Introduction to Three Dimensional Geometry Test

Result

Introduction to Three Dimensional Geometry Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

The coordinates of the point which divides the line segment joining the points ( 5 , 4, 2 ) and ( - 1 , - 2 , 4 ) in the ratio 2 : 3 externally is

A
( 13/5 , 8/5 , 14/5 )

B
( 17 , 16 , -2 )

C
( 17/5 , 16/5 , - 2/5 )

D
none of these

Solution

for external division ration must be greater than 1
Question 2
1 / -0

Lines OA , OB are drawn from 0 with direction cosines proportional to <1, - 2, - 1> and <3, - 2, 3> respectively . The direction ratios of the normal to the plane AOB are

A
<- 4, - 3, 2>

B
<-4, 3, 2>

C
<4, 3, -2>

D
<4, 3, 2>

Solution

let the direction ratios of normal to plane be a, b, c

the line in the plane with drs 1, -2, -1 and 3,-2,3 must be perpendicular to normal

then a₁a₂+b₁b₂+c₁c₂ = 0

thus a-2b-c =0

3a-2b +3c = 0

solving by cross multiplication a = -8k, b = -6k c = 4k

drs of normal will be 4,3,-2
Question 3
1 / -0

The points A ( 5 , - 1 , 1 ) , B ( 7 , - 4 , 7 ) , C ( 1 , - 6 , 10 ) and D ( - 1 , - 3 , 4 ) are the vertices of

A
square

B
rhombus

C
rectangle

D
none of these

Solution

distance between AB=7, BC =7 CD= 7 AND DA=7

whereas distance betweeen AC and BD IS NOT EQUAL
Question 4
1 / -0

The centre of sphere passing through four points ( 0 , 0 , 0 ) , ( 0 , 2 , 0 ) , ( 1 , 0 , 0 ) and ( 0 , 0 , 4 ) is

A
( 1 , 1/2, 2 )

B
( -1/2 , 1 , 2 )

C
( 1/2, 1 , 2 )

D
( 1/2, 1 , - 2 )

Solution

Equation of sphere is given by x² + y² + z² + 2ux + 2vy + 2wz + d = 0

Since sphere passing through four points (0,0,0), (0,2,0), (1,0,0) and (0,0,4)

Substituting in the given equation we get d =0

4 + 4v = 0

v = -1

1 + 2u = 0

u = -1/2

16+8w = 0

w = -2

Centre of sphere (-u, -v, -w) = (1/2, 1, 2)