The Plane

If the given points are then the equation of the plane passing through these three points is given by the following equation.

Now substitute the values given

Now apply the determinant

– 21x + 42 – 9y + 9 + 3z = 0

– 21x – 9y + 3z + 51 = 0

– 3(7x + 3y – z – 17 = 0)

7x + 3y – z – 17 = 0

This is the equation of the plane.

If the given points are then the equation of the plane passing through these three points is given by the following equation.

Now substitute the values given

Now apply the determinant

(x + 5)(0 + 18) – y(0 + 9) + (z + 6)(12 – 30) = 0

(x + 5)(18) – y(9) + (z + 6)(– 18) = 0

18x + 90 – 9y – 18z – 108 = 0

Now divide both sides 9 then we get the plane equation as

2x – y – 2z – 2 = 0

If the given points are then the equation of the plane passing through these three points is given by the following equation.

Now substitute the values given

Now apply the determinant

(x – 1)(– 2 + 3) – (y – 1)(0 + 3) + (z – 1)(0 – 6) = 0

(x – 1)1 – (y – 1)3 + (z – 1)(– 6) = 0

x – 3y – 6z + 8 = 0

this is the equation of plane.

If the given points are then the equation of the plane passing through these three points is given by the following equation.

Now substitute the values given

Now apply the determinant

(x – 2)(– 4 – 12) – (y – 3)(10 + 6) + (z – 4)(20 – 4) = 0

(x – 2)(– 16) – (y – 3)(16) + (z – 4)(16) = 0

– 16x + 32 – 16y + 48 + 16z – 64 = 0

– 16x – 16y + 16z + 16 = 0

(x + y – z – 1) × – 16 = 0

The equation of plane is

If the given points are then the equation of the plane passing through these three points is given by the following equation.

Now substitute the values given

Now apply the determinant

(x)(4 – 0) – (y + 1)(3 – 0) + z(6 – 4) = 0

4x – (y + 1)(3) + z(2) = 0

4x – 3y – 3 + 2z = 0

This is the equation of the plane.

Given that these four points are coplanar so these four points lie on the same plane.

So first let us take three points and find the equation of the plane passing through these four points and then let us substitute the fourth point in it. If it is 0 then the point lies on the plane formed by these three points then they are coplanar.

The equation of the plane passing through these three points is given by the following equation.

Now let us take (0, – 1, – 1), (4, 5, 1), (3, 9, 4) and find plane equation.

Now apply the determinant

x(30 – 20) – (y + 1)(20 – 6) + (z + 1)(40 – 18) = 0

10x – (y + 1)(14) + (z + 1)(22) = 0

10x – 14y + 22z + 8 = 0 now divide by 2 on both sides

The equation is 5x – 7y + 11z + 4 = 0

Now let us substitute fourth point (– 4, 4, 4) we get

5(– 4) – 7(4) + 11(4) + 4 = 0

– 20 – 28 + 44 + 4 = 0

– 48 + 48 = 0

0 = 0

L.H.S = R.H.S

So as said above this fourth point satisfies so this point also lies on the same plane.

Hence they are coplanar.

Given that these four points are coplanar so these four points lie on the same plane

So first let us take three points and find the equation of plane passing through these four points and then let us substitute the fourth point in it. If it is 0 then the point lies on the plane formed by these three points then they are coplanar.

the equation of the plane passing through these three points is given by the following equation.

Now let us take (0, – 1, 0), (2, 1, – 1), (1, 1, 1) and find plane equation.

Now apply the determinant

x(2 + 2) – (y + 1)(2 + 1) + z(4 – 2) = 0

4x – 3y – 3 + 2z = 0

4x – 3y + 2z – 3 = 0

Now let us substitute (3, 3, 0) in plane equation

4x – 3y + 2z – 3 = 0

4(3) – 3(3) + 2(0) – 3 = 0

12 – 9 + 0 – 3 = 0

12 – 12 = 0

0 = 0

So this point lies on the plane

Hence they are coplanar.

Given that these four points are coplanar so these four points lie on the same plane

So first let us take three points and find the equation of plane passing through these four points and then let us substitute the fourth point in it. If it is 0 then the point lies on the plane formed by these three points then they are coplanar.

the equation of the plane passing through these three points is given by the following equation.

Now let us take (0, 4, 3), (– 1, – 5, – 3), (– 2, – 2, 1) and find plane equation.

Now apply the determinant

x(18 – 36) – (y – 4)(2 – 12) + (z – 3)(6 – 18) = 0

x(– 18) – (y – 4)(– 10) + (z – 3)(– 12) = 0

– 18x + 10y – 40 – 12z + 36 = 0

– 18x + 10y – 12z – 4 = 0

Now let us substitute (1, 1, – 1) in plane equation

– 18x + 10y – 12z – 4 = 0

– 18(1) + 10(1) – 12(– 1) – 4 = 0

– 18 + 10 + 12 – 4 = 0

– 22 + 22 = 0

0 = 0

Lhs = rhs

So this point lies on the plane

Hence they are coplanar.

We know that the equation passing through two point (a, b, c) and (d, e, f) is given by

the line through A(3, – 4, – 5) and B(2, – 3, 1) is

Now let us see how a point is going to be on the line

X = – k + 3, y = k – 4, z = 6k – 5

So now let a point P be the point of the intersection of a line and the plane so let the coordinates of P = (– k + 3, k – 4, 6k – 5)

Now let us find the equation of the plane passing through L(2, 2, 1), M(3, 0, 1) and N(4, – 1, 0).

The equation of the plane passing through these three points is given by the following equation.

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

2x – 4 + y – 2 – z + 1 = 0

2x + y – z = 5

Now point P lies on this plane so

2(3 – k) + (k – 4) – (6k – 5) = 5

6 – 2k + k – 4 – 6k + 5 = 5

– 7k = – 2

So point P is

Now we have to find the ration in which this P divides AB

Let the ratio b m:1

We know the section formula

That is if a line AB is divided by P in ratio m:1 then

Solving

We get 19m + 19 = 14m + 21

5m = – 2

So point P divides line in ration 2:5 externally since ratio is negative.

Let P(x₁,y₁,z₁) be any point on 2x – y + 3z – 4 = 0.

⟹ 2x_{1 –} y₁ + 3z₁ – 4 = 0

⟹ 2x_{1 –} y₁ + 3z₁ = 4 eq(i)

Distance between (x₁,y₁,z₁) and the plane

6x – 3y + 9z + 13 = 0:

We know, the distance of point (x₁,y₁,z₁) from the plane

is given by:

Putting the necessary values,

⟹

⟹

⟹ (using eq (i) )

⟹

∴ the distance between the parallel planes 2x – y + 3z – 4 = 0 and

6x – 3y + 9z + 13 = 0 is units.

Since the plane is parallel to 2x – 3y + 5z + 7 = 0, it must be of the form:

2x – 3y + 5z + θ = 0

According to question,

The plane passes through (3, 4, –1)

⟹ 2(3) – 3(4) +5(–1) + θ = 0

⟹ θ = 11

So, the equation of the plane is as follows:

2x – 3y + 5z + 11 = 0

Distance of the plane 2x – 3y + 5z + 7 = 0 from (3, 4, –1):

We know, the distance of point (x₁,y₁,z₁) from the plane

is given by:

Putting the necessary values,

⟹

⟹

∴ the distance of the plane 2x – 3y + 5z + 7 = 0 from (3, 4, –1) is

units

Given:

* Equation of planes: π₁= 2x – 2y + z + 3 = 0
π₂= 2x – 2y + z + 9 = 0

Let the equation of the plane mid–parallel to these planes be:

π₃: 2x – 2y + z + θ = 0

Now,

Let P(x₁,y₁,z₁) be any point on this plane,

⟹ 2(x₁) – 2(y₁) + (z₁) + θ = 0 eq(i)

We know, the distance of point (x₁,y₁,z₁) from the plane

is given by:

⟹ Distance of P from π₁:

⟹ (using eq(i) )

Similarly

⟹ Distance of P from π₂ :

⟹ (using eq(i) )

As π₃ is mid–parallel to π₁ and π_{2 :}

p = q

⟹

Squaring both sides,

⟹

⟹ (3 – θ)² = (9 – θ)²

⟹ 9 – 6θ + θ² = 81 – 18θ + θ²

⟹ θ = 6

∴ equation of the mid–parallel plane is 2x – 2y + z + 6 = 0

Let be the position vector of any point P on the plane

:

⟹ eq(i)

We know, the distance of from the plane is given by:

Putting the values of and :

⟹

⟹

⟹

⟹

∴ the distance between the planes and is units.

Equation of line is

_{And the equation of the plane is}

As we know that the angle θ between the line and a plane a₂x + b₂y + c₂z + d₂ = 0 is given by

Here, a₁ = 1, b₁ = 1 and c₁ = 1

and a₂ = 2, b₂ = 3 and c₂ = 4

_{The angle between them is given by}

As we know that the angle θ between the line and a plane a₂x + b₂y + c₂z + d₂ = 0 is given by

……(1)

Now, given equation of the line is

So, a₁ = 1 , b₁ = – 1 and c₁ = 1

Equation of plane is 2x + y – z – 4 = 0

So, a₂ = 2, b₂ = 1, c₂ = – 1 and d₂ = – 4

∴

⇒

⇒

⇒ sinθ = 0

∴ the angle between the plane and the line is 0°

As we know that the angle θ between the line and a plane a₂x + b₂y + c₂z + d₂ = 0 is given by

……(1)

Given that the line is passing through A(3, – 4, – 2) and (12, 2, 0)

∴ the direction ratios of line AB are

= (12 – 3, 2 – (– 4), 0 – (– 2))

= (12 – 3, 2 + 4, 0 + 2)

= (9, 6, 2)

So,

a₁ = 9, b₁ = 6 and c₁ = 2 ……(2)

Given equation of plane is 3x – y + z = 1

So,

a₂ = 3, b₂ = – 1 and c₂ = 1 ……(3)

∴

⇒

⇒

⇒ θ = sin ^{– 1} ()

Therefore the required angle is sin ^{– 1} ()

We know that line is parallel to the plane if …… (1)

Given the equation of the line is

and the equation of the plane is

So,

Putting the values in equation (1)

⇒ 2m – 3m – 3 = 0

⇒ – m = 3

⇒ m = – 3

We know that line and plane is parallel if

……(1)

Given, the equation of the line
and equation of plane is the

,

So, and

Now,

=

= 1 + 3 – 4 = 0

So, the line and the plane are parallel

We know that the distance (D) of a plane from a point is given by

We take the mod value

So,

The required line is perpendicular to the plane

So line is parallel to normal vector
of plane.

And it is also passing through

We know that the equation of the line passing through and parallel to is ……(1)

Hence equation of the required line is

We know that the equation of plane passing through (x₁,y₁,z₁) is given by

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0 ……(1)

So, equation of plane passing through (2,3, – 4) is

a(x – 2) + b(y – 3) + c(z + 4) = 0 ……(2)

It also passes through (1, – 1, – 3)

So, equation (2) must satisfy the point (1, – 1, – 3)

∴ a(1 – 2) + b(– 1 – 3) + c(– 3 + 4) = 0

⇒ – a – 4b + c = 0

⇒ a + 4b – 7c = 0 ……(3)

We know that line is parallel to plane

a₂x + b₂y + c₂z + d₂ = 0 if a₁a₂ + b₁b₂ + c₁c₂ = 0 ……(4)

Here, equation(2) is parallel to x axis,

……(5)

Using (2) and (5) in equation (4)

a×1 + b×0 + c×0 = 0

⇒ a = 0

Putting the value of a in equation (3)

a – 4b + 7c = 0

⇒ 0 – 4b + 7c = 0

⇒ – 4b = – 7c

⇒ b =

Now, putting the value of a and b in equation (2)

a(x – 2) + b(y – 3) + c(z + 4)

⇒ 0(x – 2) + (y – 3) + c(z + 4) = 0

⇒

⇒ 7cy – 21c + 4cz + 16c = 0

Dividing by c we have,

7y – 21 + 4z + 16 = 0

⇒ 7y + 4z – 5 = 0

Equation of required plane is 7y + 4z – 5 = 0

We know that the equation of plane passing through (x₁,y₁,z₁) is given by

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0 ……(1)

So, equation of plane passing through (0,0,0) is

a(x – 0) + b(y – 0) + c(z – 0) = 0

ax + by + cz = 0 ……(2)

It also passes through (3, – 1,2)

So, equation (2) must satisfy the point (3, – 1,2)

∴ 3a – b + 2c = 0 ……(3)

We know that line is parallel to plane a₂x + b₂y + c₂z + d₂ = 0 if a₁a₂ + b₁b₂ + c₁c₂ = 0 ……(4)

Here, the plane is parallel to line,

So,

a×1 + b× – 4 + c×7 = 0

⇒ a – 4b + 7c = 0 ……(5)

Solving equation (3) and (5) by cross multiplication we have,

⇒

⇒

∴ a = k, b = – – 19k and c = – 11k

Putting the value in equation (2)

ax + by + cz = 0

kx – 19ky – 11kz = 0

Dividing by k we have

x – 19y – 11z = 0

The required equation is x – 19y – 11z = 0

We know that the equation of line passing through (1,2,3) is given by

……(1)

We know that line is parallel to plane a₂x + b₂y + c₂z + d₂ = 0 if a₁a₂ + b₁b₂ + c₁c₂ = 0 ……(2)

Here, line (1) is parallel to plane,

x – y + 2z = 5

So,

a×1 + b× – 1 + c×2 = 0

⇒ a – b + 2c = 0 ……(3)

Also, line (1) is parallel to plane,

3x + y + z = 6

So,

a×3 + b×1 + c×1 = 0

⇒ 3a + b + c = 0 ……(4)

Solving equation (3) and (4) by cross multiplication we have,

⇒

⇒

∴ a = – 3k, b = 5k and c = 4k

Putting the value in equation (1)

Multiplying by k we have

The required equation is

Let a₁,b₁ and c₁ be the direction ratios of the line 5x + 2y – 4z + 2 = 0 and 2x + 8y + 2z – 1 = 0.

As we know that if two planes are perpendicular with direction ratios as a₁, b₁ and c₁ and a₂ , b₂ and c₂ then

a₁a₂ + b₁b₂ + c₁c₂ = 0

Since, line lies in both the planes, so it is perpendicular to both planes

5a₁ + 2b₁ – 4c₁ = 0 ……(1)

2a₁ + 8b₁ + 2c₁ = 0 ……(2)

Solving equation (1) and (2) by cross multiplication we have,

⇒

⇒

⇒

∴ a = 2k, b = – k and c = 2k

We know that line is parallel to plane a₂x + b₂y + c₂z + d₂ = 0 if a₁a₂ + b₁b₂ + c₁c₂ = 0 ……(3)

Here, line with direction ratios a₁,b₁ and c₁ is parallel to plane,

4x – 2y – 5z = 5

So,

2×4 + (– 1)× – 2 + 2× – 5 = 0

⇒ 8 + 2 – 10 = 0

Therefore, the line of section is parallel to the plane.

Equation of line passing through and parallel to is given by ……(1)

Given that the line passes through (1, – 1,2) is

…… (2)

Since , line (1) is perpendicular to the plane 2x – y + 3z – 5 = 0, so normal to plane is parallel to the line.

In vector form,

is parallel to

So, as l is any scalar

Thus, the equation of required line,

We know that the equation of plane passing through (x₁,y₁,z₁) is given by

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0 ……(1)

So, equation of plane passing through (2,2, – 1) is

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

It also passes through (3,4,2)

So, equation (2) must satisfy the point (3,4,2)

∴ a(3 – 2) + b(4 – 2) + c(2 + 1) = 0

⇒ a + 2b + 3c = 0 ……(3)

We know that line is parallel to plane a₂x + b₂y + c₂z + d₂ = 0 if a₁a₂ + b₁b₂ + c₁c₂ = 0 ……(4)

Here, the plane (2) is parallel to line having direction ratios 7,0,6 ,

So,

a×7 + b×0 + c×6 = 0

⇒ 7a + 6c = 0

⇒ ……(5)

Putting the value of a in equation (3)

a + 2b + 3c = 0

⇒ + 2b + 3c = 0

⇒ – 6c + 14b + 21c = 0

⇒ 14b + 15c = 0

⇒

Putting the value of a and b in equation (2)

a(x – 2) + b(y – 2) + c(z + 1) = 0

⇒ (x – 2) + (y – 2) + c(z + 1) = 0

⇒

Multiplying by we have,

– 12x + 24 – 15y + 30 + 14z + 14 = 0

⇒ – 12x + 15y + 14z + 68 = 0

⇒ 12x – 15y – 14z – 68 = 0

Equation of required plane is 12x – 15y – 14z – 68 = 0

We know that line is parallel to plane a₂x + b₂y + c₂z + d₂ = 0 if a₁a₂ + b₁b₂ + c₁c₂ = 0 is given by

……(1)

Now, given equation of line is

So, a1 = 3 , b1 = – 1 and c1 = 2

Equation of plane is 3x + 4y + z + 5 = 0

So, a2 = 3, b2 = 4, c2 = 1 and d2 = – 5

∴

⇒

⇒

⇒ sinθ = ⇒ θ = sin ^{– 1}(

∴ the angle between the plane and the line is sin ^{– 1}(

We know that equation of plane passing through the intersection of planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given by

(a₁x + b₁y + c₁z + d₁) + k(a₂x + b₂y + c₂z + d₂) = 0

So, equation of plane passing through the intersection of planes

x – 2y + z – 1 = 0 and 2x + y + z – 8 = 0 is

(x – 2y + z – 1) + k(2x + y + z – 8) = 0 ……(1)

⇒ x(1 + 2k) + y(– 2 + k) + z(1 + k) + (– 1 – 8k) = 0

We know that line is parallel to plane a₂x + b₂y + c₂z + d₂ = 0 if a₁a₂ + b₁b₂ + c₁c₂ = 0

Given the plane is parallel to line with direction ratios 1,2,1

1×(1 + 2k) + 2×(– 2 + k) + 1×(1 + k) = 0

⇒ 1 + 2k – 4 + 2k + 1 + k = 0

⇒ k =

Putting the value of k in equation (1)

⇒

⇒

⇒ 9x – 8y + 7z – 21 = 0

We know that the distance (D) of point (x₁,y₁,z₁) from plane ax + by + cz – d = 0 is given by

o, distance of point (1,1,1) from plane (1) is

⇒

⇒

⇒

Taking the mod value we have

We know that line and plane is parallel if

……(1)

Given, equation of line
and equation of plane is

,

So, and

Now,

=

= – 2 + 2 = 0

So, the line and the plane are parallel

We know that the distance (D) of a plane from a point is given by

We take the mod value

So,

We know that line and plane is parallel if

……(1)

Given, equation of line
and equation of plane is

,

So, and

Now,

=

= 2 + 2 – 4 = 0

So, the line and the plane are parallel

We know that the distance (D) of a plane from a point is given by

We take the mod value

So,

We know that equation of plane passing through the intersection of planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given by

(a₁x + b₁y + c₁z + d₁) + k(a₂x + b₂y + c₂z + d₂) = 0

So, equation of plane passing through the intersection of planes

3x – 4y + 5z – 10 = 0 and 2x + 2y – 3z – 4 = 0 is

(3x – 4y + 5z – 10) + k(2x + 2y – 3z – 4) = 0 ……(1)

⇒ x(3 + 2k) + y(– 4 + 2k) + z(5 – 3k) + (– 10 – 4k) = 0

We know that line is parallel to plane a₂x + b₂y + c₂z + d₂ = 0 if a₁a₂ + b₁b₂ + c₁c₂ = 0

Given the plane is parallel to line

6×(3 + 2k) + 3× (– 4 + 2k) + 2×(5 – 3k) = 0

⇒ 18 + 12k – 12 + 6k + 10 – 6k = 0

⇒ k =

Putting the value of k in equation (1)

⇒

⇒

⇒ x – 20y + 27z – 14 = 0

The required equation is x – 20y + 27z – 14 = 0

The plane passes through the point (1,2, – 4)

A vector in a direction perpendicular to

and

is

Equation of the plane is (

⇒

Substituting , we get the Cartesian form as

– 9x + 8y – z = 11

The distance of the point (9, – 8, – 10) from the plane

We know that the equation of plane passing through (x₁,y₁,z₁) is given by

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0 …… (1)

So, equation of plane passing through (3,4,1) is

a(x – 3) + b(y – 4) + c(z – 1) = 0 ……(2)

It also passes through (0,1,0)

So, equation (2) must satisfy the point (0,1,0)

∴ a(0 – 3) + b(1 – 4) + c(0 – 1) = 0

⇒ – 3a – 3b – c = 0

⇒ 3a + 3b + c = 0 ……(3)

We know that line is parallel to plane a₂x + b₂y + c₂z + d₂ = 0 if a₁a₂ + b₁b₂ + c₁c₂ = 0 …… (4)

So,

a×2 + b×7 + c×5 = 0

⇒ 2a + 7b + 5c = 0 ……(5)

Solving equation (3) and (5) by cross multiplication we have,

⇒

⇒

∴ a = 8k, b = – – 13k and c = 15k

Putting the value in equation (2)

8k(x – 3) – 13k(y – 4) + 15k(z – 1) = 0

8kx – 24k – 13ky + 52k + 15kz – 15k = 0

Dividing by k we have

8x – 13y + 15z + 13 = 0

Equation of required plane is 8x – 13y + 15z + 13 = 0

Given line

Let

⇒ x = 3r + 2, y = 4r – 1, z = 2r + 2

Substituting in the equation of the plane x – y + z – 5 = 0

We get , (3r + 2) – (4r – 1) + (2r + 2) – 5 = 0

⇒ 3r + 2 – 4r + 1 + 2r + 2 – 5 = 0

⇒ r = 0

∴ x = 3×0 + 2, y = 4×0 – 1 , z = 2×0 + 2

⇒ x = 2, y = – 1, z = 2

Direction ratios of the line are 3, 4, 2

Direction ratios of the line perpendicular to the plane are 1, – 1, 1

∴

⇒

⇒

⇒ sinθ = ⇒ θ = sin ^{– 1}(

∴ the angle between the plane and the line is sin ^{– 1}(

We know that equation of line passing through point and parallel to vector is given by

……………..(1)

Given that , the line is passing through (1,2,3)

So,

It is given that line is perpendicular to plane

So, normal to plane ( is parallel to

So, let

Putting and in (1) , equation of line is

⇒

Direction ratios of the line are (2,3,6)

Direction ratio of a line perpendicular to the plane

10x + 2y – 11z = 3 are 10, 2, – 11

As we know that the angle θ between the line and a plane a₂x + b₂y + c₂z + d₂ = 0 is given by

If θ is the angle between the line and the plane, then

∴

⇒

⇒

⇒ sinθ = ⇒ θ = sin ^{– 1}(

∴ the angle between the plane and the line is sin ^{– 1}(

We know that the equation of line passing through (1,2,3) is given by

……(1)

We know that line is parallel to plane a₂x + b₂y + c₂z + d₂ = 0 if a₁a₂ + b₁b₂ + c₁c₂ = 0 ……(2)

Here, line (1) is parallel to plane,

x – y + 2z = 5

So,

a×1 + b× – 1 + c×2 = 0

⇒ a – b + 2c = 0 ……(3)

Also, line (1) is parallel to plane,

3x + y + z = 6

So,

a×3 + b×1 + c×1 = 0

⇒ 3a + b + c = 0 ……(4)

Solving equation (3) and (4) by cross multiplication we have,

⇒

⇒

∴ a = – 3k, b = 5k and c = 4k

Putting the value in equation (1)

Multiplying by k we have

The required equation is

The vector equation is

Here, given midline is perpendicular to plane 3x – y – 2z = 7.

We know that line is perpendicular to plane a₂x + b₂y + c₂z + d₂ = 0 if

So, normal vector of plane is parallel to line .

So, direction ratios of normal to plane are proportional to the direction ratios of line .

Here,

By cross multiplying the last two we have

– 2k = 4

⇒ k = – 2

We know that the equation of plane passing through (x₁,y₁,z₁) is given by

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0 ……(1)

So, equation of plane passing through (– 1,2,0) is

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

It also passes through (2,2, – 1)

So, equation (2) must satisfy the point (2,2, – 1)

∴ a(2 + 1) + b(2 – 2) + c(– 1) = 0

⇒ 3a – c = 0 ……(3)

We know that line is parallel to plane a₂x + b₂y + c₂z + d₂ = 0 if a₁a₂ + b₁b₂ + c₁c₂ = 0 ……(4)

Here, the plane is parallel to line,

So,

a×1 + b×2 + c×1 = 0

⇒ a + 2b + c = 0 ……(5)

Solving equation (3) and (5) by cross multiplication we have,

⇒

⇒

∴ a = k, b = – 2k and c = 3k

Putting the value in equation (2)

k(x + 1) – 2k(y – 2) + 3k(z – 0) = 0

kx + k – 2ky + 4k + 3kz = 0

Dividing by k we have

x – 2y + 3z + 5 = 0

The required equation is x – 2y + 3z + 5 = 0

(a) Direction ratio of given line are (5 – 3, 1 – 4, 6 – 1) = (2, – 3, 5)Hence, equation of line isFor any point on the yz –plane x = 0x = 2r + 5 = 0r = y = – 3() + 1 = z = 5() + 6 = Hence, the coordinates of this point are (0, ).

(b): Direction ratio of given line are (5 – 3, 1 – 4, 6 – 1) = (2, – 3, 5)

Hence, equation of line is

For any point on zx –plane y = 0

y = – 3r + 1 = 0

r =

x = 2() + 5 =

z = 5() + 6 =

Hence, the coordinates of this point are (, 0, ).

Let the coordinates of the points A and B be (3, – 4, – 5) and (2, – 3, 1) respectively.

The equation of the line joining the points ) is

where r is a constant

Thus, the equation of AB is

Any point on the line AB is the form

– r + 3, r – 4, 6r – 5

Let p be the point of intersection of the line AB and the plane 2x + y + z = 7

Thus, we have,

2(– r + 3) + r – 4 + 6r – 5 = 7

⇒ – 2r + 6 + r – 4 + 6r – 5 = 7

⇒ 5r = 10

⇒ r = 2

Substituting the value of r in – r + 3, r – 4, 6r – 5, the coordinates of P are:

(– 2 + 3, 2 – 4, 12 – 5) = (1, – 2, 7)

The equation of the given line is

…… (1)

The equation of the given plane is

…… (2)

Substituting the value of from equation (1) in equation (2), We obtain

(3 + 2) – (4 – 1) + (2 + 2) = 5

= 0

Substituting the value of equation (1), We obtain the equation of the line as

This means that the position vector of the point of intersection of the line and the

plane is

This shows that the point of intersection of the given plane and line is given by the coordinates, (2, – 1, 2). The point is (– 1, – 5, – 10) .

The distance d between the points, (2, – 1, 2) and (– 1, – 5, – 10) is

d =

d =

d = 13

To find the point of intersection of the line

and the plane

= 0

We are substituting of line in the plane.

⇒ 2 + 3 + 8 – 8 + 2 + 2 = 0

⇒ 3

⇒

Hence, the distance of the point from

Equation of line through the point A(2, – 1, 2) and B (5, 3, 4) is

⇒

⇒

Substituting these in the plane equation, We get

(3r + 2) – (4r – 1) + (2r + 2) = 5

⇒ r = 0

⇒ x = 2, y = – 1, z = 2

Distance of (2, – 1, 2) from (– 1, – 5, – 10) is

=

= 13

Equation of line through the point A (3, – 4, – 5) and B (2, – 3, 1) is

⇒

⇒

Substituting these in the plane equation, We get

2(– r + 3) + (r – 4) + (6r – 5) = 7

⇒ r = 2

⇒ x = 1, y = – 4, z = 7

Distance of (1, – 2, 7) from (3, 4, 4) is

=

= 7

The given plane is x – y + z = 5 …… (1)

We have to find the distance of the point (1, – 5, 9) from this plane measured along a line to

x = y = z

So, the direction ratio of the line from the point (1, – 5, 9) to the given plane will be the same as that

of given line.

The equation of line passing through (1, – 5, 9) and having direction ratio is

x = r + 1, y = r – 5, z = r + 9

put in equation (1)

r + 1 – r + 5 + r + 9 = 5

⇒ r + 15 = 5

⇒ r = – 10

Coordinates are (–10 + 1, –10 – 5, –10 + 9) is (–9, –8, –1)

Distance of (–9, –15, –1) from (1, – 5, 9)

=

We know that the lines,

and are coplanar if

And the equation of the plane containing them is

Here,

And

Since , the lines are coplanar Now the equation of the plane containing the given lines is

we know that line and are coplanar if

And equation of the plane containing them is

Here, equation of lines are

and

So, x₁ = – 1, y₁ = 3, z₁ = – 2, l₁ = – 3, m₁ = 2, n₁ = 1

x₂ = 0, y₂ = 7, z₂ = – 7, l₂ = 1, m₂ = – 3, n₂ = 2

so,

= 1(4 + 3) – 4(– 6 – 1) – 5(9 – 2)

= 7 + 28 – 35

= 0

So, lines are coplanar

Equation of plane containing line is

(x + 1)(4 + 3) – (y – 3)(– 6 – 1) + (z + 2)(9 – 2) = 0

7x + 7 + 7y – 21 + 7z + 14 = 0

7x + 7y + 7z = 0

X + y + z = 0

we know that equation of plane passing through (x₁, y₁, z₁) is given by

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0 …… (1)

Required plane is passing through (0, 7, – 7) so

a(x – 0) + b(y – 7) + c(z + 7) = 0

ax + b(y – 7) + c(z + 7) = 0 …… (2)

plane (2) also contain line so, it passes through point (– 1, 3, – 2),

a(– 1) + b(3 – 7) + c(– 2 + 7) = 0

– a – 4b + 5c = 0 …… (3)

Also plane (2) will be parallel to line so,

a₁a₂ + b₁b₂ + c₁c₂ = 0

(a)(– 3) + (b)(2) + (c)(1) = 0

– 3a + 2b + c = 0 …… (4)

Solution (3) and (4) by cross – multiplication,

a = – 14, b = – 14, c = – 14

put a, b, c in equation (2),

ax + b(y – 7) + c(z + 7) = 0

(– 14)x + (– 14)(y – 7) + (– 14)(z + 7) = 0

Dividing by (– 14) we get

x + y – 7 + z + 7 = 0

x + y + z = 0

so, equation of plane containing the given point and line is x + y + z = 0

the other line is

so, a₁a₂ + b₁b₂ + c₁c₂ = 0

(1)(1) + (1)(– 3) + (1)(2) = 0

1 – 3 + 2 = 0

0 = 0

LHS = RHS

So, lie on plane x + y + z = 0

we know that equation of plane passing through (x₁, y₁, z₁) is given by

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0 …… (1)

since, required plane contain lines

and

So, required plane passes through (4, 3, 2) and (3, – 2, 0) so equation of required plane is

a(x – 4) + b(y – 3) + c(z – 2) = 0 …… (2)

plane (2) also passes through (3, – 2, 0), so

a(3 – 4) + b(– 2 – 3) + c(0 – 2)

– a – 5b – 2c = 0

a + 5b + 2c = 0 …… (3)

now plane (2) is also parallel to line with direction ratios 1, – 4, 5 so,

a₁a₂ + b₁b₂ + c₁c₂ = 0

(a)(1) + (b)(– 4) + (c)(5) = 0

a – 4b + 5c = 0 …… (4)

solving equation (3) and (4) by cross – multiplication,

Multiplying by 3,

a = 11 λ, b = – λ, c = – 3 λ

put a, b, c in equation (2),

a(x – 4) + b(y – 3) + c(z – 2) = 0

(11 λ)(x – 4) + (– λ)(y – 3) + (– 3 λ)(z – 2) = 0

11 λx – 44 λ – λy + 3 λ – 3 λz + 6 λ = 0

11 λx – λy – 3 λz – 35 λ = 0

Dividing by λ,

11x – y – 3z – 35 = 0

So, equation of required plane is 11x – y – 3z – 35 = 0

we have, equation of the line is

General point on the line is given by (3 – 4, 5 – 6, – 2 + 1) …… (1)

Another equation of line is

3x – 2y + z + 5 = 0

2x + 3y + 4z – 4 = 0

Let a, b, c be the direction ratio of the line so, it will be perpendicular to normal of 3x – 2y + z + 5 = 0 and 2x + 3y + 4z – 4 = 0

So, using a1a2 + b1b2 + c1c2 = 0

(3)(a) + (– 2)(b) + (1)(c) = 0

3a – 2b + c = 0 …… (2)

Again, (2)(a) + (3)(b) + (4)(c) = 0

2a + 3b + 4c = 0 …… (3)

Solving (2) and (3) by cross – multiplication,

Direction ratios are proportional to – 11, – 10, 13

Let z = 0 so

3x – 2y = – 5 …… (i)

2x + 3y = 4 …… (ii)

Solving (i) and (ii) by eliminations method,

– 13y = – 22

Put y in equation (i)

3x – 2y = – 5

3x – 2 = – 5

3x – = – 5

3x = – 5 +

3x =

x =

so, the equation of the line (2) in symmetrical form,

Put the general point of a line from equation (1)

The equation of the plane is 45x – 17y + 25z + 53 = 0

Their point of intersection is (2, 4, – 3)

we know that plane contains the line if

Given, equation of plane and equation of line

So

d = 3

= (2)(1) + (1)(2) + (4)(– 1)

= 2 + 2 – 4

= (1)(1) + (1)(2) + (0)(– 1)

= 1 + 2 – 0

= 3

= d

Since, and

So, Given line is on the given plane.

Hence, Proved.

Let L ₁: and

L₂ : be the equations of two lines

Let the plane be ax + by + cz + d = 0 …… (1)

Given that the required plane through the intersection of the lines L₁ and L₂

Hence the normal to the plane is perpendicular to the lines L₁ and L₂

3a – 2b + 6c = 0

a – 3b + 2c = 0

Using cross multiplication, we get

Let the equation of the plane be …… (1)

Plane is passing through (3, 4, 2) and (7, 0, 6)

Required plane is perpendicular to 2x – 5y – 15 = 0

2b = 5a

Solving the above equations

a = 3.4 = , b = 8.5 = and

substituting the values in (1)

5x + 2y – 3z = 17

Vector equation of the plane is

The line passes through B(1, 3, – 2)

5(1) + 2(3)—3(– 2) = 17

The point B lies on the plane .

the line lies on the plane

The direction ratio of the line is r₁ = (– 3, – 2k, 2)

The direction ratio of the line is r₂ = (k, 1, 5)

Since the line and are perpendicular so

r₁.r₂ = 0

(– 3, – 2k, 2). (k, 1, 5) = 0

– 3k – 2k + 10 = 0

– 5k = – 10

k = 2

the equation of the line are and

The equation of the plane containing the perpendicular lines and is

(– 20 – 2)x – y(– 15 – 4) + z(– 3 + 8) = d

– 22x + 19y + 5z = d

The line pass through the point (1, 2, 3) so putting x = 1, y = 2, z = 3 in the equation – 22x + 19y + 5z = d we get

– 22(1) + 19(2) + 5(3) = d

d = – 22 + 38 + 15

d = 31

The equation of the plane containing the lines is – 22x + 19y + 5z = 31

Any point on the line is of the form,

(3k + 2, 4k – 1, 2k + 2)

If the point p(3k + 2, 4k – 1, 2k + 2) lies in the plane x – y + z – 5 = 0, we have,

(3k + 2) – (4k – 1) + (2k + 2) – 5 = 0

3k + 2 – 4k + 1 + 2k + 2 – 5 = 0

k = 0

thus, the coordinates of the point of intersection of the line and the plane are :

{3(0) + 2, 4(0) – 1, 2(0) + 2}

P(2, – 1,, 2)

Let be the angle between the line and the plane . thus

, where l, m and n are the direction ratios of the line and a, b and c are the direction ratios of the normal to the plane

Here, l = 3, m = 4, n = 2, a = 1, b = – 1, and c = 1 hence,

Let A, B and C be three point with position vector and

Thus,

As we know that cross product of two vectors gives a perpendicular vector so

So, the equation of the required plane is

Also we have to find the coordinates of the point of intersection of this plane and the line

Any point on the line is of the form, p(3 + 2, – 1 – 2, – 1 + )

Point p(3 + 2, – 1 – 2, – 1 + ) lies in the plane,

so,

9(3 + 2) – 3(– 1 – 2) – (– 1 + ) = 14

27 + 18 – 3 – 6 + 1 – = 14

11 = – 11

= – 1

Thus the required point of intersection is

p(3 + 2, – 1 – 2, – 1 + )

put value of in this equation

p[3 + 2(– 1), – 1 – 2(– 1), – 1 + (– 1)]

p(1, 1, – 2)

……(1)

……(2)

a₁ = 4, b₁ = 4, c₁ = – 5

a₂ = 7, b₂ = 1, c₂ = 3

x₁ = 5, y₁ = 7, z₁ = – 3

x₂ = 8, y₂ = 4, z₂ = 5

the condition for two line to be coplanar,

= 0

=

=

= 3(12 + 5) + 3(12 + 35) + 8(4 – 28)

= 3 × 17 + 3 × 47 + 8 × (–24)

= 51 + 141 – 192

= 192 – 192

= 0

The lines are coplanar to each other .

Given that, a plane is passes through the point (3, 2, 0) so equation will be

a(x – 3) + b(y – 2) + c(z – 0) = 0

a(x – 3) + b(y – 2) + cz = 0 ……(1)

plane also contains the line

so it pass through the point (3, 2, 0)

a(3 – 3) + b(6 – 2) + c(4) = 0

4b + 4c = 0 …… (2)

Also plane will be parallel to,

a(1) + b(5) + c(4) = 0

a + 5b + 4c = 0 ……(3)

solving (2) and (3) by cross multiplication,

a = – k, b = k, c = – k

put a = – k, b = k, c = – k in equation (i) we get

(– k)(x – 3) + (k)(y – 2) + (– k)z = 0

– x + 3 + y – 2 – z = 0

x – y + z – 1 = 0

We know that the lines are coplanar if

Here,

x₁ = –3, x₁ = –1, y₁ = 1, y₂ = 2, z₁ = 5, z₂ = 5

l₁ = –3, l₂ = –1, m₁ = 1, m₂ = 2, n₁ = 5, n₂ = 5

= 2(–5) – 1(–10) = – 10 + 10

= 0

So the given line are coplanar .

The equation of plane contains lines is

(x + 3)(5 – 10) – (y – 1)(– 15 – (– 5)) + (z – 5)(– 6 – (– 1)) = 0

– 5x – 15 + 10y – 10 – 5z + 25 = 0

– 5x + 10y – 5z = 0

Divided by – 5

x – 2y + z = 0

We know that the lines lies in plane ax + by + cz + d = 0, then

a

Here,

x₁ = 3, y₁ = –2, z₁ = –4 and l = 2, m = –1, n = 3

a = l, b = m, c = –1, d = –9

i.e, 3l + (– 2)m + (– 4)(– 1) – 9 = 0 and 2l – m – 3 = 0

3l – 2m = 5 and 2l – m = 3

3l – 2m = 5 …… (1)

2l – m = 3 ……(2)

Multiply eq.(1) by 2 and eq.(2) by 3 and then subtract we get

m = –1

l = 1

l² + m² = 2

We know that the lines are coplanar if

Here,

Let

(

Put value of t

is neglected because direction cosine can not be imaginary

λ =

We know that the lines are coplanar if

Here,

(α – 5)[(3 – α)(2 – α) – 2] = 0

(α – 5)(6 – 3α – 2α + α² – 2) = 0

(α – 1)(α – 4)(α – 5) = 0

α = 1, 4, 5

We know that the lines are coplanar if

Here,

k = ±2

The equation of plane contains lines is

when k = 2

(x – 1)(4 – 4) – (y + 1)(4 – 10) + (z)(4 – 10) = 0

6y – 6z + 6 = 0

y – z + 1 = 0

The equation of plane contains lines is

When k = – 2

(x – 1)(4 – 4) – (y + 1)(– 4 – 10) + (z)(4 + 10) = 0

14y + 14z + 14 = 0

y + z + 1 = 0

Let the two lines be l₁ and l₂.

So, and

We need to find the shortest distance between l₁ and l₂.

Recall the shortest distance between the lines: and is given by

Here, (x₁, y₁, z₁) = (2, 5, 0) and (x₂, y₂, z₂) = (0, –1, 1)

Also (a₁, b₁, c₁) = (–1, 2, 3) and (a₂, b₂, c₂) = (2, –1, 2)

We will evaluate the numerator first.

Let

⇒ N = (–2)[(2)(2) – (–1)(3)] – (–6)[(–1)(2) – (2)(3)] + (1)[(–1)(–1) – (2)(2)]

⇒ N = –2(4 + 3) + 6(–2 – 6) + (1 – 4)

⇒ N = –14 – 48 – 3

∴ N = –65

Now, we will evaluate the denominator.

Let

b₁c₂ – b₂c₁ = (2)(2) – (–1)(3) = 4 – (–3) = 7

c₁a₂ – c₂a₁ = (3)(2) – (2)(–1) = 6 – (–2) = 8

a₁b₂ – a₂b₁ = (–1)(–1) – (2)(2) = 1 – 4 = –3

So, shortest distance =

Thus, the required shortest distance is units.

Let the two lines be l₁ and l₂.

So, and

We need to find the shortest distance between l₁ and l₂.

Recall the shortest distance between the lines: and is given by

Here, (x₁, y₁, z₁) = (–1, –1, –1) and (x₂, y₂, z₂) = (3, 5, 7)

Also (a₁, b₁, c₁) = (7, –6, 1) and (a₂, b₂, c₂) = (1, –2, 1)

We will evaluate the numerator first.

Let

⇒ N = (4)[(–6)(1) – (–2)(1)] – (6)[(7)(1) – (1)(1)] + (8)[(7)(–2) – (1)(–6)]

⇒ N = 4(–6 + 2) – 6(7 – 1) + 8(–14 + 6)

⇒ N = –16 – 36 – 64

∴ N = –116

Now, we will evaluate the denominator.

Let

b₁c₂ – b₂c₁ = (–6)(1) – (–2)(1) = –6 + 2 = –4

c₁a₂ – c₂a₁ = (1)(1) – (1)(7) = 1 – 7 = –6

a₁b₂ – a₂b₁ = (7)(–2) – (1)(–6) = –14 + 6 = –8

So, shortest distance =

Thus, the required shortest distance is units.

Let the two lines be l₁ and l₂.

So, and l₂: 3x – y – 2z + 4 = 0 = 2x + y + z + 1

We need to find the shortest distance between l₁ and l₂.

The equation of a plane containing the line l₂ is given by

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

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

Direction ratios of l₁ are 2, 4, 1 and those of the line containing the shortest distance are proportional to 3 + 2λ, λ – 1 and λ – 2.

We know that if two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂) are perpendicular to each other, then a₁a₂ + b₁b₂ + c₁c₂ = 0.

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

⇒ 6 + 4λ + 4λ – 4 + λ – 2 = 0

⇒ 9λ = 0

∴ λ = 0

Thus, the plane containing line l₂ is 3x – y – 2z + 4 = 0.

We have

When α = 0, (x, y, z) = (1, 3, –2)

So, the point (1, 3, –2) lies on the line l₁.

Hence, the shortest distance between the two lines is same as the distance of the perpendicular from (1, 3, –2) on to the plane 3x – y – 2z + 4 = 0.

Recall the length of the perpendicular drawn from (x₁, y₁, z₁) to the plane Ax + By + Cz + D = 0 is given by

Here, (x₁, y₁, z₁) = (1, 3, –2) and (A, B, C, D) = (3, –1, –2, 4)

Thus, the required shortest distance is units.

Let point P = (0, 0, 0) and M be the image of P in the plane 3x + 4y – 6z + 1 = 0.

Direction ratios of PM are proportional to 3, 4, – 6 as PM is normal to the plane.

Recall the equation of the line passing through (x₁, y₁, z₁) and having direction ratios proportional to l, m, n is given by

Here, (x₁, y₁, z₁) = (0, 0, 0) and (l, m, n) = (3, 4, –6)

Hence, the equation of PM is

⇒ x = 3α, y = 4α, z = –6α

Let M = (3α, 4α, –6α).

As M is the image of P in the given plane, the midpoint of PM lies on the plane.

Using the midpoint formula, we have

This point lies on the given plane, which means this point satisfies the plane equation.

We have M = (3α, 4α, –6α)

Thus, the image of (0, 0, 0) in the plane 3x + 4y – 6z + 1 = 0 is.

Let point P = (1, 2, –1) and M be the image of P in the plane 3x – 5y + 4z = 5.

Direction ratios of PM are proportional to 3, –5, 4 as PM is normal to the plane.

Recall the equation of the line passing through (x₁, y₁, z₁) and having direction ratios proportional to l, m, n is given by

Here, (x₁, y₁, z₁) = (1, 2, –1) and (l, m, n) = (3, –5, 4)

Hence, the equation of PM is

⇒ x = 3α + 1, y = –5α + 2, z = 4α – 1

Let M = (3α + 1, –5α + 2, 4α – 1).

As M is the image of P in the given plane, the midpoint of PM lies on the plane.

Using the midpoint formula, we have

This point lies on the given plane, which means this point satisfies the plane equation.

We have M = (3α + 1, –5α + 2, 4α – 1)

Thus, the image of (1, 2, –1) in the plane 3x – 5y + 4z = 5 is.

Let point P = (5, 4, 2) and Q be the foot of the perpendicular drawn from to P the line.

Q is a point on the given line. So, for some α, Q is given by

⇒ x = 2α – 1, y = 3α + 3, z = –α + 1

Thus, Q = (2α – 1, 3α + 3, –α + 1)

Now, we find the direction ratios of PQ.

Recall the direction ratios of a line joining two points (x₁, y₁, z₁) and (x₂, y₂, z₂) are given by (x₂ – x₁, y₂ – y₁, z₂ – z₁).

Here, (x₁, y₁, z₁) = (5, 4, 2) and (x₂, y₂, z₂) = (2α – 1, 3α + 3, –α + 1)

⇒ Direction Ratios of PQ are ((2α – 1) – (5), (3α + 3) – (4), (–α + 1) – (2))

⇒ Direction Ratios of PQ are (2α – 6, 3α – 1, –α – 1)

PQ is perpendicular to the given line, whose direction ratios are (2, 3, –1).

We know that if two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂) are perpendicular to each other, then a₁a₂ + b₁b₂ + c₁c₂ = 0.

⇒ (2)(2α – 6) + (3)(3α – 1) + (–1)(–α – 1) = 0

⇒ 4α – 12 + 9α – 3 + α + 1 = 0

⇒ 14α – 14 = 0

⇒ 14α = 14

∴ α = 1

We have Q = (2α – 1, 3α + 3, –α + 1)

⇒ Q = (2×1 – 1, 3×1 + 3, –1 + 1)

∴ Q = (1, 6, 0)

Using the distance formula, we have

Thus, the required foot of perpendicular is (1, 6, 0) and the length of the perpendicular is units.

Let P be the point with position vector and M be the image of P in the plane.

In addition, let Q be the foot of the perpendicular from P on to the given plane. So, Q is the midpoint of PM.

Direction ratios of PM are proportional to 2, –1, 1 as PM is normal to the plane and parallel to.

Recall the vector equation of the line passing through the point with position vector and parallel to vector is given by

Here, and

Hence, the equation of PM is

Let the position vector of M be. As M is a point on this line, for some scalar α, we have

Now, let us find the position vector of Q, the midpoint of PM.

Let this be.

Using the midpoint formula, we have

This point lies on the given plane, which means this point satisfies the plane equation.

We have the image

Therefore, image is (1, 2, 1)

Foot of the perpendicular

Thus, the position vector of the image is and that of the foot of perpendicular is.

Let point P = (1, 1, 2) and Q be the foot of the perpendicular drawn from P to the plane 2x – 2y + 4z + 5 = 0.

Direction ratios of PQ are proportional to 2, –2, 4 as PQ is normal to the plane.

Recall the equation of the line passing through (x₁, y₁, z₁) and having direction ratios proportional to l, m, n is given by

Here, (x₁, y₁, z₁) = (1, 1, 2) and (l, m, n) = (2, –2, 4)

Hence, the equation of PQ is

⇒ x = 2α + 1, y = –2α + 1, z = 4α + 2

Let Q = (2α + 1, –2α + 1, 4α + 2).

This point lies on the given plane, which means this point satisfies the plane equation.

⇒ 2(2α + 1) – 2(–2α + 1) + 4(4α + 2) + 5 = 0

⇒ 4α + 2 + 4α – 2 + 16α + 8 + 5 = 0

⇒ 24α + 13 = 0

⇒ 24α = –13

We have Q = (2α + 1, –2α + 1, 4α + 2)

Recall the length of the perpendicular drawn from (x₁, y₁, z₁) to the plane Ax + By + Cz + D = 0 is given by

Here, (x₁, y₁, z₁) = (1, 1, 2) and (A, B, C, D) = (2, –2, 4, 5)

Thus, the required foot of perpendicular is and the length of the perpendicular is units.

Let point P = (1, –2, 3).

We need to find distance from P to the plane x – y + z = 5 measured along a line parallel to.

Let the line drawn from P parallel to the given line meet the plane at Q.

Direction ratios of PQ are proportional to 2, 3, –6 as PQ is parallel to the given line.

Recall the equation of the line passing through (x₁, y₁, z₁) and having direction ratios proportional to l, m, n is given by

Here, (x₁, y₁, z₁) = (1, –2, 3) and (l, m, n) = (2, 3, –6)

Hence, the equation of PQ is

⇒ x = 2α + 1, y = 3α – 2, z = –6α + 3

Let Q = (2α + 1, 3α – 2, –6α + 3).

This point lies on the given plane, which means this point satisfies the plane equation.

⇒ (2α + 1) – (3α – 2) + (–6α + 3) = 5

⇒ 2α + 1 – 3α + 2 – 6α + 3 = 5

⇒ –7α + 6 = 5

⇒ –7α = –1

We have Q = (2α + 1, 3α – 2, –6α + 3)

Using the distance formula, we have

Thus, the required distance is 1 unit.

Let point P = (2, 3, 7) and Q be the foot of the perpendicular drawn from P to the plane 3x – y – z = 7.

Direction ratios of PQ are proportional to 3, –1, –1 as PQ is normal to the plane.

Recall the equation of the line passing through (x₁, y₁, z₁) and having direction ratios proportional to l, m, n is given by

Here, (x₁, y₁, z₁) = (2, 3, 7) and (l, m, n) = (3, –1, –1)

Hence, the equation of PQ is

⇒ x = 3α + 2, y = 3 – α, z = 7 – α

Let Q = (3α + 2, 3 – α, 7 – α).

This point lies on the given plane, which means this point satisfies the plane equation.

⇒ 3(3α + 2) – (3 – α) – (7 – α) = 7

⇒ 9α + 6 – 3 + α – 7 + α = 7

⇒ 11α – 4 = 7

⇒ 11α = 11

∴ α = 1

We have Q = (3α + 2, 3 – α, 7 – α)

⇒ Q = (3×1 + 2, 3 – 1, 7 – 1)

∴ Q = (5, 2, 6)

Using the distance formula, we have

Thus, the required foot of perpendicular is (5, 2, 6) and the length of the perpendicular is units.

Let point P = (1, 3, 4) and M be the image of P in the plane 2x – y + z + 3 = 0.

Direction ratios of PM are proportional to 2, –1, 1 as PM is normal to the plane.

Recall the equation of the line passing through (x₁, y₁, z₁) and having direction ratios proportional to l, m, n is given by

Here, (x₁, y₁, z₁) = (1, 3, 4) and (l, m, n) = (2, –1, 1)

Hence, the equation of PM is

⇒ x = 2α + 1, y = 3 – α, z = α + 4

Let M = (2α + 1, 3 – α, α + 4).

As M is the image of P in the given plane, the midpoint of PM lies on the plane.

Using the midpoint formula, we have

This point lies on the given plane, which means this point satisfies the plane equation.

We have M = (2α + 1, 3 – α, α + 4)

⇒ M = (2(–2) + 1, 3 – (–2), (–2) + 4)

∴ M = (–3, 5, 2)

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

Let P be the point with position vector and Q be the point of intersection of the given line and the plane.

We have the line equation as

Let the position vector of Q be. As Q is a point on this line, for some scalar α, we have

This point Q also lies on the given plane, which means this point satisfies the plane equation.

⇒ (2 + 3α)(1) + (–1 + 4α)(–1) + (2 + 12α)(1) = 5

⇒ 2 + 3α + 1 – 4α + 2 + 12α = 5

⇒ 11α + 5 = 5

⇒ 11α = 0

∴ α = 0

We have

Using the distance formula, we have

Thus, the required distance is 13 units.

Let point P = (1, 1, 2) and Q be the foot of the perpendicular drawn from to P the plane.

Direction ratios of PQ are proportional to 1, –2, 4 as PQ is normal to the plane and parallel to.

Recall the vector equation of the line passing through the point with position vector and parallel to vector is given by

Here, and

Hence, the equation of PQ is

Let the position vector of Q be. As Q is a point on this line, for some scalar α, we have

This point lies on the given plane, which means this point satisfies the plane equation.

⇒ (1 + α)(1) + (1 – 2α)(–2) + (2 + 4α)(4) = –5

⇒ 1 + α – 2 + 4α + 8 + 16α = –5

⇒ 21α + 7 = –5

⇒ 21α = –12

Foot of the perpendicular

Thus,

Using the distance formula, we have

Thus, the required foot of perpendicular is and the length of the perpendicular is units.

Let point P = (3, 2, 1) and M be the image of P in the plane 2x – y + z + 1 = 0.

In addition, let Q be the foot of the perpendicular from P on to the given plane so that Q is the midpoint of PM.

Direction ratios of PM are proportional to 2, –1, 1 as PM is normal to the plane.

Recall the equation of the line passing through (x₁, y₁, z₁) and having direction ratios proportional to l, m, n is given by

Here, (x₁, y₁, z₁) = (3, 2, 1) and (l, m, n) = (2, –1, 1)

Hence, the equation of PM is

⇒ x = 2α + 3, y = 2 – α, z = α + 1

Let M = (2α + 3, 2 – α, α + 1).

Now, we will find Q, the midpoint of PM.

Using the midpoint formula, we have

This point Q lies on the given plane, which means Q satisfies the plane equation 2x – y + z + 1 = 0.

We have M = (2α + 3, 2 – α, α + 1)

⇒ M = (2(–2) + 3, 2 – (–2), (–2) + 1)

∴ M = (–1, 4, –1)

We have

∴ Q = (1, 3, 0)

Using the distance formula, we have

Thus, the required foot of perpendicular is (1, 3, 0) and the length of the perpendicular is units. Also, the image of the given point is (–1, 4, –1)

The given plane equation is

Now, we calculate the magnitude of the vector.

On dividing both sides of the plane equation by 7, we get

Recall that the equation of the plane in normal form is given by where is a unit vector perpendicular to the plane through the origin.

So, here

This is a unit vector normal to the plane.

Thus, the direction cosines of the unit vector perpendicular to the given plane are.

Let point P = (0, 0, 0) and Q be the foot of the perpendicular drawn from P to the plane 2x – 3y + 4z – 6 = 0.

Direction ratios of PQ are proportional to 2, –3, 4 as PQ is normal to the plane.

Recall the equation of the line passing through (x₁, y₁, z₁) and having direction ratios proportional to l, m, n is given by

Here, (x₁, y₁, z₁) = (0, 0, 0) and (l, m, n) = (2, –3, 4)

Hence, the equation of PQ is

⇒ x = 2α, y = –3α, z = 4α

Let Q = (2α, –3α, 4α).

This point lies on the given plane, which means this point satisfies the plane equation.

⇒ 2(2α) – 3(–3α) + 4(4α) – 6 = 0

⇒ 4α + 9α + 16α – 6 = 0

⇒ 29α = 6

We have Q = (2α, –3α, 4α)

Thus, the required foot of perpendicular is.

Let point P = (1, 3/2, 2) and Q be the foot of the perpendicular drawn from P to the plane 2x – 2y + 4z + 5 = 0.

Direction ratios of PQ are proportional to 2, –2, 4 as PQ is normal to the plane.

Recall the equation of the line passing through (x₁, y₁, z₁) and having direction ratios proportional to l, m, n is given by

Here, (x₁, y₁, z₁) = (1,, 2) and (l, m, n) = (2, –2, 4)

Hence, the equation of PQ is

Let.

This point lies on the given plane, which means this point satisfies the plane equation.

⇒ 4α + 2 – (–4α + 3) + 16α + 8 + 5 = 0

⇒ 20α + 4α – 3 + 15 = 0

⇒ 24α = –12

We have

Using the distance formula, we have

Thus, the required foot of perpendicular is and the length of the perpendicular is units.

Let the position vector of P be so that and M be the image of P in the plane.

In addition, let Q be the foot of the perpendicular from P on to the given plane so that Q is the midpoint of PM.

Direction ratios of PM are proportional to 2, 1, 3 as PM is normal to the plane and parallel to.

Recall the vector equation of the line passing through the point with position vector and parallel to vector is given by

Here, and

Hence, the equation of PM is

Let the position vector of M be. As M is a point on this line, for some scalar α, we have

Now, let us find the position vector of Q, the midpoint of PM.

Let this be.

Using the midpoint formula, we have

This point lies on the given plane, which means this point satisfies the plane equation.

We have the image

Foot of the perpendicular

Using the distance formula, we have

Thus, the position vector of the image of the given point is and that of the foot of perpendicular is. Also, the length of this perpendicular is units.

Equation of XOY plane is z=0.

Since the required plane should pass through the point (2, - 3, 5).

We know, the vector equation of a plane perpendicular to a given direction and passing through a given point() is given by,

i.e.

⟹ (z - 5)=0

⟹ z=5

hereis given byas it is perpendicular to XOY plane, and here

Hence, the desired equation for the plane is z=5.

Equation of YOZ plane is x=0.

Since the required plane should pass through the point ( - 4, 1, 0).

We know, the vector equation of a plane perpendicular to a given direction and passing through a given point() is given by,

i.e.

⟹ (x + 4)=0

⟹ x= - 4

hereis given by as it is perpendicular to YOZ plane, and here

Hence, the desired equation for the plane is x= - 4.

We know that the general equation of a plane is given by,

Ax + By + Cz + D=0, where ……… (1)

Here, A, B, C are the co - ordinates of a normal vector to the plane, while (x, y, z) are the co - ordinates of any general point through which the plane passes.

Now let us say, this plane is making intercepts at points P, Q, and R on the x, y, and z - axes respectively at (a, 0, 0), (0, b, 0) and (0, 0, c).

So, the plane cuts the x - axis, y - axis and z - axis at three points P(a, 0, 0), Q(0, b, 0) and R(0, 0, c) respectively.

Since the plane also passes through each of these three points, we can substitute them into equation (1) i.e. general equation of the plane and we have,

(i) Aa + D=0

(ii) Bb + D=0

(iii) Cc + D=0

Substituting these values of A, B, and C in equation (1) of the plane, we shall get the equation of a plane in intercept form, which is given by,

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

The required plane is parallel to X - axis i.e. the normal of the plane is perpendicular to X - axis so, the component of the normal vector along X - axis is zero (0).

We know that the general equation of a plane is given by,

Ax + By + Cz + D=0, where ……… (1)

Here, A, B, C are the coordinates of a normal vector to the plane, while (x, y, z) are the co - ordinates of any point through which the plane passes.

Putting A=0 [∵ the component of the normal vector along X - axis is zero (0)] in the general equation i.e. in equation (1) of plane we get,

By + Cz + D=0, where ……… (2)

Hence, By + Cz + D=0 is the general equation of a plane parallel to X - axis.

Equation of the first plane is given as,

x–2y + kz=4 …………… (1)

and the equation of the second plane is given as,

2x + 5y–z=9 ……………… (2)

So, the normal vector of plane (1) is given by,

Similarly, the normal vector of plane (2) is given by,

When the two planes are perpendicular to each other, we should have,

2 - 10 - k=0

k= - 8.

Hence, the planes x–2y + kz=4 and 2x + 5y–z=9 will be perpendicular to each other if k= - 8.

We know, that the general equation of a plane is given by,

Ax + By + Cz + D=0, where ……… (1)

Here, A, B, C are the coordinates of a normal vector to the plane, while (x, y, z) are the co - ordinates of any point through which the plane passes.

Again, we know the intercept form of plane, which is given by,

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

The equation of the plane is given as,

2x–3y + 4z=12

i.e. 2x–3y + 4z - 12=0 ………………… (2)

Comparing equation (2) with in the general equation i.e. in equation (1) of plane we get,

A=2, B= - 3 and C=4 and D= - 12.

=6,

=−4

The given plane (given by equation (2)) makes intercepts at (6, 0, 0), (0, - 4, 0) and (0, 0, 3) with the x - , y - and z - axes respectively.

We know that, the ratio in which the plane Ax + By + Cz + D=0 (where ) divides the line segment joining (x₁, y₁, z₁) and (x₂, y₂, z₂) then is given as,

Here, the equation of the given plane is, 4x + 5y–3z=8 i.e. 4x + 5y–3z - 8=0 and the co - ordinates of the two points are (–2, 1, 5) and (3, 3, 2).

Comparing with the general formula, we get,

A=4, B=5, C= - 3, D= - 8, x₁= - 2, y₁=1, z₁=5and x₂=3, y₂=3 and z₂=2.

So, the required ratio is

Hence, the plane 4x + 5y–3z=8 divides the line segment joining points (–2, 1, 5) and (3, 3, 2) in 2:1 ratio.

We know that, distance between two parallel planes:

Ax + By + Cz + D₁=0…… (1) and Ax + By + Cz + D₂=0…… (2) is given by,

Here, the two parallel planes are given as,

2x–y + 3z=4 i.e. 2x–y + 3z - 4=0 …………… (3)

and 2x–y + 3z=18 i.e. 2x–y + 3z - 18=0 ………… (4)

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

A=2, B= - 1, C=3, D₁= - 4 and D₂= - 18.

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

Hence, the distance between the parallel planes 2x–y + 3z=4 and 2x–y + 3z=18 is .

The plane is given as, .

We can write the equation of the plane in general form as,

2x + 3y - 6z - 14=0 …………… (1)

Now, to get the normal form of a plane given in general form as, Ax + By + Cz + D=0 where …… (2), we have to divide the equation (1) by , where is the normal vector given as,

Now,

Comparing equation (1) with equation (2), we get, D= - 14

=2

[where p is the distance between the plane and the origin]

Normal form of the equation is given as,

Here, normal form of the given plane is,

The plane is given as, .

We can write the equation of the plane in general form as,

2x - y + 2z - 12=0 …………… (1)

Now, to get the normal form of a plane given in general form as, Ax + By + Cz + D=0 where …… (2), we have to divide the equation (1) by , where is the normal vector given as,

Now,

Comparing equation (1) with equation (2), we get, D= - 12

So, the distance between the plane and the origin (using formula)

p=4

The distance between the plane and the origin is 4 units.

The given plane is .

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

So, 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,

.

Hence, the equation of the plane in scalar product form is given as, or, .

We have the plane as,

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

Hence, a vector normal to the plane is

The required plane is parallel to 3x + 2y - z=7, so required plane and the given plane must have the same normal vector.

Vector normal to the plane 3x + 2y - z=7 is

We know that, equation of plane perpendicular to a given direction & passing through a given point is given by,

Here, it is given that, the plane passes through (2, - 1, 1) so in this case, in vector form, can be denoted as,

Equation of the required plane is,

3(x - 2) + 2(y + 1) - (z - 1)=0

3x - 6 + 2y + 2 - z + 1=0

3x + 2y - z=3

Hence, the equation of the plane passing through (2, –1, 1) and parallel to the plane 3x + 2y–z=7 is 3x + 2y - z=3.

The, required plane should contain the lines, and

So, it is clear from the given equation of lines, that both the lines are passing through a point and one of the line is parallel to and the other one is parallel to .

So, 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,

.

Hence, the equation of the plane containing the lines and is given as, or,

Let, the the position vector of the point where the line meets the plane be .

As is the position vector of the point of intersection of the line and the plane, so it must satisfy both of the equation of line and the equation of plane.

Substituting, in place of in both the equations, we

get,

……………………… (1), and

…………………….. (2)

Putting in equation (2) we get,

Substituting this value of λ in equation (1) we get,

The position vector of the point where the line meets the plane is .

Equation of the line in Cartesian form is given as,

So, the direction cosines of the line are given as,

The equation of the plane is, so, we have the vector normal to the plane as,

It is required that, the line should be perpendicular to the normal to the plane , so, we should have,

(2⨯2) + (3⨯3) + (k⨯4)=0

4 + 9 + 4k=0

13 + 4k=0

4k= - 13

Hence, for the line will be perpendicular to the normal to the plane

.

We know, the angle between the line and the plane Ax + By + Cz + D=0 is given as,

In this case, l=2, m=1, n= - 2, A=1, B=1, C=0 and D=4.

Putting these values in equation (1) we get,

=45°

Hence, the angle between the line and the plane x + y + 4=0 is 45̊

We know, that the general equation of a plane is given by,

Ax + By + Cz + D=0, where ……… (1)

Here, A, B, C are the coordinates of a normal vector to the plane, while (x, y, z) are the co - ordinates of any point through which the plane passes.

Again, we know the intercept form of plane which is given by,

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

The equation of the plane is given as,

2x + y - z=5

i.e. 2x + y - z - 5=0 ………………… (2)

Comparing equation (2) with in the general equation i.e. in equation (1) of plane we get,

A=2, B=1 and C= - 1 and D= - 5.

=2.5

Hence, the intercept cut off by the plane 2x + y–z=5 on x - axis is of 2.5 units.

We know the distance of a point (x₀, y₀, z₀) from a plane Ax + By + Cz + D=0 …………… (1) is

On comparing, the equation of the given plane i.e.

2x - 3y + 6z + 21=0 with equation (1) we get,

A=2, B= - 3, C=6, D=21.

Again, we know that, the co - ordinates of the origin are

(0, 0, 0).

So, the length of the perpendicular drawn from the origin is

=3

Hence, the length of the perpendicular drawn from the origin to the plane 2x–3y + 6z + 21=0 is = 3 units.

Equation of the given plane is,

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

As the required line should be normal to the given plane, so, the line should be parallel to the normal vector i.e. .

The line should pass through the point (1, –2, –3).

We can write the position vector of the point as,

So, the vector equation of the line passing through the point (1, –2, –3) and normal to the plane i.e. parallel to is given by,

, where is a scalar(constant).

Hence, equation of the required line is

where, is a scalar(constant)

The required plane is parallel to , so required plane and the given plane must have the same normal vector.

Vector normal to the plane is

The required plane is passing through a given point

(a, b, c), so can write the position vector of the point as

Now, the equation of the required plane is given by,

(x - a) + (y - b) + (z - c)=0

x + y + z - (a + b + c)=0

x + y + z = a + b + c

Hence, the equation of the plane passing through (a, b, c) and parallel to the plane is i.e. (in vector form), or, in general form x + y + z = a + b + c.

From the given vector normal to the required plane, we can write the equation of the plane as,

[where, d is a constant]

…………………… (1)

We know, that the distance of a point (x₀, y₀, z₀) from a plane Ax + By + Cz + D=0 …………… (2) is

On comparing, equation (1) i.e. 2x - 3y + 6z + D=0 with

equation (2) we get,

A=2, B= - 3, C=6, D= - d.

Again, we know that, the co - ordinates of the origin are

(0, 0, 0).

So, the length of the perpendicular drawn from the origin is

Here, it is given that, the plane is at a distance of 5 units from the origin, so, we have,

|D|=35

D=±35

∴d=± 35 [∵ D= - d]

Hence, the vector equation of a plane which is at a distance of 5 units from the origin and whose normal vector is is, 2x - 3y + 6z - ( - 35)=0 i.e. 2x - 3y + 6z + 35=0 or 2x - 3y + 6z - 35=0.

Hence, required equation of the plane, is i.e. 2x - 3y + 6z + 35=0 or, i.e. 2x - 3y + 6z - 35=0.

Given, the plane is at a distance of units form the origin and the normal to the plane is equally inclined with the co - ordinates axis, so, its direction cosines are

We know, for a plane having direction cosines as l, m and n, and p be the distance of the plane from the origin, the equation of the plane is given as, lx + my + nz=p

So, in this problem, the equation of the required equation of the plane is given by,

=15

Hence, the equation of the required plane which is at a distance of units from the origin and the normal to which is equally inclined to co - ordinate axes is

x + y + z=15.

The given equation plane is,

We can rewrite the equation of the given plane as,

2x–(1 + λ)y + 3λz=0

2x - y - λ(y - 3z)=0

So, the given plane passes through the intersection of

the planes 2x - y=0 and y - 3z=0.

We know, the angle between two planes,

a₁x + b₁y + c₁z + d₁=0 and a₂x + b₂y + c₂z + d₂=0 is,

Here, a₁=2, b₁= - 1, c₁=1, d₁= - 6 and a₂=1, b₂=1, c₂=2,

d₂= - 3.

So, the acute angle between the planes 2x–y + z=6 and

x + y + 2z = 3 is

θ=60 ̊

the acute angle between the planes 2x–y + z=6 and

x + y + 2z = 3 is 60 ̊.

The equation of the plane through the intersection of

the planes x + 2y + 3z=4 or, x + 2y + 3z - 4=0 and

2x + y–z=–5 or, 2x + y–z + 5=0 is given as,

(x + 2y + 3z - 4) + λ(2x + y–z + 5)=0

[where λ is a scalar]

x(1 + 2λ) + y(2 + λ) + z(3 - λ) - 4 + 5λ=0

Given, that the required plane is perpendicular to the plane 5x + 3y + 6z + 8=0 so, we should have,

5(1 + 2λ) + 3(2 + λ) + 6(3 - λ)=0

5 + 10λ + 6 + 3λ + 18 - 6λ=0

29 + 7λ=0

Therefore, the equation of the required plane is,

7(x + 2y + 3z - 4) - 29(2x + y–z + 5)=0

7x + 14y + 21z - 28 - 58x - 29y + 29z - 145=0

- 51x - 15y + 50z - 173=0

51x + 15y - 50z + 173=0

We know that, distance between two parallel planes:

Ax + By + Cz + D₁=0…… (1) and Ax + By + Cz + D₂=0…… (2) is given by,

Here, the two parallel planes are given as,

2x + 2y - z + 2=0 …………… (3)

and 4x + 4y - 2z + 5=0 i.e. 2x + 2y - z + =0 ………… (4)

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

A=2, B=2, C= - 1, D₁=2 and D₂=.

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

Hence, the distance between the parallel planes 2x + 2y - z + 2=0 and 4x + 4y - 6z + = is .

We know, if the image of a point P (x₀, y₀, z₀) on a plane Ax + By + Cz + D=0……… (1) is Q (x₁, y₁, z₁) then,

The given plane is, 2x–y + z + 3=0 …………. (2)

Comparing equation (2) with equation (1) we get,

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

And, here x₀=1, y₀=3, z₀=4

So,

x_1=(2⨯( - 2)) + 1

= - 4 + 1

x_1= - 3,

y_1=(( - 1)⨯( - 2)) + 3

=2 + 3

y_1=5 and

z_1=(1⨯( - 2)) + 4

= - 2 + 4

z_1=2

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

We know, the two lines given as,

will be co - planar if

Here, x₁=1, y₁= - 1, z₁=0, x₂=0, y _�2=2, z₂= - 1 and, A₁=2, B �₁= - 1, C₁=3, A₂= - 2, B �₂= - 3, C₂= - 1.

=[{(( - 1)×( - 1)×( - 1)) + (3×3×( - 2)) + (( - 1)×2×( - 3))} - {(( - 1)×( - 1)×( - 2)) + (3×2×( - 1)) + (( - 1)×( - 3)×3)}]

=[{( - 1) - 18 + 6} - {( - 2) + ( - 6) + 9}]

= - 13 - 1

= - 14

≠ 0

Hence, the two lines are not co - planar.

The given plane is .

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

So, 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,

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

.

We have the, straight line given as,

and the plane as,

i.e. x - 5y + z=5 x - 5y + z - 5=0

Let us, check whether the plane and the straight line are parallel using the scalar product between the governing vector of the straight line, , and the normal vector of the plane given as, . If the straight line and the plane are parallel the scalar product will be zero.

=1 - 5 + 4

=0

From the given equation of the line, it is clear that, (2, - 2, 3) is a point 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 a 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 from point (2, - 2, 3) to the plane

Equation of line passing through the line x + y + z + 3=0 and 2x–y + 3z + 1=0 is given by,

(x + y + z + 3) + k(2x–y + 3z + 1)=0 …………………….(1)

x(1 + 2k) + y(1 - k) + z(1 + 3k) + 3 + k=0 [k is a constant]

Again, the required plane is parallel to the line

So, we should have,

[1×(1 + 2k)] + [2×(1 - k)] + [3×(1 + 3k)]=0

1 + 2k + 2 - 2k + 3 + 9k=0

9k= - 6

Putting in equation (1) we get,

3(x + y + z + 3) - 2(2x–y + 3z + 1)=0

3x + 3y + 3z + 9 - 4x + 2y - 6z - 2=0

- x + 5y - 3z + 7=0

x - 5y + 3z - 7=0

x - 5y + 3z=7

∴The equation of the plane through the line x + y + z + 3 = 0 = 2x – y + 3z + 1 and parallel to the line

is x - 5y + 3z=7.

The plane contains the line and the point

As, the plane contains the line,

so, the plane contains the point also.

On putting λ=1, we get another point on the plane which is i.e.

So, we got three points on the plane, they are, , and

Let,

and

So, and

Now, the normal of these two vectors i.e. and is,

The general equation of plane is,

7(x - 1) + 21(z - 3)=0

7x - 7 + 21z - 63=0

7x + 21z=70

x + 3z=10

or,

Hence, the vector equation of the plane containing the line and the point is .

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

Let, the co - ordinates of the point A (α, 0, 0), B (0, β, 0) and C (0, 0, γ).

According to the centroid formula,

We know the intercept form of a plane is given as,

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

Here, p=α=3a, q=β=3b and r=γ=3c

So, the equation of the plane is,

Equation of the given plane is

On comparing, we get, k=3.

Let, the point of intersection of the line

and the plane x + y + z=17 be (x₀, y₀, z₀).

As (x₀, y₀, z₀) is the point of intersection of the line and

the plane, so it must satisfy both of the equation of

line and the equation of plane.

Substituting, (x₀, y₀, z₀) in place of (x, y, z) in both the equations, we get,

i.e. x₀=k + 3,

y₀=2k + 4 and

z₀=2k + 5

Putting this values in the equation of plane we get,

x₀ + y₀ + z₀=17

(k + 3) + (2k + 4) + (2k + 5)=17

5k + 12=17

5k=5

k=1

∴ x₀=k + 3

=1 + 3

=4

y₀=2k + 4

=(2×1) + 4

=6

z₀=2k + 5

=(2×1) + 5

=7

Hence, the point of intersection is, (4, 6, 7).

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

Hence, the distance between the point (3, 4, 5) the point where the line meets the plane x + y + z = 17, is 3 units.

The two planes are, and

The line of intersection of planes and

is parallel to

The line of intersection of planes and

is parallel to

Alternative:

The two planes are, and

or, 3x - y + z=1 and x - 4y - 2z=2

Putting, z=k, we get,

3x - y + k=1 ………………………. (1)

and x - 4y - 2k=2 …………………… (2)

Multiplying equation (2) by 3 and then subtracting equation (1) from it, we get,

3(x - 4y - 2k) - (3x - y + k)=(3×2) - 1

3x - 12y - 6k - 3x + y - k=6 - 1

- 11y - 7k=5

Substituting y, in equation (1) we get,