Three-dimensional Geometry Test

Result

Three-dimensional Geometry Test - 8

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0.25

The equation of the plane passing through the point (3, –3, 1) and perpendicular to the line joining the points (3, 4, –1) and (2, –1, 5) is:

A
–x –5y + 6z + 18 = 0

B
x –5y + 6z + 18 = 0

C
x + 5y –6z + 18 = 0

D
–x –5y –6z + 18 = 0

Solution

The equation of the plane passing through the point (3, –3, 1) is:
a(x –3) + b(y + 3) + c(z –1) = 0 and the direction ratios of the line joining the points
(3, 4, –1) and (2, –1, 5) is 2 –3, –1 –4, 5 + 1, i.e., –1, –5, 6.
Since the plane is perpendicular to the line whose direction ratios are –1, –5, 6, therefore, direction ratios of the normal to the plane is –1, –5, 6.
So, required equation of plane is: –1(x –3) –5(y + 3) + 6(z –1) = 0
i.e., x + 5y –6z + 18 = 0.
Question 2
1 / -0.25

The equation of the plane, which is at a distance of 5 unit from the origin and has as a normal vector, is:

A

B

C

D

Solution

x = 3i - 2j - 6k
|x| = ((3)² + (2)² + (6)² )
|x| = (49)^½
|x| = 7
x = x/|x|
= (3i - 2j - 6k)/7
The required equation of plane is r.x = d
⇒ r.(3i - 2j - 6k)/7 = 5
⇒ r.(3i - 2j - 6k) = 35
Question 3
1 / -0.25

The Cartesian form of the equation of the plane is:

A
2x+3y-z=10

B
2x+3y-z= √14

C
2x+3y-z+√14=0

D
2x+3y-z+10=0

Solution

Let r = xi + yj + zk
(xi + yj + zk) . (2i + 3j - k) = 10
2x + 3y - k = 10
Question 4
1 / -0.25

The length of the perpendicular from the origin to the plane 3x + 2y –6z = 21 is:

A
3

B
14

C
21

D
7

Solution

The given point is P(0,0,0) and the given line is 3x+2y-6z-21=0
Let d be the length of the perpendicular from P(0,0) to the line 3x+2y-6z-21=0
Then, d= |3*0 + 2*0 + (-6)*0 -21|/[(3)^2 + (2)² + (-6)² ]^1/2
= |-21|/7
= 3
Question 5
1 / -0.25

The slope intercept form of the linear equation is

A
y = c ⁄b - ax ⁄b

B
y = b ⁄c - ax ⁄b

C
y = c ⁄b - ab ⁄x

D
xy = bx ⁄c - ay ⁄b
Question 6
1 / -0.25

If l, m, n are the direction cosines of the normal to the plane and p be the perpendicular distance of the plane from the origin, then the equation of the plane is:

A

B

C

D

Solution

Let P(x, y, z) be any point on the plane.
OP = r = xi + yj +zk
Let l,m,n be the direction of cosine of n, then
n - li + mj + nk
As we know that r.n = d
(xi + yj + zk)(li + mj + nk) = d
i.e. lx + my + nz = d
this is the cartesian equation of plane in normal form.
Question 7
1 / -0.25

If is the normal from the origin to the plane, and is the unit vector along . P(x, y, z) be any point on the plane and is perpendicular to . Then

A

B

C

D

Solution

NP is perpendicular to ON
NP.ON = |NP||ON| cos θ (θ= 90deg)
|NP||ON|cos90deg
⇒0
Question 8
1 / -0.25

The length of the perpendicular from the origin to the plane 2x –3y + 6z = 21 is:

A
14 unit

B
21 unit

C
7 unit

D
3 unit
Question 9
1 / -0.25

The angle between two lines whose direction ratios are 1,2,1 and 2,-3,4 is:

A
30 °

B
60 °

C
90 °

D
45 °

Solution

(1,2,1) and (2, -3, 4)
Cos θ= (a1b1+a2b2+a3b3]/[(a1)² +(a2)² + (a3)² )]^1/2 [(b1)² + (b2)² + (b3)² ]^1/2
Cos θ= [(1)(2) + (2)(-3) + (1)(4)]/ [(1)^2 + (2)^2 + (1)² ]^½ [(2)² + (-3)² + (4)² ]^½
Cos θ= [2 - 6 + 4]/[1 + 4 + 1]^½ [4 + 9 + 16]^½
Cos θ= 0
θ= 90deg
Question 10
1 / -0.25

The equation of the plane passing through the points (2, 1, 0), (3, –2, –2) and (3, 1, 7) is:

A
7x + 3y –z = 17

B
7x + 3y –2z = 21

C
7x + 3y –z = 21

D
-7x –3y + z = 17

Solution

We know that, the equation of a plane passing through three non-collinear points (x1. y1, z1), (x2 , y2 , z2) and (x3, y3, z3)
{(x-x1) (y-y1) (z-z1)} {(x2-x1) (y2-y1) (z2-z1)} {(x3-x2) (y3-y2) (z3 - z2)}
= {(x-2) (y-1) (z-0)} {(3-2) (-2-1) (-2-0)} {(3-2) (1-1) (7-0)} = 0
= {(x-2) (y-1) (z)} {(1) (-3) (-2)} {(1) (0) (7)}
=>(x-2)(-21) - (y-1)(9) + z(3) = 0
7x + 3y - z = 17