The Plane

(i) A(2, 2, -1), B(3, 4, 2) and C(7, 0, 6)

Given Points :

A = (2, 2, -1)

B = (3, 4, 2)

C = (7, 0, 6)

To Find : Equation of plane passing through points A, B & C

Formulae :

1) Position vectors :

If A is a point having co-ordinates (a₁, a₂, a₃), then its position vector is given by,

2) Vector :

If A and B be two points with position vectors respectively, where

then,

3) Cross Product :

If are two vectors

then,

4) Dot Product :

If are two vectors

then,

5) Equation of Plane :

If A = (a₁, a₂, a₃), B = (b₁, b₂, b₃), C = (c₁, c₂, c₃) are three non-collinear points,

Then, the vector equation of the plane passing through these points is

Where,

For given points,

A = (2, 2, -1)

B = (3, 4, 2)

C = (7, 0, 6)

Position vectors are given by,

Now, vectors are

Therefore,

Now,

= 40 + 16 + 12

= 68

………eq(1)

And

= 20x + 8y – 12z

………eq(2)

Vector equation of the plane passing through points A, B & C is

From eq(1) and eq(2)

20x + 8y – 12z = 68

This is 5x + 2y – 3z = 17 vector equation of required plane.

(ii) Given Points :

A = (0, -1, -1)

B = (4, 5, 1)

C = (3, 9, 4)

To Find : Equation of plane passing through points A, B & C

Formulae :

1) Position vectors :

If A is a point having co-ordinates (a₁, a₂, a₃), then its position vector is given by,

2) Vector :

If A and B be two points with position vectors respectively, where

then,

3) Cross Product :

If are two vectors

then,

4) Dot Product :

If are two vectors

then,

5) Equation of Plane :

If A = (a₁, a₂, a₃), B = (b₁, b₂, b₃), C = (c₁, c₂, c₃) are three non-collinear points,

Then, vector equation of the plane passing through these points is

Where,

For given points,

A = (0, -1, -1)

B = (4, 5, 1)

C = (3, 9, 4)

Position vectors are given by,

Now, vectors are

Therefore,

Now,

= 0 + 14 – 22

= - 8

………eq(1)

And

= 10x - 14y + 22z

………eq(2)

Vector equation of plane passing through points A, B & C is

From eq(1) and eq(2)

10x - 14y + 22z = - 8

This is 5x - 7y + 11z = - 4 vector equation of required plane

(iii) Given Points :

A = (-2, 6, -6)

B = (-3, 10, 9)

C = (-5, 0, -6)

To Find : Equation of plane passing through points A, B & C

Formulae :

1) Position vectors :

If A is a point having co-ordinates (a₁, a₂, a₃), then its position vector is given by,

2) Vector :

If A and B be two points with position vectors respectively, where

then,

3) Cross Product :

If are two vectors

then,

4) Dot Product :

If are two vectors

then,

5) Equation of Plane :

If A = (a₁, a₂, a₃), B = (b₁, b₂, b₃), C = (c₁, c₂, c₃) are three non-collinear points,

Then, vector equation of the plane passing through these points is

Where,

For given points,

A = (-2, 6, -6)

B = (-3, 10, 9)

C = (-5, 0, -6)

Position vectors are given by,

Now, vectors are

Therefore,

Now,

= - 180 - 270 – 108

= - 558

………eq(1)

And

= 90x - 45y + 18z

………eq(2)

Vector equation of plane passing through points A, B & C is

From eq(1) and eq(2)

90x - 45y + 18z = - 558

This is 10x - 5y + 2z = - 62 vector equation of required plane

Given Points :

A = (3, 2, -5)

B = (-1, 4, -3)

C = (-3, 8, -5)

D = (-3, 2, 1)

To Prove : Points A, B, C & D are coplanar.

To Find : Equation of plane passing through points A, B, C & D.

Formulae :

1) Position vectors :

If A is a point having co-ordinates (a₁, a₂, a₃), then its position vector is given by,

2) Equation of line

If A and B are two points having position vectors then equation of line passing through two points is given by,

3) Cross Product :

If are two vectors

then,

4) Dot Product :

If are two vectors

then,

5) Coplanarity of two lines :

If two lines are coplanar then

6) Equation of plane :

If two lines are coplanar then equation of the plane containing them is

Where,

For given points,

A = (3, 2, -5)

B = (-1, 4, -3)

C = (-3, 8, -5)

D = (-3, 2, 1)

Position vectors are given by,

Equation of line passing through points A & B is

Let,

Where,

&

And the equation of the line passing through points C & D is

Let,

Where,

&

Now,

Therefore,

= 72 + 48 – 120

= 0

……… eq(1)

And

= - 72 + 192 – 120

= 0

……… eq(2)

From eq(1) and eq(2)

Hence lines are coplanar

Therefore, points A, B, C & D are also coplanar.

As lines are coplanar therefore equation of the plane passing through two lines containing four given points is

Now,

= 24x + 24y + 24z

From eq(1)

Therefore, equation of required plane is

24x + 24y + 24z = 0

x + y + z = 0

Given Points :

A = (0, -1, 0)

B = (2, 1, -1)

C = (1, 1, 1)

D = (3, 3, 0)

To Prove : Points A, B, C & D are coplanar.

To Find : Equation of plane passing through points A, B, C & D.

Formulae :

1) Position vectors :

If A is a point having co-ordinates (a₁, a₂, a₃), then its position vector is given by,

2) Equation of line

If A and B are two points having position vectors then equation of line passing through two points is given by,

3) Cross Product :

If are two vectors

then,

4) Dot Product :

If are two vectors

then,

5) Coplanarity of two lines :

If two lines are coplanar then

6) Equation of plane :

If two lines are coplanar then equation of the plane containing them is

Where,

For given points,

A = (0, -1, 0)

B = (2, 1, -1)

C = (1, 1, 1)

D = (3, 3, 0)

Position vectors are given by,

Equation of line passing through points A & D is

Let,

Where,

&

And equation of line passing through points B & C is

Let,

Where,

&

Now,

Therefore,

= 0 + 6 + 0

= 6

……… eq(1)

And

= 16 – 6 – 4

= 6

……… eq(2)

From eq(1) and eq(2)

Hence lines are coplanar

Therefore, points A, B, C & D are also coplanar.

As lines are coplanar therefore equation of the plane passing through two lines containing four given points is

Now,

= 8x - 6y + 4z

From eq(1)

Therefore, equation of required plane is

8x - 6y + 4z = 6

4x - 3y + 2z = 3

Given :

X – intercept, a = 2

Y – intercept, b = - 4

Z – intercept, c = 5

To Find : Equation of plane

Formula :

If a, b & c are the intercepts made by plane on X, Y & Z axes respectively, then equation of the plane is given by,

Multiplying above equation throughout by 40,

20x – 10y + 8z = 40

10x – 5y + 4z = 20

This the equation of the required plane.

Given :

Equation of plane : 4x – 3y + 2z = 12

To Find :

1) Equation of plane in intercept form

2) Intercepts made by the plane with the co-ordinate axes.

Formula :

If

is the equation of a plane in intercept form then intercept made by it with co-ordinate axes are

X-intercept = a

Y-intercept = b

Z-intercept = c

Given the equation of plane:

4x – 3y + 2z = 12

Dividing the above equation throughout by 12

This is the equation of a plane in intercept form.

Comparing the above equation with

We get,

a = 3

b = -4

c = 6

Therefore, intercepts made by plane with co-ordinate axes are

X-intercept = 3

Y-intercept = -4

Z-intercept = 6

Equation of the plane making a, b & c intercepts with X, Y & Z axes respectively is

But, the plane makes equal intercepts on the co-ordinate axes

Therefore, a = b = c

Therefore the equation of the plane is

x + y + z = a

As plane passes through the point (2, -3, 7),

Substituting x = 2, y = -3 & z = 7

2 – 3 + 7 = a

Therefore, a = 6

Hence, required equation of plane is

x + y + z = 6

Given :

X-intercept = A

Y-intercept = B

Z-intercept = C

Centroid of ∆ABC = (1, -2, 3)

To Find : Equation of a plane

Formulae :

1) Centroid Formula :

For ∆ABC if co-ordinates of A, B & C are

A = (x₁, x₂, x₃)

B = (y₁, y₂, y₃)

C = (z₁, z₂, z₃)

Then co-ordinates of the centroid of ∆ABC are

2) Equation of plane :

Equation of the plane making a, b & c intercepts with X, Y & Z axes respectively is

As the plane makes intercepts at points A, B & C on X, Y & Z axes respectively, let co-ordinates of A, B, C be

A = (a, 0, 0)

B = (0, b, 0)

C = (0, 0, c)

By centroid formula,

The centroid of ∆ABC is given by

But, Centroid of ∆ABC = (1, -2, 3) …… given

Therefore, a = 3, b = - 6, c = 9

Therefore,

X-intercept = a = 3

Y-intercept = b = - 6

Z-intercept = c = 9

Therefore, equation of required plane is

Given :

A = (1, 2, 3)

Direction ratios of perpendicular vector = (2, 3, -4)

To Find : Equation of a plane

Formulae :

1) Position vectors :

If A is a point having co-ordinates (a₁, a₂, a₃), then its position vector is given by,

2) Dot Product :

If are two vectors

then,

3) Equation of plane :

If a plane is passing through point A, then the equation of a plane is

Where,

For point A = (1, 2, 3), position vector is

Vector perpendicular to the plane with direction ratios (2, 3, -4) is

Now,

= 2 + 6 – 12

= - 4

Equation of the plane passing through point A and perpendicular to vector is

As

= 2x + 3y – 4z

Therefore, equation of the plane is

2x + 3y – 4z = -4

Or

2x + 3y – 4z + 4 = 0

Given :

P = (1, 2, -3)

O = (0, 0, 0)

To Find : Equation of a plane

Formulae :

1) Position vectors :

If A is a point having co-ordinates (a₁, a₂, a₃), then its position vector is given by,

2) Vector :

If A and B be two points with position vectors respectively, where

then,

3) Dot Product :

If are two vectors

then,

4) Equation of plane :

If a plane is passing through point A, then the equation of a plane is

Where,

For points,

P = (1, 2, -3)

O = (0, 0, 0)

Position vectors are

Vector

Now,

= 1 + 4 + 9

= 14

And

= x + 2y + 3z

Equation of the plane passing through point A and perpendicular to the vector is

But,

Therefore, the equation of the plane is

x + 2y + 3z = 14

x + 2y + 3z – 14 = 0

Given :

Equation of plane : x + 2y – 3z = 5

To Find : direction ratios of normal

Answer :

Given equation of plane : x + 2y – 3z = 5

It can be written as

Comparing with

Therefore, normal vector is

Hence, direction ratios of normal are (1, 2, -3).

Given :

Equation of plane : 2x + 3y – z = 4

To Find : Direction cosines of the normal i.e.

Formula :

1) Direction cosines :

If a, b & c are direction ratios of the vector then its direction cosines are given by

Answer :

For the given equation of plane

2x + 3y – z = 4

Direction ratios of normal vector are (2, 3, -1)

Therefore, direction cosines are

Given :

Equation of plane : y = 3

To Find : Direction cosines of the normal i.e.

Formula :

1) Direction cosines :

If a, b & c are direction ratios of the vector then its direction cosines are given by

Answer :

For the given equation of plane

y = 3

Direction ratios of normal vector are (0, 1, 0)

= 1

Therefore, direction cosines are

Given :

Equation of plane : 3x + 4 = 0

To Find : Direction cosines of the normal i.e.

Formula :

1) Direction cosines :

If a, b & c are direction ratios of the vector then its direction cosines are given by

Answer :

For the given equation of plane

-3x = 4

Direction ratios of normal vector are (-3, 0, 0)

= 3

Therefore, direction cosines are

Given :

Point : (4, -2, 3)

To Find : equation of plane

Formula :

1) Equation of plane :

Equation of plane passing through point A with position vector and perpendicular to vector is given by,

Where,

Answer :

Position vector for given point A ≡ (4, -2, 3) is

As required plane is parallel to XY plane, therefore Z-axis is perpendicular to the plane.

Therefore, equation of plane is

This is required equation of plane.

Given :

Point : (-3, 2, 0)

To Find : equation of plane

Formula :

1) Equation of plane :

Equation of plane passing through point A with position vector and perpendicular to vector is given by,

Where,

Answer :

Position vector for given point A ≡ (-3, 2, 0) is

As required plane is parallel to YZ plane, therefore X-axis is perpendicular to the plane.

Therefore, equation of plane is

This is required equation of plane.

Let, normal vector of plane be

Equation of plane is given by,

As the required plane is parallel to the given plane, hence normal vector of plane is perpendicular to x-axis.

a = 0

Therefore, equation of plane is

by + cz = d

Given :

Equation of plane : 2x + y – z = 5

To Find : Intercept made by the plane with the X-axis.

Formula :

If

is the equation of plane in intercept form then intercept made by it with co-ordinate axes are

X-intercept = a

Y-intercept = b

Z-intercept = c

Answer :

Given equation of plane:

2x + y – z = 5

Dividing above equation throughout by 5

Comparing above equation with

We get,

a = 5/2

Therefore, intercepts made by plane with X-axis are

X-intercept = 5/2

Given :

Equation of plane : 4x – 3y + 2z = 12

To Find :

1) Equation of plane in intercept form

2) Intercepts made by the plane with the co-ordinate axes.

Formula :

If

is the equation of plane in intercept form then intercept made by it with co-ordinate axes are

X-intercept = a

Y-intercept = b

Z-intercept = c

Answer :

Given equation of plane:

4x – 3y + 2z = 12

Dividing above equation throughout by 12

This is the equation of plane in intercept form.

Comparing above equation with

We get,

a = 3

b = -4

c = 6

Therefore, intercepts made by plane with co-ordinate axes are

X-intercept = 3

Y-intercept = -4

Z-intercept = 6

Given :

Equation of plane : 2x – 3y + 5z + 4 = 0

To Find :

1) Equation of plane in intercept form

2) Intercepts made by the plane with the co-ordinate axes.

Formula :

If

is the equation of plane in intercept form then intercept made by it with co-ordinate axes are

X-intercept = a

Y-intercept = b

Z-intercept = c

Answer :

Given equation of plane:

2x – 3y + 5z = -4

Dividing above equation throughout by -4

This is the equation of plane in intercept form.

Comparing above equation with

We get,

a = -2

Therefore, intercepts made by plane with co-ordinate axes are

X-intercept = -2

Given : Plane is passing through points

A ≡ (a, 0, 0)

B ≡ (0, b, 0)

C ≡ (0, 0, c)

To Find : Equation of plane

Formulae :

Equation of plane making intercepts (a, b, c) on X, Y & Z axes respectively is given by,

Answer : As plane is passing through points A ≡ (a, 0, 0),

B ≡ (0, b, 0) & C ≡ (0, 0, c)

Therefore, intercepts made by it on X, Y & Z axes respectively are

a, b & c.

hence, equation of plane is

Given : equations of perpendicular planes-

2x – 5y + kz = 4

x + 2y – z = 6

To Find : k

Formulae :

Normal vector to the plane :

If equation of the plane is ax + by + cz = d then,

Vector normal to the plane is given by,

Answer :

For given planes –

2x – 5y + kz = 4

x + 2y – z = 6

normal vectors are

As given vectors are perpendicular, hence their normal vectors are also perpendicular to each other.

(2×1) + (-5×2) + (k×(-1)) = 0

2 – 10 – k = 0

- 8 – k = 0

k = -8

Given : equations of planes-

2x + y – 2z = 5

3x – 6y – 2z = 7

To Find : angle between two planes

Formulae :

1) Normal vector to the plane :

If equation of the plane is ax + by + cz = d then,

Vector normal to the plane is given by,

2) Angle between two planes :

The angle Ө between the planes and is given by

Answer :

For given planes

2x + y – 2z = 5

3x – 6y – 2z = 7

Normal vectors are

and

Therefore, angle between two planes is

Given : equations of planes-

To Find : angle between two planes

Formulae :

Angle between two planes :

The angle Ө between the planes and is given by

Answer :

For given planes

Normal vectors are

and

Therefore, angle between two planes is

Given : equations of planes-

To Find : angle between two planes

Formulae :

Angle between two planes :

The angle Ө between the planes and is given by

Answer :

For given planes

Normal vectors are

and

Therefore, angle between two planes is

Given :

Equation of line :

Equation of plane : 10x + 2y – 11z = 3

To Find : angle between line and plane

Formulae :

1) Parallel vector to the line :

If equation of the line is then,

Vector parallel to the line is given by,

2) Normal vector to the plane :

If equation of the plane is ax + by + cz = d then,

Vector normal to the plane is given by,

3) Angle between a line and a plane :

If Ө is a angle between the line and the plane , then

Where, is vector parallel to the line and

is the vector normal to the plane.

Answer :

For given equation of line,

Parallel vector to the line is

For given equation of plane,

10x + 2y – 11z = 3

normal vector to the plane is

Therefore, angle between given line and plane is

Given :

Equation of line :

Equation of plane :

To Find : angle between line and plane

Formulae :

1) Angle between a line and a plane :

If Ө is a angle between the line and the plane , then

Where, is vector parallel to the line and

is the vector normal to the plane.

Answer :

For given equation of line,

Parallel vector to the line is

For given equation of plane,

normal vector to the plane is

Therefore, angle between given line and plane is

Given :

Equation of line :

Equation of plane : 3x – y – 2z = 7

To Find :

Formulae :

1) Parallel vector to the line :

If equation of the line is then,

Vector parallel to the line is given by,

2) Normal vector to the plane :

If equation of the plane is ax + by + cz = d then,

Vector normal to the plane is given by,

3) Cross Product :

If are two vectors

then,

Answer :

For given equation of line,

Parallel vector to the line is

For given equation of plane,

3x – y – 2z = 7

normal vector to the plane is

As given line and plane are perpendicular to each other.

Comparing coefficients of on both sides

3λ = -6

λ = -2

Given :

A ≡ (a, b, c)

Equation of plane parallel to required plane

To Find : Equation of plane

Formulae :

1) Position vectors :

If A is a point having co-ordinates (a₁, a₂, a₃), then its position vector is given by,

2) Dot Product :

If are two vectors

then,

3) Equation of plane :

If a plane is passing through point A, then equation of plane is

Where,

Answer :

For point A ≡ (a, b, c), position vector is

As plane is parallel to the required plane, the vector normal to required plane is

Now,

= a + b + c

Equation of the plane passing through point A and perpendicular to vector is

Given :

Equation of plane : 2x – 3y + 6z + 21 = 0

To Find :

Length of perpendicular drawn from origin to the plane = d

Formulae :

1) Distance of the plane from the origin :

Distance of the plane from the origin is given by,

Answer :

For the given equation of plane

2x – 3y + 6z = -21

Direction ratios of normal vector are (2, -3, 6)

Therefore, equation of normal vector is

= 7

From given equation of plane,

p = -21

Now, distance of the plane from the origin is

d = 3 units

Given :

Equation of plane :

To Find :

Direction cosines of the normal i.e.

Formulae :

1) Direction cosines :

If a, b & c are direction ratios of the vector then its direction cosines are given by

Answer :

For the given equation of plane

Equation of normal vector is

= 7

Therefore, direction cosines are

Given :

Equation of plane : :

Equation of line :

To Prove : Given line is parallel to the given plane.

Answer :

Comparing given plane i.e.

with , we get,

This is the vector perpendicular to the given plane.

Now, comparing given equation of line i.e.

with , we get,

Now,

= 20 – 8 – 12

= 0

Therefore, vector normal to the plane is perpendicular to the vector parallel to the line.

Hence, the given line is parallel to the given plane.

Given :

Equation of plane :

To Find : Length of perpendicular = d

Formulae :

1) Unit Vector :

Let be any vector

Then unit vector of is

Where,

2) Length of perpendicular :

The length of the perpendicular from the origin to the plane

is given by,

Answer :

Given equation of the plane is

Comparing above equation with

We get,

& p = 14

Therefore,

= 7

The length of the perpendicular from the origin to the given plane is

Given :

Equation of line :

Equation of plane :

To Find : λ

Formulae :

1) Parallel vector to the line :

If equation of the line is then,

Vector parallel to the line is given by,

2) Angle between a line and a plane :

If Ө is a angle between the line and the plane , then

Where, is vector parallel to the line and

is the vector normal to the plane.

Answer :

For given equation of line,

Parallel vector to the line is

For given equation of plane,

normal vector to the plane is

Therefore, angle between given line and plane is

As given line is parallel too the given plane, angle between them is 0.

4 + 9 + 4 λ = 0

13 + 4λ = 0

4λ = -13

Given :

Equation of line :

Equation of plane : x + y + 4 = 0

To Find : angle between line and plane

Formulae :

1) Parallel vector to the line :

If equation of the line is then,

Vector parallel to the line is given by,

2) Normal vector to the plane :

If equation of the plane is ax + by + cz = d then,

Vector normal to the plane is given by,

3) Angle between a line and a plane :

If Ө is a angle between the line and the plane , then

Where, is vector parallel to the line and

is the vector normal to the plane.

Answer :

For given equation of line,

Parallel vector to the line is

For given equation of plane,

x + y + 4 = 0

normal vector to the plane is

Therefore, angle between given line and plane is

Given :

A ≡ (2, -1, 1)

Plane parallel to the required plane : 3x + 2y – z = 7

To Find : Equation of plane

Formulae :

1) Position vectors :

If A is a point having co-ordinates (a₁, a₂, a₃), then its position vector is given by,

2) Dot Product :

If are two vectors

then,

3) Equation of plane :

If a plane is passing through point A, then equation of plane is

Where,

Answer :

For point A ≡ (2, -1, 1), position vector is

As required plane is parallel to 3x + 2y – z = 7.

Therefore, normal vector of given plane is also perpendicular to required plane

Now,

= 6 – 2 – 1

= 3

Equation of the plane passing through point A and perpendicular to vector is

As

= 3x + 2y – z

Therefore, equation of the plane is

3x + 2y – z = 3

3x + 2y – z - 3 = 0

Given: Equation of plane is

Formula Used: Equation of a plane is where is the unit vector normal to the plane, represents a point on the plane and is the distance of the plane from the origin.

Explanation:

The equation of the given plane is … (1)

Now,

= 7

is a unit vector.

(1) can be rewritten as

which is of the form

Perpendicular vector from the origin to the plane is

So, direction cosines of the vector perpendicular from the origin to the plane is

Given: Equation of plane is 5y + 4 = 0

Formula Used: Equation of a plane is lx + my + nz = p where (l, m, n) are the direction cosines of the normal to the plane and (x, y, z) is a point on the plane and p is the distance of plane from origin.

Explanation:

Given equation is 5y = -4

Dividing by -5,

which is of the form lx + my + nz = p where l = 0, m = -1, n = 0

Therefore, direction cosines of the normal to the plane is (0, -1, 0)

Given: Equation of plane is

Formula Used: Equation of a plane is where is the unit vector normal to the plane, represents a point on the plane and is the distance of the plane from the origin.

Explanation:

Given equation is … (1)

Now,

= 13

Dividing (1) by 13 and multiplying by -1,

which is of the form

Therefore, length of perpendicular from origin to plane is 3 units.

Given: A(2, -3, 7) is a point on the plane making equal intercepts on the axes.

Formula Used: Equation of plane is where (x, y, z) is a point on the plane and a, b, c are intercepts on x-axis, y-axis and z-axis respectively.

Explanation:

Let the equation of the plane be

… (1)

Here a = b = c = p (let’s say)

Since (2, -3, 7) is a point on the plane,

(1) becomes

p = 6

Therefore equation of the plane is

x + y + z = 6

Given: Plane makes intercepts 3, -4 and 6 with the coordinate axes.

Formula Used: Equation of plane is where (x, y, z) is a point on the plane and a, b, c are intercepts on x-axis, y-axis and z-axis respectively.

Normal Form of a plane ⇒ lx + my + nz = p where (l, m, n) is the direction cosines and p is the distance of perpendicular to the plane from the origin.

Explanation:

Equation of the given plane is

i.e., 4x – 3y + 2z = 12 … (1)

which is of the form ax + by + cz = d

Direction ratios are (4, -3, 12)

So,

= √29

Dividing (1) by 13,

which is in the normal form

Therefore length of perpendicular from the origin is units

Given:

1. Equation of line is

2. Equation of plane is 2x – 3y + kz = 0

Formula Used: If two direction ratios are perpendicular, then

a �₁a₂ + b₁b₂ + c₁c₂ = 0

Explanation:

Direction ratios of given line is (3, 4, 5)

Direction ratios of given plane is (2, -3, k)

Since the given line is parallel to the plane, the normal to the plane is perpendicular to the line.

So direction ratio of line is perpendicular to direction ratios of plane.

⇒ 3× 2 + 4× -3+ 5× k = 0

⇒ 6 – 12 + 5k = 0

Therefore,

Given: P(1, 2, -3) is a point on the plane. OP is perpendicular to the plane.

Explanation:

Let equation of plane be ax + by + cz = d … (1)

Substituting point P,

⇒ a + 2b -3c = d … (2)

Since OP is perpendicular to the plane, direction ratio of the normal is (1, 2, -3)

Substituting in (2)

1 + 4 + 9 = d

d = 14

Substituting the direction ratios and value of ‘d’ in (1), we get

x + 2y – 3z = 14

Therefore equation of plane is x + 2y – 3z = 14

Given: Equation of plane is 2x – 4y + z = 7

Line lies on the given plane.

Formula Used: Equation of a line is

Where (x₁, y₁, z₁) is a point on the line and b₁, b₂, b₃ : direction ratios of line.

Explanation:

Let

So the given line passes through the point (4, 2, k)

Since the line lies on the given plane, (4, 2, k) is a point on the plane.

Therefore, substituting the point on the equation for the plane,

⇒ 8 – 8 + k = 7

⇒ k = 7

Given: The plane 2x + 3y + 4z = 12 meets coordinate axes at A, B and C.

To find: Centroid of ∆ABC

Formula Used: Equation of plane is where (x, y, z) is a point on the plane and a, b, c are intercepts on x-axis, y-axis and z-axis respectively.

Centroid of a triangle

Explanation:

Equation of given plane is 2x + 3y + 4z = 12

Dividing by 12,

Therefore the intercepts on x, y and z-axis are 6, 6 and 3 respectively.

So, the vertices of ∆ABC are (6, 0, 0), (0, 4, 0) and (0, 0, 3)

Centroid

= (2, 4/3, 1)

Therefore, the centroid of ∆ABC is (2, 4/3, 1)

Given: Centroid of ∆ABC is (1, 2, 4)

To find: Equation of plane.

Formula Used: Equation of plane is where (x, y, z) is a point on the plane and a, b, c are intercepts on x-axis, y-axis and z-axis respectively.

Centroid of a triangle

Explanation:

Let the equation of plane be

… (1)

Therefore, A = 3a, B = 3b, C = 3c where (a, b, c) is the centroid of the triangle with vertices (A, 0, 0), (0, B, 0) and (0, 0, C)

Substituting in (1),

Here a = 1, b = 2 and c = 4

Multiplying by 12,

4x + 2y + z = 12

Therefore equation of required plane is 4x + 2y + z = 12

Given: Plane passes through the point A(1, 0, -1).

Plane is perpendicular to the line

To find: Equation of the plane.

Formula Used: Equation of a plane is ax + by + cz = d where (a, b, c) are the direction ratios of the normal to the plane.

Explanation:

Let the equation of the plane be

ax + by + cz = d … (1)

Substituting point A,

a – z = d

Since the given line is perpendicular to the plane, it is the normal.

Direction ratios of line is 2, 4, -3

Therefore, 2 + 3 = d

d = 5

So the direction ratios of perpendicular to plane is 2, 4, -3 and d = 5

Substituting in (1),

2x + 4y – 3z = 5

Therefore, equation of plane is 2x + 4y – 3z = 5

Given: Line meets plane 2x + 3y – z = 14

To find: Point of intersection of line and plane.

Explanation:

Let the equation of the line be

Therefore, any point on the line is (2λ + 1, 4λ +2, -3λ +3)

Since this point also lies on the plane,

2(2λ + 1) + 3(4λ +2) – (-3λ +3) =14

4λ + 2 + 12λ + 6 + 3λ – 3 = 14

19λ + 5 = 14

Therefore the required point is (3, 5, 7).

Given: Plane passes through A(2, 2, 1) and B(9, 3, 6). Plane is perpendicular to 2x + 6y + 6z = 1

To find: Equation of the plane

Formula Used: Equation of a plane is

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

where a:b:c is the direction ratios of the normal to the plane.

(x₁, y₁, z₁) is a point on the plane.

Explanation:

Let the equation of plane be a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

Since (2, 2, 1) is a point in the plane,

a(x – 2) + b(y – 2) + c(z – 1) = 0 … (1)

Since B(9, 3, 6) is another point on the plane,

a(9 – 2) + b(3 – 2) + c(6 – 1) = 0

7a + b + 5c = 0 … (1)

Since this plane is perpendicular to the plane 2x + 6y + 6z = 1, the direction ratios of the normal to the plane will also be perpendicular.

So, 2a + 6b + 6c = 0 ⇒ a + 3b + 3c = 0 … (2)

Solving (1) and (2),

a : b : c = 3 : 4 : -5

Substituting in (1),

3x – 6 + 4y – 8 – 5z + 5 = 0

3x + 4y – 5z – 9 = 0

Therefore the equation of the plane is 3x + 4y – 5z – 9 = 0

Given: Plane passes through the intersection of planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0. Point A(2, 2, 1) lies on the plane.

To find: Equation of the plane.

Formula Used: Equation of plane passing through the intersection of 2 planes P �₁ and P₂ is given by P₁ + λP₂ = 0

Explanation:

Equation of plane is

3x – y + 2z – 4 + λ (x + y + z - 2) = 0 … (1)

Since A(2, 2, 1) lies on the plane,

6 – 2 + 2 – 4 + λ (2 + 2 + 1 – 2) = 0

2 + 3λ = 0

Substituting in (1) and multiplying by 3,

9x – 3y + 6z – 12 – 2 (x + y + z - 2) = 0

9x – 3y + 6z – 12 – 2x – 2y – 2z + 4 = 0

7x – 5y + 4z – 8 = 0

Therefore the equation of the plane is 7x – 5y + 4z – 8 = 0

Given: Plane passes through A(0, -1, 0), B(2, 1, -1) and C(1, 1, 1)

To find: Equation of the plane

Formula Used: Equation of a plane is

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

where a:b:c is the direction ratios of the normal to the plane.

(x₁, y₁, z₁) is a point on the plane.

Explanation:

Let the equation of plane be a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

Substituting point A,

ax + b(y + 1) + cz = 0 … (1)

Substituting points B and C,

2a + 2b – c = 0 and a + 2b +c = 0

Solving,

Therefore, a : b : c = 4 : -3 : 2

Substituting in (1),

4x – 3 (y + 1) + 2z = 0

4x – 3y + 2z – 3 = 0

Therefore equation of plane is 4x – 3y + 2z – 3 = 0

Given: Plane 2x – y + z = 0 is parallel to the line

To find: value of a

Formula Used: If two lines with direction ratios a₁:a₂:a₃ and b₁:b₂:b₃ are perpendicular, then

a₁b₁ + a₂b₂ + a₃b₃ = 0

Explanation:

Since the plane is parallel to the line, the normal to the plane will be perpendicular to the line.

Equation of the line can be rewritten as

Direction ratio of the normal to the plane is 2 : -1 : 1

Direction ratio of line is 1 : -2 : a

Therefore,

2 + 2 + a = 0

a = -4

Therefore, a = -4

Given: Equation of line is

_{Equation of plane is x – y + z = 0}

To find: Angle between a line and the normal to a plane.

Formula Used: If θ is the angle between two lines with direction ratios b₁:b₂:b₃ and c₁:c₂:c₃, then

Explanation:

Direction ratios of given line is 1 : 2 : 1

Direction ratios of the normal to the plane is 1 : -1 : 1

Therefore,

cos θ = 0

θ = 90°

Therefore angle between them is 90°

Given: Line meets plane 2x – y + 3z – 1 = 0

To find: Point of intersection of line and plane.

Explanation:

Let the equation of the line be

Therefore, any point on the line is (3λ + 1, 4λ - 2, -2λ + 3)

Since this point also lies on the plane,

2(3λ + 1) - (4λ -2) + 3(-2λ +3) = 1

6λ + 2 – 4λ + 2 – 6λ + 9 = 1

-4λ = -12

λ = 3

Therefore required point is (10, 10, -3)

Given: Plane passes through the points A(a, 0, 0), B(0, b, 0) and C(0, 0, c)

To find: Equation of plane.

Explanation:

The given points lie on the co-ordinate axes.

Therefore, the plane makes intercepts of a, b and c on the x, y and z-axis respectively.

Equation of the plane is

Given: Equation of two planes are 2x – y + 2z = 3 and 6x – 2y + 3z = 5

To find: cos θ where θ: angle between the planes

Formula Used: Angle between two planes a₁x + b₁y + c₁z = 0 and a₂x + b₂y + c₂z = 0 is

where θ : angle between the planes,

Explanation:

Here a₁ = 2, b₁ = -1, c₁ = 2

a₂ = 6, b₂ = -2, c₂ = 3

Therefore,

Given: Equation of two planes are 2x – y + z = 6 and x + y + 2z = 7

To find: Angle between the two planes

Formula Used: Angle between two planes a₁x + b₁y + c₁z = 0 and a₂x + b₂y + c₂z = 0 is

where θ : angle between the planes,

Explanation:

Here a₁ = 2, b₁ = -1, c₁ = 1

a₂ = 1, b₂ = 1, c₂ = 2

Therefore angle between the planes is

Given: Equation of two planes are and

To find: Angle between the two planes

Formula Used: Angle between two planes a₁x + b₁y + c₁z = 0 and a₂x + b₂y + c₂z = 0 is

where θ : angle between the planes,

Explanation:

Since , the given equation of planes can rewritten as:

3x – 6y + 2z = 4 and 2x – y + 2z = 3

Here a₁ = 3, b₁ = -6, c₁ = 2

a₂ = 2, b₂ = -1, c₂ = 2

Therefore angle between the planes is

Given: Plane passes through the points A(2, 3, 1) and B(4, -5, 3) and is parallel to x-axis

To find: Equation of plane

Formula Used: Equation of a plane parallel o x-axis is

b(y – y₁) + c(z – z₁) = 0

Explanation:

Let the equation of the plane be

b(y – y₁) + c(z – z₁) = 0

Since A(2, 3, 1) lies on the plane,

b(y – 3) + c(z – 1) = 0 … (1)

Since B(4, -5, 3) lies on the plane,

b(-5 – 3) + c(3 – 1) = 0

-8b + 2c = 0 or –4b + c = 0

b : c = 1 : 4

Substituting in (1),

y – 3 + 4z – 4 = 0

y + 4z = 7

The equation of the plane is y + 4z = 7

Given: Variable plane moves so that the sum of the reciprocals of its intercepts on the coordinate axes is (1/2)

Formula Used: Equation of a plane is

Explanation:

Let the intercepts made by the plane on the co-ordinate axes be a, b and c.

Let the equation of the plane be

On solving for each of the given options,

(0, 0, 0) ⇒ LHS ≠ RHS

(1, 1, 1) ⇒ LHS ≠ RHS

Therefore, plane passes through the point (2, 2, 2)

Given: Plane is perpendicular to and is at a distance of 5 units from origin.

To find: Equation of plane

Formula Used: Equation of a plane is lx + my + nz = p where p is the distance from the origin and l, m and n are the direction cosines of the normal to the plane

Explanation:

Direction ratio of normal to the plane is 2:-3:1

Therefore, direction cosines of the normal to the plane is

l , m , n

Since the equation of a plane is lx + my + nz = p where p is the distance from the origin,

2x – 3y + z = 5√14

Therefore, equation of the plane is 2x – 3y + z = 5√14

Given: Point A(2, 3, 4) lies on a plane which is parallel to 5x – 6y + 7z = 3

To find: Equation of the plane

Formula Used: Equation of a plane is

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

where a:b:c is the direction ratios of the normal to the plane

(x₁, y₁, z₁) is a point on the plane.

Explanation:

Since the plane (say P₁) is parallel to the plane 5x – 6y + 7z = 3 (say P₂), the direction ratios of the normal to P₁ is same as the direction ratios of the normal to P₂.

i.e., direction ratios of P₁ is 5 : -6 : 7

Let the equation of the required plane be

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

Here a = 5, b = -6 and c = 7

Since (2, 3, 4) lies on the plane,

5(x - 2) – 6 (y - 3) + 7 (z - 4) = 0

5x – 6y + 7z – 10 + 18 – 28 = 0

5x – 6y + 7z = 20

The equation of the plane is 5x – 6y + 7z = 20

Given: Perpendicular dropped from A(7, 14, 5) on to the plane 2x + 4y – z = 2

To find: co-ordinates of the foot of perpendicular

Formula Used: Equation of a line is

Where b₁:b₂:b₃ is the direction ratio and (x₁, x₂, x₃) is a point on the line.

Explanation:

Let the foot of the perpendicular be (a, b, c)

Since this point lies on the plane,

2a + 4b – c = 2 … (1)

Direction ratio of the normal to the plane is 2 : 4 : -1

Direction ratio perpendicular = direction ratio of normal to the plane

So, equation of the perpendicular is

Since (a, b, c) is a point on the perpendicular,

(7, 14, 5) is a point on the perpendicular.

So, a = 7 - 2λ, b = 14 – 4λ, c = 5 + λ

Substituting in (1),

14 – 4λ + 56 – 16λ – 5 - λ = 2

21λ = 70 – 7 = 63

λ = 3

Therefore, foot of the perpendicular is (1, 2, 8)

Given: Centroid of triangle is (α, β, γ)

To find: Equation of plane.

Formula Used: Equation of plane is where (x, y, z) is a point on the plane and a, b, c are intercepts on x-axis, y-axis and z-axis respectively.

Centroid of a triangle

Explanation:

Let the equation of plane be

… (1)

Therefore, A = 3α, B = 3β, C = 3γ where (a, b, c) is the centroid of the triangle with vertices (A, 0, 0), (0, B, 0) and (0, 0, C)

Substituting in (1),

Here a = α, b = β and c = γ

Therefore equation of required plane is

Given: Equation of plane is

To find: Intercepts made by the plane.

Formula Used: Equation of plane is where (x, y, z) is a point on the plane and a, b, c are intercepts on x-axis, y-axis and z-axis respectively.

Explanation:

The equation of the plane can be written as

2x – 3y + 4z = 12

Dividing by 12,

which is of the form

Therefore the intercepts made by the plane are 6, -4, 3

Given: Equation of line is

_{Equation of the plane is 2x – 3y + z = 5}

_{To find: angle between line and plane}

Formula Used: If θ is the angle between a line with direction ratio b₁:b₂:b₃ and a plane with direction ratio of normal n₁:n₂:n₃, then

Explanation:

Here direction ratio of the line is 1 : -2 : -3

Direction ratio of normal to the plane is 2 : -3 : 1

Therefore,

Therefore, angle between the line and plane is

Given: Equation of line is

Equation of plane is

_{To find: angle between line and plane}

Formula Used: If θ is the angle between a line with direction ratio b₁:b₂:b₃ and a plane with direction ratio of normal n₁:n₂:n₃, then

Explanation:

Here direction ratio of the line is 2 : 2 : 1

Direction ratio of normal to the plane is 6 : -3 : 2

Therefore,

Therefore, angle between the line and plane is

Given: Point is at and equation of plane is

To find: distance of point from plane

Formula Used: Perpendicular distance from (x₁, y₁, z₁) to the plane ax + by + cz + d = 0 is

Explanation:

The point is at (1, 2, 5) and equation of plane is x + y + z + 17 = 0

Distance

Therefore, distance units

Given: The equations of the parallel planes are 2x – 3y + 6z = 5 and 6x – 9y + 18z + 20 = 0

To find: distance between the planes

Formula Used: Distance between two parallel planes ax + by + cz + d₁ = 0 and ax₁ + by₁ + cz₁ + d₁ = 0 is

Explanation:

The equations of the parallel planes are:

2x – 3y + 6z – 5 = 0

Therefore distance between them is

Therefore distance between the planes is units

Given: The equations of the planes are x + 2y – 2z + 1 = 0 and 2x + 4y – 4z – 4z + 5 = 0

To find: distance between the planes

Formula Used: Distance between two parallel planes ax + by + cz + d₁ = 0 and ax₁ + by₁ + cz₁ + d₁ = 0 is

Explanation:

The equations of the planes are:

x + 2y – 2z + 1 = 0 and 2x + 4y – 4z – 4z + 5 = 0

Multiplying the equation of first plane by 2,

2x + 4y – 4z + 2 = 0

Therefore distance between them is

Therefore distance between the planes is units

Given: Equation of plane is 2x – y + z + 3 = 0. P is at (1, 3, 4)

To find: image of P

Explanation:

From the figure, P’ is the image of P and B is the midpoint of PP’

If B is (a, b, c), then

B lies on the plane.

So, 2a – b + c + 3 = 0 … (1)

Also, equation of line PB is

So any point on the line PB will be of the form

x = 2λ +1, y = -λ + 3, z = λ + 4

Since (a, b, c) will also be of this form we can substitute these values in (1)

⇒ 2(2λ + 1) – (-λ + 3) + (λ + 4) + 3 = 0

⇒ 4λ + 2 + λ – 3 + λ + 4 + 3 = 0

⇒ 6λ = -6

⇒ λ = -1

So (a, b, c) = (-1, 4, 3)

Substituting these values in the equations of a, b and c,

Therefore,

x₁ = -3, y₁ = 5, z₁ = 2

Therefore, the image is (-3, 5, 2)

Given :

d = 5

To Find : Equation of a plane

Formulae :

1) Dot Product :

If are two vectors

then,

2) Equation of plane :

Equation of plane which is at a distance of 5 units from the origin and having as a unit vector normal to it is

Where,

For given d = 5 and ,

Equation of plane is

This is a vector equation of the plane

Now,

= z

Therefore, the equation of the plane is

This is - the Cartesian z = 5 equation of the plane.

Given :

d = 7

To Find : Equation of plane

Formulae :

1) Unit Vector :

Let be any vector

Then unit vector of is

Where,

2) Dot Product :

If are two vectors

then,

3) Equation of plane :

Equation of plane which is at a distance of 5 units from the origin and having as a unit vector normal to it is

Where,

For given normal vector

Unit vector normal to the plane is

Equation of the plane is

This is a vector equation of the plane.

Now,

= (x × 3) + (y × 5) + (z × (-6))

= 3x + 5y – 6z

Therefore equation of the plane is

This is the Cartesian equation of the plane.

Given :

To Find : Equation of a plane

Formulae :

1) Unit Vector :

Let be any vector

Then the unit vector of is

Where,

2) Dot Product :

If are two vectors

then,

3) Equation of plane :

Equation of plane which is at a distance of 5 units from the origin and having as a unit vector normal to it is

Where,

For given normal vector

Unit vector normal to the plane is

Equation of the plane is

This is a vector equation of the plane.

Now,

= (x × 2) + (y × (-3)) + (z × 4)

= 2x - 3y + 4z

Therefore equation of the plane is

2x - 3y + 4z = 6

This is the Cartesian equation of the plane.

Given :

d = 6

direction ratios of are (2, -1, -2)

To Find : Equation of plane

Formulae :

1) Unit Vector :

Let be any vector

Then the unit vector of is

Where,

2) Dot Product :

If are two vectors

then,

3) Equation of plane :

Equation of plane which is at a distance of 5 units from the origin and having as a unit vector normal to it is

Where,

For given normal vector

Unit vector normal to the plane is

Equation of the plane is

This is vector equation of the plane.

Now,

= (x × 2) + (y × (-1)) + (z × (-2))

= 2x - y – 2z

Therefore equation of the plane is

This is Cartesian equation of the plane.

Given :

A = (1, 4, 6)

To Find : Equation of plane.

Formulae :

1) Position Vector :

If A is a point having co-ordinates (a₁, a₂, a₃), then its position vector is given by,

2) Dot Product :

If are two vectors

then,

3) Equation of plane :

Equation of plane passing through point A and having as a unit vector normal to it is

Where,

Position vector of point A = (1, 4, 6) is

Now,

= (1×1) + (4×(-2)) + (6×1)

= 1 – 8 + 6

= - 1

Equation of plane is

This is vector equation of the plane.

As

Therefore

= (x × 1) + (y × (-2)) + (z × 1)

= x - 2y + z

Therefore equation of the plane is

This is Cartesian equation of the plane.

Given :

Equation of plane :

To Find :

i) Length of perpendicular = d

ii) Unit normal vector =

Formulae :

1) Unit Vector :

Let be any vector

Then unit vector of is

Where,

2) Length of perpendicular :

The length of the perpendicular from the origin to the plane

is given by,

Given the equation of the plane is

Comparing the above equation with

We get,

& p = 39

Therefore,

= 13

The length of the perpendicular from the origin to the given plane is

Vector normal to the plane is

Therefore, the unit vector normal to the plane is

Given :

Vector equation of the plane is

To Find : Cartesian equation of the given plane.

Formulae :

1) Dot Product :

If are two vectors

then,

Given the equation of the plane is

Here,

= (x × 3) + (y × 5) + (z × (-9))

= 3x + 5y – 9z

Therefore equation of the plane is

3x + 5y – 9z = 8

This is the Cartesian equation of the given plane.

Given :

Cartesian equation of the plane is

5x – 7y + 2z + 4 = 0

To Find : Vector equation of the given plane.

Formulae :

1) Dot Product :

If are two vectors

then,

Given the equation of the plane is

5x – 7y + 2z + 4 = 0

⇒ 5x – 7y + 2z = - 4

The term (5x – 7y + 2z) can be written as

But

Therefore the equation of the plane is

or

This is Vector equation of the given plane.

Given :

Equation of plane : x – 2y + 2z = 6

To Find : unit normal vector

Formula :

Unit Vector :

Let be any vector

Then the unit vector of is

Where,

From the given equation of a plane

x – 2y + 2z = 6

direction ratios of vector normal to the plane are (1, -2, 2).

Therefore, the equation of normal vector is

Therefore unit normal vector is given by

Given :

Equation of plane : 3x – 6y + 2z = 7

To Find : Direction cosines of the normal, i.e.

Formula :

1) Direction cosines :

If a, b & c are direction ratios of the vector, then its direction cosines are given by

For the given equation of a plane

3x – 6y + 2z = 7

Direction ratios of normal vector are (3, -6, 2)

Therefore, direction cosines are

(i) 2x + 3y – z = 5

Given :

Equation of plane : 2x + 3y – z = 5

To Find :

Direction cosines of the normal i.e.

Distance of the plane from the origin = d

Formulae :

1) Direction cosines :

If a, b & c are direction ratios of the vector then its direction cosines are given by

2) The distance of the plane from the origin :

Distance of the plane from the origin is given by,

For the given equation of plane

2x + 3y – z = 5

Direction ratios of normal vector are (2, 3, -1)

Therefore, equation of normal vector is

Therefore, direction cosines are

Now, the distance of the plane from the origin is

(ii) Given :

Equation of plane : z = 3

To Find :

Direction cosines of the normal, i.e.

The distance of the plane from the origin = d

Formulae :

3) Direction cosines :

If a, b & c are direction ratios of the vector, then its direction cosines are given by

4) The distance of the plane from the origin :

Distance of the plane from the origin is given by,

For the given equation of a plane

z = 3

Direction ratios of normal vector are (0, 0, 1)

Therefore, equation of normal vector is

Therefore, direction cosines are

Now, the distance of the plane from the origin is

(iii) Given :

Equation of plane : 3y + 5 = 0

To Find :

Direction cosines of the normal, i.e.

The distance of the plane from the origin = d

Formulae :

1) Direction cosines :

If a, b & c are direction ratios of the vector, then its direction cosines are given by

2) Distance of the plane from the origin :

Distance of the plane from the origin is given by,

For the given equation of a plane

3y + 5 = 0

⇒-3y = 5

Direction ratios of normal vector are (0, -3, 0)

Therefore, equation of normal vector is

Therefore, direction cosines are

Now, distance of the plane from the origin is

Given :

A = (2, -1, 1)

Direction ratios of perpendicular vector = (4, 2, -3)

To Find : Equation of a plane

Formulae :

1) Position vectors :

If A is a point having co-ordinates (a₁, a₂, a₃), then its position vector is given by,

2) Dot Product :

If are two vectors

then,

3) Equation of plane :

If a plane is passing through point A, then the equation of a plane is

Where,

For point A = (2, -1, 1), position vector is

Vector perpendicular to the plane with direction ratios (4, 2, -3) is

Now,

= 8 – 2 – 3

= 3

Equation of the plane passing through point A and perpendicular to vector is

As

= 4x + 2y – 3z

Therefore, the equation of the plane is

4x + 2y – 3z = 3

Or

4x + 2y – 3z - 3 = 0

(i) 2x + 3y + 4z - 12 = 0

Given :

Equation of plane : 2x + 3y + 4z + 12 = 0

To Find :

coordinates of the foot of the perpendicular

Note :

If two vectors with direction ratios (a₁, a₂, a₃) & (b₁, b₂, b₃) are parallel then

From the given equation of the plane

2x + 3y + 4z – 12 = 0

⇒ 2x + 3y + 4z = 12

Direction ratios of the vector normal to the plane are (2, 3, 4)

Let, P = (x, y, z) be the foot of perpendicular perpendicular drawn from origin to the plane.

Therefore perpendicular drawn is

Let direction ratios of are (x, y, z)

As normal vector and are parallel

⇒x = 2k, y = 3k, z = 4k

As point P lies on the plane, we can write

2(2k) + 3(3k) + 4(4k) = 12

⇒ 4k + 9k + 16k = 12

⇒ 29k = 12

,

Therefore co-ordinates of the foot of perpendicular are

P(x, y, z) =

P =

(ii) Given :

Equation of plane : 5y + 8 = 0

To Find :

coordinates of the foot of the perpendicular

Note :

If two vectors with direction ratios (a₁, a₂, a₃) & (b₁, b₂, b₃) are parallel then

From the given equation of the plane

5y + 8 = 0

⇒ 5y = - 8

Direction ratios of the vector normal to the plane are (0, 5, 0)

Let, P = (x, y, z) be the foot of perpendicular perpendicular drawn from origin to the plane.

Therefore perpendicular drawn is

Let direction ratios of are (x, y, z)

As normal vector and are parallel

⇒x = 0, y = 5k, z = 0

As point P lies on the plane, we can write

5(5k) = -8

⇒ 25k = -8

,

Therefore co-ordinates of the foot of perpendicular are

P(x, y, z) =

P =

Given :

Equation of plane : 3x – y – z = 7

A = (2, 3, 7)

To Find :

i) Length of perpendicular = d

ii) coordinates of the foot of the perpendicular

Formulae :

1) Unit Vector :

Let be any vector

Then unit vector of is

Where,

2) Length of perpendicular :

The length of the perpendicular from point A with position vector to the plane is given by,

Note :

If two vectors with direction ratios (a₁, a₂, a₃) & (b₁, b₂, b₃) are parallel then

Given equation of the plane is

3x – y – z = 7 ………..eq(1)

Therefore direction ratios of normal vector of the plane are

(3, -1, -1)

Therefore normal vector of the plane is

From eq(1), p = 7

Given point A = (2, 3, 7)

Position vector of A is

Now,

= (2×3) + (3×(-1)) + (7×(-1))

= 6 – 3 – 7

= -4

Length of the perpendicular from point A to the plane is

Let P be the foot of perpendicular drawn from point A to the given plane,

Let P = (x, y, z)

As normal vector and are parallel

⇒x = 3k+2, y = - k+3, z = -k+7

As point P lies on the plane, we can write

3(3k+2) - (- k+3) - (-k+7) = 7

⇒ 9k + 6 + k – 3 + k – 7 = 7

⇒ 11k = 11

,

Therefore co-ordinates of the foot of perpendicular are

P(x, y, z) =

P = (5, 2, 6)

Given :

Equation of plane :

A = (1, 1, 2)

To Find :

i) Length of perpendicular = d

ii) coordinates of the foot of the perpendicular

Formulae :

1) Unit Vector :

Let be any vector

Then unit vector of is

Where,

2) Length of perpendicular :

The length of the perpendicular from point A with position vector to the plane is given by,

Note :

If two vectors with direction ratios (a₁, a₂, a₃) & (b₁, b₂, b₃) are parallel then

Given equation of the plane is

………..eq(1)

As

Therefore equation of plane is

2x – 2y + 4z = -5 ……… eq(2)

From eq(1) normal vector of the plane is

From eq(1), p = -5

Given point A = (1, 1, 2)

Position vector of A is

Now,

= (1×2) + (1×(-2)) + (2×4)

= 2 – 2 + 8

= 8

Length of the perpendicular from point A to the plane is

Let P be the foot of perpendicular drawn from point A to the given plane,

Let P = (x, y, z)

As normal vector and are parallel

⇒x = 2k+1, y = -2k+1, z = 4k+2

As point P lies on the plane, we can write

2(2k+1) – 2(-2k+1) + 4(4k+2) = -5

⇒ 4k + 2 + 4k – 2 + 16k + 8 = -5

⇒ 24k = -13

,

Therefore co-ordinates of the foot of perpendicular are

P(x, y, z) =

P ≡

Given :

Equation of plane : 2x + y – 2z + 3 = 0

P = (1, 2, 4)

To Find :

i) Equation of perpendicular

ii) Length of perpendicular = d

iii) coordinates of the foot of the perpendicular

Formulae :

1) Unit Vector :

Let be any vector

Then unit vector of is

Where,

2) Length of perpendicular :

The length of the perpendicular from point A with position vector to the plane is given by,

Note :

If two vectors with direction ratios (a₁, a₂, a₃) & (b₁, b₂, b₃) are parallel then

Given equation of the plane is

2x + y – 2z + 3 = 0

⇒2x + y – 2z = -3 ………..eq(1)

From eq(1) direction ratios of normal vector of the plane are

(2, 1, -2)

Therefore, equation of normal vector is

= 3

From eq(1), p = -3

Given point P = (1, 2, 4)

Position vector of A is

Here,

Now,

= (1×2) + (2×1) + (4×(-2))

= 2 + 2 - 8

= -4

Length of the perpendicular from point A to the plane is

Let Q be the foot of perpendicular drawn from point P to the given plane,

Let Q = (x, y, z)

As normal vector and are parallel, we can write,

This is the equation of perpendicular.

⇒x = 2k+1, y = k+2, z = -2k+4

As point Q lies on the plane, we can write

2(2k+1) + (k+2) - 2(-2k+4) = -3

⇒ 4k + 2 + k + 2 + 4k - 8 = -3

⇒ 9k = 1

,

Therefore co-ordinates of the foot of perpendicular are

Q(x, y, z) =

Q ≡

Given :

Equation of plane : 2x - y + z + 1 = 0

P = (3, 2, 1)

To Find :

i) Length of perpendicular = d

ii) coordinates of the foot of the perpendicular

iii) Image of point P in the plane.

Formulae :

1) Unit Vector :

Let be any vector

Then unit vector of is

Where,

2) Length of perpendicular :

The length of the perpendicular from point A with position vector to the plane is given by,

Note :

If two vectors with direction ratios (a₁, a₂, a₃) & (b₁, b₂, b₃) are parallel then

Given equation of the plane is

2x - y + z + 1 = 0

⇒2x - y + z = -1 ………..eq(1)

From eq(1) direction ratios of normal vector of the plane are

(2, -1, 1)

Therefore, equation of normal vector is

From eq(1), p = -1

Given point P = (3, 2, 1)

Position vector of A is

Here,

Now,

= (3×2) + (2×(-1)) + (1×1)

= 6 - 2 + 1

= 5

Length of the perpendicular from point A to the plane is

Let Q be the foot of perpendicular drawn from point P to the given plane,

Let Q = (x, y, z)

As normal vector and are parallel, we can write,

⇒x = 2k+3, y = -k+2, z = k+1

As point A lies on the plane, we can write

2(2k+3) - (-k+2) + (k+1) = -1

⇒ 4k + 6 + k – 2 + k + 1 = -1

⇒ 6k = -6

,

Therefore, co-ordinates of the foot of perpendicular are

Q(x, y, z) =

Q ≡

Let, R(a, b, c) be image of point P in the given plane.

Therefore, the power of points P and R in the given plane will be equal and opposite.

2a – b + c + 1 = - (2(3) – 2 + 1 + 1)

⇒2a – b + c + 1 = - 6

⇒2a – b + c = - 7 ………eq(2)

Now,

As are parallel