# NCERT Solutions class 12 Maths Exercise 11.3 (Ex 11.3) Chapter 11 Three Dimensional Geometry

## NCERT Solutions for Class 12 Maths Exercise 11.3 Chapter 11 Three Dimensional Geometry – FREE PDF Download

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.3 (Ex 11.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks.

# NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry (Ex 11.3) Exercise 11.3

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 Therefore, the direction ratios of the normal to the plane are 0, 0, 1.

a=0,b=0,c=1a=0,b=0,c=1

a2+b2+c2−−−−−−−−−√=(0)2+(0)2+(1)2−−−−−−−−−−−−−−√=1a2+b2+c2=(0)2+(0)2+(1)2=1

Therefore, 0x1+0y1+z1=210x1+0y1+z1=21

Comparing with lx+my+nz=p,lx+my+nz=p, we get p=21p=21

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

a=1,b=1,c=1a=1,b=1,c=1

a2+b2+c2−−−−−−−−−√a2+b2+c2 =

Therefore, x3+y3+z3=13x3+y3+z3=13

Comparing with lx+my+nz=p,lx+my+nz=p, we get p=13p=13

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

a=2,b=3,c=1a=2,b=3,c=−1

a2+b2+c2−−−−−−−−−√a2+b2+c2 =

Therefore, 2x14+3y14z14=5142×14+3y14−z14=514

Comparing with lx+my+nz=p,lx+my+nz=p, we get

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

a=0,b=5,c=0a=0,b=−5,c=0

a2+b2+c2−−−−−−−−−√a2+b2+c2 =

Therefore, 0x5+5y5+0z5=850x5+−5y5+0z5=85

Comparing with lx+my+cz=p,lx+my+cz=p, we get

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 .

### 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

### 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.

### 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

2(2λ)+3(3λ)+4(4λ)12=02(2λ)+3(3λ)+4(4λ)−12=0

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

3(3λ)+4(4λ)6=03(3λ)+4(4λ)−6=0

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 .

### 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

### 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

(say)

Again direction ratios of line BC are

(say)

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 i.e. no unique plane can be drawn.

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

Now direction ratios of line AB are

(say)

Again direction ratios of line BC are

(say)

Now    a1a2=0,b1b2=,=c1c2=12a1a2=0,b1b2=∞,=c1c2=−12

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 .

### 7. Find the intercepts cut off by the plane

Ans. Equation of the plane is

Comparing with intercept form , we have a=52,b=5,c=5a=52,b=5,c=−5  which are intercepts cut off by the plane on axis, axis and axis respectively.

### 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

### 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

### 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

### 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

3x+3y+3z32x3y4z+5=03x+3y+3z−3−2x−3y−4z+5=0

### 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

### 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 2xy+3z+3=02x−y+3z+3=0

(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., x2y+0z+5=0x−2y+0z+5=0

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 y+z4=0y+z−4=0 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.

=  =  =

### 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 2xy+2z+3=02x−y+2z+3=0

(c) Point

Plane

(d) Point

Plane

Ans. (a) Distance (of course perpendicular) of the point (0, 0, 0) from the plane 3x4y+12z=33x−4y+12z=3      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

=