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JEE Questions for Maths Three Dimensional Geometry Quiz 8 - MCQExams.com
JEE
Maths
Three Dimensional Geometry
Quiz 8
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The plane 2
x
- 2y + z + 5 = 0 is a tangent to the sphere (
x
- 2)
2
+ (y - 2)
2
+ (z - 1)
2
= r
2
, if r equals
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1
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2
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4
0%
None of these
The centre and radius of the sphere
x
2
+ y
2
+ z
2
+ 3
x
- 4z + 1 = 0 are
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2)
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The shortest distance from the plane 12
x
+ 4y + 3z = 327 to the sphere
x
2
+ y
2
+ z
2
+ 4
x
- 2y - 6z = 155 is
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26
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2)
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13
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39
The radius of the sphere
x
2
+ y
2
+ z
2
=
x
+ 2y + 3z is
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√14/2
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√7
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7/2
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√7/2
If (2, 3,is one end of a diameter of the sphere
x
2
+ y
2
+ z
2
- 6
x
- 12y - 2z + 20 = 0, then the coordinates of the other end of the diameter are
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(4, 9, -3)
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(4, -3, 3)
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(4, 3, 5)
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(4, 3, -3)
A plane
x
+ y + z = -a√3 touches the sphere 2
x
2
+ 2y
2
+ 2z
2
- 2
x
+ 4y - 4z + 3 = 0, then the value of a is
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1/√3
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1/2√3
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1 - (1/√3)
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1 + (1/√3)
The plane
x
+ 2y - z = 4 cuts the sphere
x
2
+ y
2
+ z
2
-
x
+ z - 2 = 0 in a circle of radius
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√2
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2
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1
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3
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(-1, -1, 0)
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(1, 1, 1)
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2)
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0%
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(0, 6, -and (1, -2, -1)
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(0, 6, -and (-1, -4, -2)
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(1, -2, -and (1, 4, -2)
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(1, -2, -and (0, -6, 1)
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1/4
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- (1/4)
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1/8
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-(1/8)
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hyperbola
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circle
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straight line
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ellipse
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parabola
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2)
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Find the equation of the perpendicular drawn from the origin to the plane 2
x
+ 4y - 5z = 10
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r = (2k, 5k, 4k), k ϵ R
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r = (2k, 4k, -5k), k ϵ R
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r = (3k, 4k, 5k), k ϵ R
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None of the above
The vector equation of the plane passing through the origin and the line of intersection of the plane r ∙a = λ and r ∙b = μ is
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r ∙(λa - μb) = 0
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r ∙(λb - μa) = 0
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r ∙(λa + μb) = 0
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r ∙(λb + μa) = 0
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2)
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The vector equation of the line passing through the points (3, 2,and (-2, 1,is
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2)
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0%
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10/9
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3/10
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10/3√3
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10/9
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(3, 1, 4)
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(1, 3, 4)
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(1, 2, 4)
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None of these
Though the point P(α, β, γ), a plane is drawn at right angle to OP to meet the coordinate axes at A, B, C respectively. IF OP = p, then equation of the plane ABC is
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αx + βy + γz = p
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2αx + 2βy + 2γz = p2
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αx + βy + γz = p2
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0%
2)
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0%
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The equation of the plane passing through three non - collinear points a, b, c is
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r∙(b × c + c × a + a × b) = 0
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r∙(b × c + c × a + a × b) = [a b c]
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r∙[ a ×(b × c)] = [a b c]
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r∙(a + b + c) = 0
The vector equation of the sphere whose centre is the point (1, 0,and radius is 4, is
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2)
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0%
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If the points (–1, 2, – 3), (4, a,and (b, 8,are collinear, then a and b are respectively equal to
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5 and 5
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9 and 5
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5 and 9
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–5 and 9
The triangle formed by the points (0, 7,10), (–1, 6, 6), (–4, 9,is
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Equilateral
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Isosceles
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Right-angled
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Right-angled isosceles
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2)
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20
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10
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30
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40
The direction cosines of a line equally inclined to all the three rectangular co-ordinate axis are
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1, 1, 1
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None of these
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