Conic 3D MCQ Quiz - Objective Question with Answer for Conic 3D - Download Free PDF

Explanation:

The Centre of the sphere is C \((\frac{-3}{4}, \frac{-3}{4}, \frac{-3}{4})\)

4x + 8y + 8z + 15 = 0,

Plane satisfy the centre of sphere – 3 –6 – 6 + 15 = 0 

⇒ 0 = 0

∴ Option (d) is correct

Explanation:

Given:

Sphere: 2x² + 2y² + 2z² + 3x + 3y + 3z – 6 = 0

⇒ \(x^2 + y^2+z^2 + \frac{3}{2}x +\frac{3}{2}y + \frac{3}{2}z\) -3 = 0

Now Radius = \(\sqrt{(\frac{3}{4})^2+ (\frac{3}{4})^2+ (\frac{3}{4})^2 + 3}\)

= \(\sqrt{\frac{27}{16} +3} = 5 \frac{\sqrt3}{4}\)

Diamter = \(5 \frac{\sqrt3}{2}\)

∴ Option (b) is correct

Explanation:

We know that the reciprocal cone of cone ax2 + by2 + cz2 = 0 is \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}=0\)

In the cone x2 + 2y2 + 3z2 = 0

a = 1, b = 2, c = 3.

Hence reciprocal cone is

\(\frac{x^2}{1}+\frac{y^2}{2}+\frac{z^2}3=0\)

i.e., 6x2 + 3y2 + 2z2 = 0

Option (1) is true.

Explanation:

λ1S1 + λ2S2 = 0

⇒ λ1(x² + y² + z² + ...) + λ2(x2 + y2 + z2 + ...) = 0

If λ1 = -λ2 then the above equation does not represent an sphere.

Hence option (3) is true. 

Concept:

A cone is a three-dimensional geometric shape that tapers smoothly from a flat

base (usually circular) to a point called the apex or vertex. It has one flat

surface and one curved surface.

Explanation:

Option 1: A plane figure is a flat, two-dimensional shape like a circle, triangle,

or square. Since a cone is three-dimensional, it cannot be classified as a plane

figure.

Option 2: A curve is a smoothly flowing, continuous line without sharp angles.

Although a cone has a curved surface, it is not solely a curve.

Option 3: Solid Figure: A solid figure is a three-dimensional object. A cone fits

this definition as it has volume and occupies space.

Option 4:  A line is a straight one-dimensional figure having no thickness and

extending infinitely in both directions. A cone is not a line as it is three-dimensional.

The correct option is option 3.

Given:

Radius of hemisphere = 7 cm

Diameter of cone = 14 cm

Formula Used:

Volume of hemisphere = 2/3 × πR³

Volume of cone = 1/3 × πr²h

Calculation:

⇒ 2/3 × πR3 = 1/3 × πr2h

⇒ 2/3 × π73 = 1/3 × π72h

⇒ 2 × 73 = 1 × 72h

⇒ h = 14

The answer is 14 cm.

Explanation:

Given:

Sphere: 2x² + 2y² + 2z² + 3x + 3y + 3z – 6 = 0

⇒ \(x^2 + y^2+z^2 + \frac{3}{2}x +\frac{3}{2}y + \frac{3}{2}z\) -3 = 0

Now Radius = \(\sqrt{(\frac{3}{4})^2+ (\frac{3}{4})^2+ (\frac{3}{4})^2 + 3}\)

= \(\sqrt{\frac{27}{16} +3} = 5 \frac{\sqrt3}{4}\)

Diamter = \(5 \frac{\sqrt3}{2}\)

∴ Option (b) is correct

 Given:

 The equation of the sphere is given.

 Concept Used:

 Comparing the given equation of sphere to the standard form of equation of spheres.

 Solution:

 We have, standard equation of sphere as: (x - a)² + (y - b)² + (z - c)² = r²

 ⇒ x² + y² + z² - 2ax - 2by - 2cz + a² + b² + c² - r² = 0

 ⇒ comparing the coefficient of x, y, z and constant.

 we have, - 2a = - 16 {coefficient of x}

 ⇒ a = 8 

 we have, - 2b = 12 {coefficient of y}

 ⇒ b = - 6

 we have, - 2c = -2\( \sqrt{d}\)

 ⇒ c = \( \sqrt{d}\)

  we have, a² + b² + c² - r² = d 

 ⇒ (8)² + (- 6)² + (\( \sqrt{d}\))² - r² = d

 ⇒ 64 + 36 + d - r² = d

 ⇒ r² =100

 ⇒ r = 10

 we know that diameter of sphere is 2 times radius of sphere.

 D = 2r 

 ⇒ D = 20

 \(\therefore\)Option 2 is correct.

Explanation:

The Centre of the sphere is C \((\frac{-3}{4}, \frac{-3}{4}, \frac{-3}{4})\)

4x + 8y + 8z + 15 = 0,

Plane satisfy the centre of sphere – 3 –6 – 6 + 15 = 0 

⇒ 0 = 0

∴ Option (d) is correct

Given:

Radius of hemisphere = 7 cm

Diameter of cone = 14 cm

Formula Used:

Volume of hemisphere = 2/3 × πR³

Volume of cone = 1/3 × πr²h

Calculation:

⇒ 2/3 × πR3 = 1/3 × πr2h

⇒ 2/3 × π73 = 1/3 × π72h

⇒ 2 × 73 = 1 × 72h

⇒ h = 14

The answer is 14 cm.

Concept:

A cone is a three-dimensional geometric shape that tapers smoothly from a flat

base (usually circular) to a point called the apex or vertex. It has one flat

surface and one curved surface.

Explanation:

Option 1: A plane figure is a flat, two-dimensional shape like a circle, triangle,

or square. Since a cone is three-dimensional, it cannot be classified as a plane

figure.

Option 2: A curve is a smoothly flowing, continuous line without sharp angles.

Although a cone has a curved surface, it is not solely a curve.

Option 3: Solid Figure: A solid figure is a three-dimensional object. A cone fits

this definition as it has volume and occupies space.

Option 4:  A line is a straight one-dimensional figure having no thickness and

extending infinitely in both directions. A cone is not a line as it is three-dimensional.

The correct option is option 3.

Concept:

General equation of sphere is x² + y² + z² + 2ux + 2vy + 2wz + c = 0

Explanation:

Now, equation of sphere is  x2 + y2 + z2 + 2ux + 2vy + 2wz + c = 0 ....... (i)

Passes through (0,0,0), (a,0,0), (0,b,0), (0,0,c) 

Now, when (i) equation passes through (0,0,0) ⇒ c = 0 

Now, when (i) equation passes through (a,0,0) ⇒ u = \(\frac{-a}{2}\) 

Similarly, For (0,b,0) we get v = \(\frac{-b}{2}\) and for (0,0,c) we get w = \(\frac{-c}{2}\) 

By substituting all value we get

 x2 + y2 + z2 - ax - by - cz = 0

Now, By completing the square we get

\((x - \frac{a}{2})^2\) + \((y - \frac{b}{2})^2\) + \((z - \frac{c}{2})^2\) =  \(\frac{1}{4}\) (a2 + b2 + c2)

Hence, Option (3) is true

Explanation:

Given:

Sphere: 2x² + 2y² + 2z² + 3x + 3y + 3z – 6 = 0

⇒ \(x^2 + y^2+z^2 + \frac{3}{2}x +\frac{3}{2}y + \frac{3}{2}z\) -3 = 0

Now Radius = \(\sqrt{(\frac{3}{4})^2+ (\frac{3}{4})^2+ (\frac{3}{4})^2 + 3}\)

= \(\sqrt{\frac{27}{16} +3} = 5 \frac{\sqrt3}{4}\)

Diamter = \(5 \frac{\sqrt3}{2}\)

∴ Option (b) is correct

 Given:

 The equation of the sphere is given.

 Concept Used:

 Comparing the given equation of sphere to the standard form of equation of spheres.

 Solution:

 We have, standard equation of sphere as: (x - a)² + (y - b)² + (z - c)² = r²

 ⇒ x² + y² + z² - 2ax - 2by - 2cz + a² + b² + c² - r² = 0

 ⇒ comparing the coefficient of x, y, z and constant.

 we have, - 2a = - 16 {coefficient of x}

 ⇒ a = 8 

 we have, - 2b = 12 {coefficient of y}

 ⇒ b = - 6

 we have, - 2c = -2\( \sqrt{d}\)

 ⇒ c = \( \sqrt{d}\)

  we have, a² + b² + c² - r² = d 

 ⇒ (8)² + (- 6)² + (\( \sqrt{d}\))² - r² = d

 ⇒ 64 + 36 + d - r² = d

 ⇒ r² =100

 ⇒ r = 10

 we know that diameter of sphere is 2 times radius of sphere.

 D = 2r 

 ⇒ D = 20

 \(\therefore\)Option 2 is correct.

Explanation:

The Centre of the sphere is C \((\frac{-3}{4}, \frac{-3}{4}, \frac{-3}{4})\)

4x + 8y + 8z + 15 = 0,

Plane satisfy the centre of sphere – 3 –6 – 6 + 15 = 0 

⇒ 0 = 0

∴ Option (d) is correct