NCERT Solutions class 12 Maths Exercise 11.3

CBSE Sample Papers 2018-19

myCBSEguide App


Install myCBSEguide mobile app for FREE sample papers, Test Papers, Revision Notes, Previous year question papers, NCERT solutions and MCQ tests. Refer myCBSEguide App to your friends and Earn upto Rs.500/-.
myCBSEguide AppmyCBSEguide app install

NCERT Solutions class 12 Maths Exercise 11.3 Class 12 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 12 Maths chapter wise NCERT solution for Maths part 1 and Maths part 2 for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

Download NCERT solutions for Three Dimensional Geometry as PDF.

NCERT Solutions class 12 Maths Exercise 11.3

NCERT Solutions class 12 Maths Three Dimensional Geometry 

Formula for equation number 1 and 2

If  is the length of perpendicular from the origin to a plane and  is a unit normal vector to the plane, then equation of the plane is  (where of course  being length is > 0)

1. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

(a)

(b)  

(c)

(d)  

Ans. (a) Given: Equation of the plane is

In vector form it is where

  (here  = 2 > 0)

 

Here,  = (Position vector of point P )

And

Now let us reduce  to

Now   [Dividing both sides by  ]

  where

  and

Therefore, direction cosines of normal to the plane are coefficients of  in , i.e., 0, 0, 1 and length of perpendicular from the origin to the plane is .

(b) Given: Equation of the plane is (here )

 

i.e.,  where

Dividing both sides by  = , we get

  where

 and

Therefore direction cosines of the normal to the plane are the coefficients of  in , i.e., and length of perpendicular from the origin to the plane is .

(c) Equation of the plane is (here  )

 

i.e.,  where

Dividing both sides by  = , we get

  where

 and

Therefore direction cosines of the normal to the plane are the coefficients of  in , i.e., and length of perpendicular from the origin to the plane is .

(d) Given: Equation of the plane is    

   (here  )

 

i.e.,  where

Dividing both sides by  = , we get

  where

 and

Therefore direction cosines of the normal to the plane are the coefficients of  in , i.e., and length of perpendicular from the origin to the plane is .


NCERT Solutions class 12 Maths Exercise 11.3

2. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector  

Ans. Here

  The unit vector perpendicular to the plane is

Also (given)

Therefore the equation of the required plane is

 

 


NCERT Solutions class 12 Maths Exercise 11.3

3. Find the Cartesian equation of the following planes:

(a)  

(b)  

(c)  

Ans. (a) Vector equation of the plane is  ……….(i)

Putting  in eq. (i) as in 3-D, Cartesian equation of the plane is

    

(b) Since,  is the position vector of any arbitrary point P on the plane.

    

which is the required Cartesian equation.

(c) Vector equation of the plane is

Since, Since,  is the position vector of any arbitrary point P on the plane.

 

  which is the required Cartesian equation.


NCERT Solutions class 12 Maths Exercise 11.3

4. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin:

(a)

(b)  

(c)

(d)  

Ans. (a) Given: Equation of the plane is …….(i) and point is O (0, 0, 0)

Let M be the foot of the perpendicular drawn from the origin (0, 0, 0) to the given plane.

Since, direction ratios of perpendicular OM to plane are coefficients of  in , i.e., 2, 3, 4 =  (say)

  Equation of the perpendicular OM is  (say)

    

 

Therefore, point M on this line OM is M ………..(ii)

But point M lies on plane (i)

 Putting  in eq. (i), we have

    

Hence, putting  in equation (ii), the coordinates of foot of the perpendicular is

(b) Given: Equation of the plane is …….(i) and point is O (0, 0, 0)

Let M be the foot of the perpendicular drawn from the origin (0, 0, 0) to the given plane.

Since, direction ratios of perpendicular OM to plane are coefficients of  in , i.e., 0, 3, 4 =  (say)

  Equation of the perpendicular OM is  (say)

    

 

Therefore, point M on this line OM is M ………..(ii)

But point M lies on plane (i)

 Putting  in eq. (i), we have

    

Hence, putting  in equation (ii), the coordinates of foot of the perpendicular is

(c) Given: Equation of the plane is …….(i) and point is O (0, 0, 0)

Let M be the foot of the perpendicular drawn from the origin (0, 0, 0) to the given plane.

Since, direction ratios of perpendicular OM to plane are coefficients of  in , i.e., 1, 1, 1 =  (say)

  Equation of the perpendicular OM is  (say)

    

 

Therefore, point M on this line OM is M ………..(ii)

But point M lies on plane (i)

 Putting  in eq. (i), we have

   

Hence, putting  in equation (ii), the coordinates of foot of the perpendicular is

(d) Given: Equation of the plane is …….(i) and point is O (0, 0, 0)

Let M be the foot of the perpendicular drawn from the origin (0, 0, 0) to the given plane.

Since, direction ratios of perpendicular OM to plane are coefficients of  in , i.e., 0, 5, 0 =  (say)

  Equation of the perpendicular OM is  (say)

    

 

Therefore, point M on this line OM is M ………..(ii)

But point M lies on plane (i)

 Putting  in eq. (i), we have

   

Hence, putting  in equation (ii), the coordinates of foot of the perpendicular is .


NCERT Solutions class 12 Maths Exercise 11.3

5. Find the vector and Cartesian equations of the planes

(a) that passes through the point  and the normal to the plane is  

(b) that passes through the point (1, 4, 6) and the normal vector to the plane is  

Ans. (a) Vector form: The given point on the plane is

  The position vector of the given point is  =

Also Normal vector to the plane is

 Vector equation of the required line is

 

Putting the values of  and ,

 

   

Cartesian form: The plane passes through the point  =

Normal vector to the plane is

 Direction ratios of normal to the plane are coefficients of  in  are

 Cartesian form of equation of plane is

    

 

(b) Vector form: The given point on the plane is

  The position vector of the given point is

Also Normal vector to the plane is

 Vector equation of the required line is

 

Putting the values of  and ,

 

   

Cartesian form: The plane passes through the point  =

Normal vector to the plane is

 Direction ratios of normal to the plane are coefficients of  in  are

 Cartesian form of equation of plane is

    

 


NCERT Solutions class 12 Maths Exercise 11.3

6. Find the equations of the planes that passes through three points:

(a)  

(b)  

Ans. We know that through three collinear points A, B, C i.e., through a straight line, we can pass an infinite number of planes.

(a) The three given points are A B and C

Now direction ratios of line AB are  

=

Again direction ratios of line BC are

=

Now   

Since,

Therefore, line AB and BC are parallel and B is their common point.

  Points A, B and C are collinear and hence an infinite number of planes can be drawn through the three given collinear points.

(b) The three given points are A B and C

Now direction ratios of line AB are  

=

Again direction ratios of line BC are

=

Now   

Since,

  Points A, B and C are not collinear and hence the unique plane can be drawn through the three given collinear points, i.e.,

 

 

Expanding along first row,

 

 

 

 

 

Hence the equation of required plane is .


NCERT Solutions class 12 Maths Exercise 11.3

7. Find the intercepts cut off by the plane  

Ans. Equation of the plane is

 

 

Comparing with intercept form , we have  which are intercepts cut off by the plane on axis, axis and axis respectively.


NCERT Solutions class 12 Maths Exercise 11.3

8. Find the equation of the plane with intercept 3 on the  axis and parallel to ZOX plane.

Ans. Since equation of ZOX plane is

  Equation of any plane parallel to ZOX plane is  ……….(i)

[ Equation of any plane parallel to the plane  is  i.e., change only the constant term]

Now, Plane (i) makes an intercept 3 on the axis ( and ) i.e., plane (i) passes through (0, 3, 0).

Putting  and  in eq. (i), 

Putting  in eq. (i), equation of required plane is


NCERT Solutions class 12 Maths Exercise 11.3

9. Find the equation of the plane through the intersection of the planes  and and the point (2, 2, 1).

Ans. Equations of given planes are  and

Since, equation of any plane through the intersection of these two planes is

  L.H.S. of plane I +  (L.H.S. of plane II) = 0

   ……….(i)

Now, required plane (i) passes through the point (2, 2, 1).

Putting  in eq. (i),

 

 

 

  

Now putting  in eq. (i) of required plane is

 

 

 


NCERT Solutions class 12 Maths Exercise 11.3

10. Find the vector equation of the plane passing through the intersection of the planes   and through the point (2, 1, 3).

Ans. Equation of first plane is

  ……….(i)

Again equation of the second plane is

  ……….(ii)

Since, equation of any plane passing through the line of intersection of two planes is

  L.H.S. of plane I +  (L.H.S. of plane II) = 0

   ……….(iii)

Now, the plane (iii) passes through the point (2, 1, 3) =

 

Putting this value of  in eq. (iii),

 

   

Putting  in eq. (iii) of required plane is

 

 

 

 


NCERT Solutions class 12 Maths Exercise 11.3

11. Find the equation of the plane through the line of intersection of the planes  and  which is perpendicular to the plane  

Ans. Equations of the given planes are  and

  and

Since, equation of any plane passing through the line of intersection of two planes is

  L.H.S. of plane I +  (L.H.S. of plane II) = 0

  ……….(i)

 

 

According to the question, this plane is perpendicular to the plane

 

 

 

   

Putting  in eq. (i) of required plane is

 

 

 


NCERT Solutions class 12 Maths Exercise 11.3

12. Find the angle between the planes whose vector equations are  and

Ans. Equation of one plane is ……….(i)

Comparing this equation with , we have

Normal vector to plane (i) is

Again, equation of second plane is  ……….(ii)

Comparing this equation with , we have

Normal vector to plane (i) is

Let  be the acute angle between plane (i) and (ii).

  angle between normals  and  to planes (i) and (ii) is also

 

=

 =

 


NCERT Solutions class 12 Maths Exercise 11.3

13. In the following cases, determine whether the given planes are parallel or perpendicular and in case they are neither, find the angle between them.

(a)  and  

(b)  and  

(c)  and  

(d)  and  

(e)  and  

Ans. (a) Equations of the given planes are   and

   

Here,

Since

Therefore, the given two planes are not parallel.

Again  = 21 – 5 – 60 = 21 – 65 = –44

Since

Therefore, the given two planes are not perpendicular.

Now let  be the angle between the two planes.

 

 =

=

=  =  =

 

(b) equations of the given planes are   and  i.e.,  

Here,

Since

Therefore, the given two planes are not parallel.

Again  =  = 2 – 2 + 0 = 0

Since

Therefore, the given two planes are perpendicular.

(c) equations of the given planes are   and  

Here,

Since

Therefore, the given two planes are parallel.

(d) equations of the given planes are   and  

Here,

Since

Therefore, the given two planes are parallel.

(e) equations of the given planes are   and  i.e.,

Here,

Since

Therefore, the given two planes are not parallel.

Again  = 4 x 0 + 8 x 1 + 1 x 1 = 0 + 8 + 1 = 9

Since

Therefore, the given two planes are not perpendicular.

Now let  be the angle between the two planes.

 

=

=  =  =  =

 


NCERT Solutions class 12 Maths Exercise 11.3

14. In the following cases find the distances of each of the given points from the corresponding given plane:

 (a) Point (0, 0, 0)

Plane  

(b) Point

Plane  

(c) Point

Plane  

(d) Point

Plane  

Ans. (a) Distance (of course perpendicular) of the point (0, 0, 0) from the plane  

   is

=

=  =

(b) Length of perpendicular from the point  on the plane  is

 

=

=  =

(c) Length of perpendicular from the point  on the plane    is

=

=  =

(d) Length of perpendicular from the point  on the plane  is

 

=

=  =

NCERT Solutions class 12 Maths Exercise 11.3

NCERT Solutions Class 12 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 13 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide

CBSE App for Students

To download NCERT Solutions for class 12 Physics, Chemistry, Biology, History, Political Science, Economics, Geography, Computer Science, Home Science, Accountancy, Business Studies and Home Science; do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through the best app for CBSE students and myCBSEguide website.


Deal of the Day
Latest Sample Papers 149/-
Test Paper & Worksheets 299/-
Click Here & Tap Shop

1 thought on “NCERT Solutions class 12 Maths Exercise 11.3”

Leave a Comment