Practice Question On 3 D For IIT-JEE , NDA and AIRFORCE

3-D

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Question no. 1

Co-ordinate of a point equidistant from the points (0,0,0) ,(a,0,0) , (0,b,0) , (0,0,c) is

looks_one

looks_two

looks_3

looks_4 (a,b,c)

Option looks_3

Solution :
is the point which is equidistant from the (0,0,0) ,(a,0,0) , (0,b,0) , (0,0,c).

Question no. 2

Let (3,4,-1) and (-1 ,2 ,3 ) are the end points of a diameter of sphere. then the radius of the sphere is equal to

looks_one 1

looks_two 2

looks_3 3

looks_4 9

Option looks_3 3

Solution :

where x₁ , y₁ , z₁ and x₂ , y₂ , z₂ are the end points of diameters.
Now, put the value in the formula .

Question no. 3

The equation of the sphere touching the three co-ordinate planes is

looks_one x² + y² + z² + 2a( x + y + z) + 2a² = 0

looks_two x² + y² + z² - 2a( x + y + z) + 2a² = 0

looks_3 x² + y² + z² ± 2a( x + y + z) + 2a² = 0

looks_4 None of these

Option looks_3 x² + y² + z² ± 2a( x + y + z) + 2a² = 0

Solution :
after finding the radius and center of sphere (consider a as any number), try to plot the sphere , you will find this equation in option 3 are touching the coordinate planes.

Question no. 4

The locus of the equation x² + y² + z² + 1 =0

looks_one An empty set

looks_two A sphere

looks_3 A degenerate set

looks_4 A pair of planes

Option looks_one An empty set

Solution :
The Given equation does not provide the real values solution (no real values of x, y and z exist for given values) , therefore locus of the equation is empty set.

Question no. 5

The equation x² + y² + z² =0 represents

looks_one (0,0,0)

looks_two A circle

looks_3 A plane

looks_4 None of these

Option looks_one (0,0,0)

Solution :
Only (0, 0, 0) real values satisfy the following equation

Question no. 6

A point moves so that the sum of the squares of its distances from two given points remains constant . The locus of the point is

looks_one A line

looks_two A plane

looks_3 A sphere

looks_4 None of these

Option looks_3 A sphere

Solution :
A point moves so that the sum of the squares of its distances from two given points remains constant . The locus of the point is sphere

Question no. 7

Equation ax² + by² + cz² + 2fyz + 2gxz + 2hxy + 2ux + 2vy + 2wz + d = 0 represent a sphere , if

looks_one a= b= c

looks_two f = g = h = 0

looks_3 v = u = w

looks_4 a = b = c and f = g = h =0

Option looks_4 a = b = c and f = g = h =0

Solution : Standard equation of sphere is (x-a)² + (y-b)² + (z-c)² = r²
it means cofficient of x , y , z must be equal and there should not be term like xy , yz , xz. therefore a = b = c and f = g = h = 0

Question no. 8

The co-ordinates of the point where the line meets the planes x- y -z = 3 are

looks_one (2,1,0)

looks_two (7,-1,-7)

looks_3 (1,2,-6)

looks_4 (5,-1,1)

Option looks_two not cross each other

very soon

Question no. 9

The co-ordinates of the point where the line through P (3 , 4 , 1) and Q (5, 1 ,6) crosses the xy plane are

looks_one

looks_two

looks_3

looks_4

Option looks_3

Solution:
let's find the equation of line :

Question no. 10

The equation of the plane through the origin containing the line is

looks_one 2x + 5y - 6z= 0

looks_two x -5y -5z = 0

looks_3 x- 5y + 3z =0

looks_4 x + y - z = 0

Option looks_3 x- 5y + 3z =0

Solution :
first find the points of the lines , the points of the given lines are (1,2,3) and (4,2,2)
Now these points must satisfy the plane equation. option 3 is satisfying both the points.

Question no. 11

The distance between the line and the plane 2x -2y -z = 6 is

looks_one 9

looks_two 1

looks_3 2

looks_4 3

Option looks_4 x² + a² = b² - y²

very soon 11

Question no. 12

The line is parallel to the plane

looks_one 2x+ 3y + 4z = 29

looks_two 3x + 4y -5z = 10

looks_3 3x + 4y + 5z = 38

looks_4 x + y + z = 0

Option looks_two 3x + 4y -5z = 10

Solution :
When the line is parallel to plane , al+ bm + cn = 0
where a,b,c are direction ratios of line and l ,m, n are components of the the normal vector of the plane
a = 3 , b = 4 , c = 5
only option 2 , components of normal vector are 3 , 4 , -5
al+bm+cn = 3(3) + 4(4) + 5(-5) = 25 -25 = 0

Question no. 13

The equation of plane through the line of intersection of planes ax + by+ cz+ d =0 , a'x + b'y+ c'z+ d' =0 and the parallel to the line y=0 , z= 0 is

looks_one (ab' - a'b)x + (bc' - b'c)y + (ad' - a'd)= 0

looks_two (ab' - a'b)x + (bc' - b'c)y + (ad' - a'd)z= 0

looks_3 (ab' - a'b)y + (bc' - b'c)z + (ad' - a'd)= 0

looks_4 None of these

Option looks_3 X=2

very son 13

Question no. 14

If line is parallel to the plane ax + by+ cz+ d =0 , then

looks_one

looks_two al + bm + cn = 0

looks_3

looks_4 None of these

Option looks_two al + bm + cn = 0

Solution:
When the line is parallel to plane , it means direction ratio of line will be normal to the normal of the plane,So al + bm + cn = 0

Question no. 15

The line and the plane 4x+ 5y + 3z -5 = 0 intersect at a point

looks_one (3,1,-2)

looks_two (3,-2 ,1)

looks_3 (2,-1 , 3)

looks_4 (-1 , -2 , -3)

Option looks_two (3,-2 ,1)

Solution :
first find the value of x,y ,z in terms of constant λ

x = 3λ - 3
y = -2λ+ 2
z = λ -1
Put the value of x, y and z in equation of plane
4x+ 5y + 3z -5 = 0
4(3λ - 3)+ 5(-2λ+ 2) +3(λ -1) - 5 = 0
12λ -12 - 10λ + 10 + 3λ - 3 -5 = 0
5λ-10 = 0
λ = 2
Now put the value of λ to find the value of x , y and z
x = 3λ - 3 = 3(2) - 3 = 3
y = -2(2) + 2 = -2
z = 2-1 = 1

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