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**Download NCERT solutions for Three Dimensional Geometry as PDF.**

## 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

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