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When it comes to preparing for board exams, RD Sharma books are the best. For many students, board exams have always been a ruse. Still, when they refer to this book in combination with Rd Sharma Class 12th Exercise MCQ solutions, they find it much easier to relax and understand the concept in a much better manner. RD Sharma solutions This exercise contains 24 questions which are based on the Equation of the plane containing two lines, Equation of the plane in scalar product form, Reflection of point, Vector and Cartesian form, equation in normal form and distance between the plane.

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The Plane exercise multiple choice question 1

Option (a).

Simplify the given equation.

The given equation plane is,

We can rewrite the equation of the given plane as,

So, the given plane passes through the intersection of the planes

.

The Plane exercise multiple choice question 2

Option(b)

Angle between two planes is 0.

We know, the angle between two planes

Here a

So, the acute angle between the planes

is

Therefore, the acute angle between the planes

The Plane exercise multiple choice question 3

Option (d)

is a scalar quantity.

are perpendicular to the plane

The equation of the plane through the intersection of the planes

is given as,

where is a scalar.

Given, that the required plane is perpendicular to the plane

so we have,

Therefore, the equation of the required plane is,

The Plane exercise multiple choice question 4

Option (c).

Distance between two parallel planes is 0.

We know that, the distance between two parallel planes:

and

is given by

Here, the two parallel planes are given as:

and

i.e.

Compiling equation (3) with equation (1), and equation (4) with equation (2), we get;

So, the distance between the given two parallel planes are,

Hence, the distance between the parallel planes

The Plane exercise multiple choice question 5

Option (b)

Solve the equation simultaneously.

(1, 3, 4) in the plane 2x - y + z + 3 = 0

We know, if the image of a point P(x

The given plane is 2x - y + z + 3 = 0 . . . . .(2)

By comparing equations (2) and (3)

A = 2, B = -1, C = 1, D = 3

And here x_0 = 1, y_0 = 3, z_0 = 4

So

Therefor,

x

=-3

y

=5

z

=2

So, the image of the (1, 3, 4), in the plane 2x - y + 3 = 0 , is (-3, 5, 2).

The Plane exercise multiple choice question 6

Option (d).

See for co-planer.

We know, the two lines given as,

Lines passes through (1,-1,0)

Hence, option a is correct.

The Plane exercise multiple choice question 7

Option (a)

Use cross product of vector.

The given plane is

So, it is clear from the given equation of plane, that the plane passing through a point and parallel to the two vectors

Therefore, the equation of the vector normal to the plane is given as

So, in scalar product form the vector equation of the plane is given as,

The Plane exercise multiple choice question 8

Option (b)

Use scalar product of vector

from the plane

We have the straight line given as,

And the plane as,

the normal vector of the plane given as,

if the straight line and the plane are parallel,

scalar product will be zero.

Hence, the point (2, -2, 3) is on the straight line. Distance from point (2, -2, 3) to the plane will be equal to the distance of the line from the plane. We know, that the distance of a point (x0, y0, z0) from plane Ax + By + Cz + D = 0 ...(2) is

On comparing, equation (1), i.e., x - 5y + z - 5 = 0 with equation (2), we get,

A = 1, B = -5, C = 1, D = -5

So, the distance is,

The Plane exercise multiple choice question 9

Option (a)

Solve the given equations simultaneously.

Equation of line passing through the line

is given by,

where k is constant

Again, the required plane is parallel to the line

So, we have

Put k in eq. (1)

Therefore, the equation of the plane through the line x + y + z + 3 = 0, 2x - y + 3z + 1 = 0 and the parallel to the line

x - 5y + 3z = 7

The Plane exercise multiple choice question 10

Option (a)

Put

The plane contains the line

so, the plane contains the point

so, we got the plane, they are:

Let,

The normal of two vectors and

The general equation of plane is,

Hence, the vector equation of the plane containing the line

and of the point

is

The Plane exercise multiple choice question 11

Option (c)

Use the centroid formula.

A plane meets the co-ordinate area of (A, B, C) such that the centroid of is the point (a,b,c).

Let the co-ordinate of the point

By the centroid formula,

The Intercept form of the plane–

If the plane makes intercepts at (p, 0, 0), (0, q, 0) and (0, 0, r) with the axes x, y, z respectively.

Here,

So, the equation of the plane is,

By comparing the equation (1) with given equation of the plane

,

we get k = 3.

The Plane exercise multiple choice question 12

Option (c)

The plane, must satisfy both the equation.

Let, the point of intersection of the line

And the plane x + y + z = 17 be (x

Put these values in the equation of plane x + y + z = 17

Hence, the distance between point (4, 6, 7) and (3, 4, 5) is,

The Plane exercise multiple choice question 13

Solve linear equations in two variables

The two planes are

The line of intersection of planes

Can be written as 3x - y + z = 1 ……………… (i)

x + 4y - 2z = 2 …………… (ii)

Solving eqn (i) & (ii)

Vector equation is

The Plane exercise multiple choice question 14

Option (c)

Let, the equation of the plane be Ax + By + Cz + D = 0, as the plane is perpendicular to, so

We have,

A = 3, B = 0 and C = 4 as the plane passes through (1, 1, 1), we have,

So, the equation of the plane becomes

3x + 4z - 7 = 0

Now, the perpendicular distance of the plane from the origin is

The Plane exercise multiple choice question 15

Option (a)

Use straight line & vector cross product.

x - 1 = 2y - 5 = 2z and 3x = 4y -11 = 3z - 4

The required plane is parallel to the lines

x - 1 = 2y - 5 = 2z and 3x = 4y - 11 = 3z - 4

Equation of the lines can be written as,

or

and,

or,

So, we know the straight lines as

And,

We have the normal vector of the plane as,

So, the equation of plane is

x - 4y + 2z + 4 = 0, (by solving further)

The Plane exercise multiple choice question 16

Option (b)

Use simultaneous equation.

Let the point of intersection of the line

And the plane

As x

Must satisfy both equations of line and the equation of plane.

Substituting in plane of in both the equations, we get,

and

Substituting, these values in equation (2)

Therefore,

Hence, the point of intersection is (2, -1, 2)

Now, the distance between the point (-1, -5, -10) and (2, -1, 2) is

Hence, the required distance between the point (-1, -5, -10) and the point where the line

And the plane

intersects is 13 units

The Plane exercise multiple choice question 17

Option (a)

is a scalar.

ax + by + cz + d = 0

The equation of the plane through the intersection of the planes

ax + by + cz + d = 0 and lx + my + nz + p = 0

Is given by

(ax + by + cz + d) + (lx + my + nz + p) = 0, [ where is a scalar ]

x (a + l ) + y (b + m ) + z (c + n ) + d + p = 0

Given that the required plane is parallel to the lines y = 0, z = 0, i.e., x-axes so, we have

1 (a + l ) + 0 ( b + m ) + 0 (c + n ) = 0

a + l = 0

(a + l )x + (b + m ) y + (c + n) z + (d + pλ ) = 0

(b l - a m) y + (c l - a n) z + d l - a p = 0

Hence, option a is correct.

The Plane exercise multiple choice question 18

Option (a)

Use simultaneous equation.

The equation of the plane which cuts equal intercepts of unit length of the co-ordinate axes.

We know that the general equation of a planes

Here,

Again, we know the intercept form of plane

A

And the plane makes intercepts at (a, 0, 0), (0, b, 0) and (0, 0, c) with the axes x, y, z respectively.

Here a = b = c = 1

Put in the equation (2)

Hence the equation is x + y + z = 1

The Plane exercise multiple choice question 19

Option (d)

Solve the equation to get the values of x, y and z

xy + yz = 0

We have

xy + yz = 0

x (y + z) = 0

therefore x = 0, y + z = 0

Above one equation of planes normal to the plane x = 0 is

y + z = 0 is

And normal to the plane

So, planes are perpendicular.

The Plane exercise multiple choice question 20

See the reflection with the graph

With the help of graph,

In xy-plane, the reflection of the point is

The Plane exercise multiple choice question 21

Option (c)

2x - 3y + 6z - 11 = 0

We have equation of plane are 2x - 3y + 6z - 11 = 0

Normal to the plane is

Also x-axis is along the vector

According to the question

The Plane exercise multiple choice question 22

Option (d)

Given line

is parallel to the vector

Equation to the plane is

2x - 2y + z = 5

Normal to the plane is

If angle between line and plane is

Then

The Plane exercise multiple choice question 23

Option (a)

Find magnitude of n vector.

Let,

N is a unit vector.

The Plane exercise multiple choice question 24

Option (b)

(-1, 5, 4)

Line passes through (-1, 5, 4) and perpendicular to the plane z = 0

Equation of plane

0. x + 0. y + 1. z = 0

Normal of plane (0, 0, 1)

Equation of line

r = (-1, 5, 4) + (0, 0, 1)

Rd Sharma Class 12 Chapter 28 Exercise MCQ is organised in a stepwise manner with hints for complex problems so that a student can find the best path to solve them. The advantages of these solutions are as follows:

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**Covers the entire syllabus**

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Download E-bookRD Sharma Class 12th Exercise MCQ material provided by Career360 contains step-by-step answers to all the questions from the syllabus. As teachers can't cover all the questions through their lectures, students can refer to this material to stay in line with their class and prepare accordingly. Moreover, as Class 12 RD Sharma Chapter 28 Exercise MCQ solutions material complies with the CBSE syllabus, students can refer to it without worrying about the difference in concepts.

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- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable

1. What is the definition of a Plane Figure?

Any geometric figure with no thickness is referred to as a plane figure.

2. Is the RD Sharma textbook a good choice for class 12?

The CBSE recommends RD Sharma as a reference book. RD Sharma's class 12 solutions include numerous solved illustrations, pictorial representations, and more than one method for problem-solving. It also teaches students mathematical tricks and concepts that help them solve problems quickly and accurately. It is an excellent choice for practice questions.

3. Is R S Agrawaal superior to RD Sharma?

RD Sharma explains the concepts more clearly. It also includes a large number of practise questions of various types. As a result, it prepares students to perform better in exams.

4. What is the main concept of the chapter ‘The Plane’?

This chapter explains the plane concept of geometry.

5. What is the best website for studying RD Sharma Solutions for Class 12 Maths?

On the Career360 website, you can find Rd Sharma Class 12th Exercise MCQ solutions, as well as step-by-step answers to all of the questions in the RD Sharma textbook. As a result, in order to understand the important topics, students in Class 12 should learn all of the concepts covered in the syllabus.

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