × + - × + ×
= - +
=
?
Ans7. Let
? ?
? ?
? ?
2 5 3
2 3 5
5 3 3
a i j k
b i j k
c i j k
= + - = - - +
= + - 
?

?

?
Vector equation is
? ? ? ? ?
( ).[( ) ( )] 0
[ (2 5 3 )].[( 4 8 8 ) (3 2 )] 0
r a b a c a
r i j k i j k i j
- - × - =
- + - - - + × - =
     

? ? ?
Ans8. Let
?
? ?
? ? ?
.( ) 2
( )( ) 2
2
r xi yi zk
r i j k
xi yi zk i j k
x y z
= + +
+ - =
+ + + - =
+ - =

? ?

?
? ? ?

Which is the required equation of plane.
Ans9. Equation of any plane through the intersection of given planes can be taken
as  (3 2 4) ( 2) 0.....( ) x y z x y z i ? - + - + + + - =
The point (2,2,1) lies in this plane
2 / 3 ? = - put in eq ….(i)
( )
2
3 2 4 ( 2) 0
3
7 5 4 8 0
x y z x y z
x y z
- + - - + + - =
- + - =
Ans10. The given plane is
? ?
? ? ? ?
.(3 4 12 ) 13 0
( ).(3 4 12 ) 13 0
3 4 12 13 0....( )
r i j k
xi y j zk i j k
x y z i
+ - + =
+ + + - + =
+ - + =

?
? ?
This plane is equidistant from the points (1,1,P) and (-3,0,1)
2 2 2 2 2 2
3(1) 4(1) 12 13 3( 3) 4(0) 12(1) 13
3 4 ( 12) 3 4 ( 12)
20 12 8
20 12 8
7
1
3
p
p
p
p or
+ - + - + - +
=
+ + - + + - - = - - = ±
= -
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