Three Dimensional Geometry

Let the direction cosines of the line making ∠ α with x-axis, β – with y axis and γ- with z axis are l, m and n

⇒ l = cos α, m = cos β and n = cos γ

Here α = 90°, β = 135° and γ = 45°

So direction cosines are

l = cos 90° = 0

m = cos 135°= cos (180° - 45°) = -cos 45° =

n = cos 45° =

⇒ Direction cosines of the line

Let the direction cosines of the line making ∠ α with x-axis, β – with y axis and γ- with z axis are l, m and n

⇒ l = cos α, m = cos β and n = cos γ

Here given α = β = γ (line makes equal angles with the coordinate axes) ……….1

Direction Cosines are

⇒ l = cos α, m = cos β and n = cos γ

We have

l² + m ² + n² = 1

cos² α + cos²β + cos²γ = 1

From 1 we have

cos² α + cos² α + cos² α = 1

3 cos² α = 1

The direction cosines are

If the direction ratios of the line are a, b and c

Then the direction cosines are

Given direction ratios are – 18, 12 and – 4

⇒ a= -18, b = 12 and c= -4

=

=

Direction cosines are

If the direction ratios of two lines segments are proportional, then the lines are collinear.

Given A(2, 3, 4), B(−1, −2, 1), C(5, 8, 7)

Direction ratio of line joining A (2,3,4) and B (−1, −2, 1), are

(−1−2), (−2−3), (1−4)

= (−3, −5, −3)

So a₁ = -3, b₁ = -5, c₁ = -3

Direction ratio of line joining B (−1, −2, 1) and C (5, 8, 7) are

(5− (−1)), (8−(−2)), (7−1)

= (6, 10, 6)

So a₂ = 6, b₂ = 10 and c₂ =6

It is clear that the direction ratios of AB and BC are of same proportions

As

and

Therefore A, B, C are collinear.

The direction cosines of the two points passing through A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is given by

(x₂ – x₁), (y₂-y₁), (z₂-z₁)

And the direction cosines of the line AB is

Where AB =

We know that

If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of two lines; and θ is the acute angle between the two lines; then cos θ = |l₁l₂ + m₁m₂ + n₁n₂|

If two lines are perpendicular, then the angle between the two is θ = 90°

⇒ For perpendicular lines, | l₁l₂ + m₁m₂ + n₁n₂ | = cos 90° = 0, i.e.

| l₁l₂ + m₁m₂ + n₁n₂ | = 0

So, in order to check if the three lines are mutually perpendicular, we compute | l₁l₂ + m₁m₂ + n₁n₂ | for all the pairs of the three lines.

Now let the direction cosines of L₁, L₂ and L₃ be l₁, m₁, n₁; l₂, m₂, n₂ and l₃, m₃, n₃.

First, consider

⇒

⇒

⇒ L₁⊥ L₂ ……(i)

Next, consider

⇒

⇒

⇒ L₂⊥ L₃ …(ii)

Now, consider

⇒

⇒

⇒ L₁⊥ L₃ …(iii)

∴ By (i), (ii) and (iii), we have

L₁, L₂ and L₃ are mutually perpendicular.

We know that

Two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular if the angle between them is θ = 90°, i.e. a₁a₂ + b₁b₂ + c₁c₂ = 0

Also, we know that the direction ratios of the line segment joining (x₁, y₁, z₁) and (x₂, y₂, z₂) is taken as x₂ – x₁, y₂ – y₁, z₂ – z₁ (or x₁ – x₂, y₁ – y₂, z₁ – z₂).

⇒ The direction ratios of the line through the points (1, –1, 2) and (3, 4, –2) is:

a₁ = 3 – 1 = 2, b₁ = 4 – (-1) = 4 + 1 = 5, c₁ = -2 –2 = -4

and the direction ratios of the line through the points (0, 3, 2) and (3, 5, 6) is:
a₂ = 3 – 0 = 3, b₂ = 5 – 3 = 2, c₂ = 6 – 2 = 4

Now, consider

a₁a₂ + b₁b₂ + c₁c₂ = 2 × 3 + 5 × 2 + (-4) × 4 = 6 + 10 + (-16) = 16 + (-16) = 0

⇒ The line through the points (1, –1, 2), (3, 4, –2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

We know that

Two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are parallel if the angle between them is θ = 0°, i. e.

⇒

Also, we know that the direction ratios of the line segment joining (x₁, y₁, z₁) and (x₂, y₂, z₂) is taken as x₂ – x₁, y₂ – y₁, z₂ – z₁ (or x₁ – x₂, y₁ – y₂, z₁ – z₂).

⇒ The direction ratios of the line through the points (4, 7, 8) and (2, 3, 4) is:

a₁ = 2 – 4 = -2, b₁ = 3 – 7 = -4, c₁ = 4 – 8 = -4

And the direction ratios of the line through the points (– 1, – 2, 1) and (1, 2, 5) is:

a₂ = 1 – (-1) = 1 + 1 = 2, b₂ = 2 – (-2) = 2 + 2 = 4, c₂ = 5 – 1 = 4

Consider

⇒

∴ The line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (–1, –2, 1), (1, 2, 5).

We know that

Vector equation of a line that passes through a given point whose position vector is and parallel to a given vector is .

So, here the position vector of the point (1, 2, 3) is given by and the parallel vector is .

∴ The vector equation of the required line is:

, where is a real number.

We know that

Vector equation of a line that passes through a given point whose position vector is and parallel to a given vector is .

Here, and

⇒ The vector equation of the required line is:

⇒

Also, we know that

The Cartesian equation of a line through a point (x₁, y₁, z₁) and having direction cosines l, m, n is .

Also, we know that if the direction ratios of the line are a, b, c, then

⇒

⇒ The Cartesian equation of a line through a point (x₁, y₁, z₁) and having direction ratios a, b, c is .

Here, x₁ = 2, y₁ = -1, z₁ = 4 and a = 1, b = 2, c = -1

⇒ The Cartesian equation of the required line is:

⇒

We know that

The Cartesian equation of a line through a point (x₁, y₁, z₁) and having direction ratios a, b, c is .

Here, The point (x₁, y₁, z₁) is (-2, 4, -5) and the direction ratios are:

a = 3, b = 5, c = 6

⇒ The Cartesian equation of the required line is:

⇒

We know that

The Cartesian equation of a line through a point (x₁, y₁, z₁) and having direction cosines l, m, n is .

Comparing this standard form with the given equation, we get

x₁ = 5, y₁ = -4, z₁ = 6 and l = 3, m = 7, n = 2

⇒ The point through which the line passes has the position vector and the vector parallel to the line is given by .

Now, ∵ Vector equation of a line that passes through a given point whose position vector is and parallel to a given vector is .

∴ The vector equation of the required line is:

⇒

We know that

The vector equation of as line which passes through two points whose position vectors are and is .

Here, the position vectors of the two points (0, 0, 0) and (5, -2, 3) are and , respectively.

So, The vector equation of the required line is:

⇒

⇒

Now, we know that

Cartesian equation of a line that passes through two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is

So, the Cartesian equation of the line that passes through the origin (0, 0, 0) and (5, – 2, 3) is

We know that

The vector equation of as line which passes through two points whose position vectors are and is .

Here, the position vectors of the two points (3, –2, –5) and (3, –2, 6) are and , respectively.

So, the vector equation of the required line is:

⇒

⇒

⇒

Now, we also know that

Cartesian equation of a line that passes through two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is

So, the Cartesian equation of the line that passes through the origin (3, -2, -5) and (3, -2, 6) is

We know that

If θ is the acute angle between and , then

……(i)

(i) and

Here and

So, from (i), we have

……(ii)

⇒

⇒

And

Now,

⇒

⇒ By (ii), we have

⇒

⇒

(ii) and

Here, and

So, from (i), we have

…(iii)

⇒

⇒

And

Now,

⇒

⇒ By (iii), we have

⇒

⇒

We know that

If and are the equations of two lines, then the acute angle between the two lines is given by

cos θ = | l₁l₂ + m₁m₂ + n₁n₂ | ……(i)

(i) and

Here, a₁ = 2, b₁ = 5, c₁ = -3 and a₂ = -1, b₂ = 8, c₂ = 4

Now, ……(ii)

Here,

And

So, from (ii), we have

⇒

And

∴ From (i), we have

⇒

⇒

(ii) and

Here, a₁ = 2, b₁ = 2, c₁ = 1 and a₂ = 4, b₂ = 1, c₂ = 8

Here,

And

So, from (ii), we have

⇒

And

∴ From (i), we have

⇒

⇒

For any two lines to be at right angles, the angle between them should be θ = 90°.

⇒ a₁a₂ + b₁b₂ + c₁c₂ = 0, where a₁, b₁, c₁ and a₂, b₂, c₂ are the direction ratios of two lines.

The standard form of a pair of Cartesian lines is:

and …(i)

Now, first we rewrite the given equations according to the standard form, i.e.

and , i.e.

and …(ii)

Now, comparing (i) and (ii), we get

and

Now, as both the lines are at right angles,

so a₁a₂ + b₁b₂ + c₁c₂ = 0

⇒

⇒

⇒

⇒

⇒ 11p = 70

⇒

We know that

Two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular if the angle between them is θ = 90°, i.e. a₁a₂ + b₁b₂ + c₁c₂ = 0

Also, here the direction ratios are:

a₁ = 7, b₁ = -5, c₁ = 1 and a₂ = 1, b₂ = 2, c₂ = 3

Now, Consider

a₁a₂ + b₁b₂ + c₁c₂ = 7 × 1 + (-5) × 2 + 1 × 3 = 7 -10 + 3 = - 3 + 3 = 0

∴ The two lines are perpendicular to each other.

We know that

Shortest distance between two lines and is

…(i)

Here, , and

,

Now,

…(ii)

Now,

⇒

……….(iii)

……….(iv)

Now,

……….(v)

Now, using (i), we have

The shortest distance between the two lines, d [From (iv) and (v)]

⇒

Rationalizing the fraction by multiplying the numerator and denominator by √2,

⇒

We know that

_{Shortest distance between the lines:}

and is

…(i)

The standard form of a pair of Cartesian lines is:

and

And the given equations are: and

_{Comparing the given equations with the standard form, we get}

_x1 = -1, y₁ = -1, z₁ = -1; x₂ = 3, y₂ = 5, z₂ = 7

_a1 = 7, b₁ = -6, c₁ = 1; a₂ = 1, b₂ = -2, c₂ = 1

_{Now, consider}

_⇒

⇒

⇒

⇒

⇒

⇒

Next, consider

⇒

⇒

⇒

⇒

_{⇒From (i), we have}

⇒

We know that

Shortest distance between two lines and is

……….(i)

Here, , and

,

Now,

……….(ii)

Now,

⇒

……….(iii)

……….(iv)

Now,

……….(v)

Now, using (i), we have

The shortest distance between the two lines,

Firstly, consider

⇒

⇒

⇒

⇒

⇒

⇒

So, we need to find the shortest distance between and .

Now, We know that

Shortest distance between two lines and is

…(i)

Here, and

⇒

Now,

…(ii)

Now,

⇒

……….(iii)

……….(iv)

Now,

……….(v)

Now, using (i), we have

The shortest distance between the two lines,

The eq. of the plane

z = 2

Direction ratio of the normal (0,0,1)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Direction cosines = 0,0,1

Distance(d) = 2

The eq. of the plane

x + y + z = 1

Direction ratio of the normal (1,1,1)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

The eq. of the plane

2x + 3y-z = 5

Direction ratio of the normal (2, 3, -1)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

The eq. of the plane

0x-5y + 0z = 8

Direction ratio of the normal (0, -5, 0)

= 5

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Direction cosine = 0, -1, 0

Vector eq. of the plane with position vector is

(Letbe the position vector of P(x,y,z)

Hence,

So, Cartesian eq. is

x + y - z = 2

(Letbe the position vector of P(x,y,z)

.

.

So, Cartesian eq. is

2x + 3y - 4z = 1

Letbe the position vector of P(x,y,z)

Hence,

So, Cartesian eq. is

(s-2t)x + (3-t)y + (2s + t)z=15

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

2x + 3y + 4z = 12

Direction ratio (2,3,4)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

0x + 3y + 4z = 6

Direction ratio (0,3,4)

.

= 5

his is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Coordinate of the foot (ld,md,nd)

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

x + y + z = 1

Direction ratio (1,1,1)

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

0x – 5y + 0z = 8

Direction ratio (0,-5,0)

= 5

This is the form of

lx + my + nz = d (∴ d = Distance of the normal from the origin.)

Coordinate of the foot (ld,md,nd)

Let the position vector of the point

Normal⊥to the plane

Vector eq. of the plane,

x – 1 + y – z – 2 = 0

x + y – z – 3 = 0

Required Cartesian eq. of the plane

x + y – z = 3

Let the position vector of the point

Vector eq. of the plane,

x – 1 – 2y + 8 + z – 6 = 0

x – 2y + z + 1 = 0

Required Cartesian eq. of the plane

x – 2y + z = - 1

The given points are (1, 1, -1), (6, 4, -5), (-4, -2, 3).

Let,

= 1(12 - 10) – 1(18 - 20) -1 (-12 + 16)

= 2 + 2 – 4

= 0

Since, the value of determinant is 0.

Therefore, these points are collinear as there will be infinite planes passing through the given 3 points.

The given points are (1, 1, 0), (1, 2, 1), (-2, 2, -1).

Let,

= 1(-2 - 2) – 1(-1 + 2)

= -4 – 1

= -5 ≠ 0

There passes a unique plane from the given 3 points.

Equation of the plane passes through the points, (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃), i.e.,

⇒ (x - 1)(-2) – (y - 1)(3) + 3z = 0

⇒ -2x + 2 – 3y + 3 + 3z = 0

⇒ 2x + 3y – 3z = 5

This is the required eq. of the plane.

We know that, the eq. of the plane in intercept form

where a, b, c are the intercepts cut-off by the plane at x, y and z axes respectively.

⇒ 2x + y – z = 5 (i)

Dividing both side of (i)eq. by 5, we get

Thus, the intercepts cut-off by the plane are 5/2, 5 and -5.

We know that the eq. of the plane ZOX is

y = 0

Eq. of plane parallel to it is of the form, y = a

Hence, the required eq. of the plane is

y = 3

Eq. of the plane passes through the intersection of the plane is given by

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

∵ Plane passes through the points (2,2,1)

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

2 + 3λ = 0

3λ = -2

(i)

Hence, the required eq. of the plane

7x – 5y + 4z – 8 = 0

This is the required eq. of the plane.

Let the vector eq. of the plane passing through the intersection of the planes

Here,

∵ Plane passes through points (2,1,3)

4 + 4λ + 2 + 5λ - 9 + 9λ - 7 - 9λ = 0

9 λ = 10

This is the required vector eq. of the plane.

Let the eq. of the plane that passes through the two-given plane x + y + z = 1 and 2x + 3y + 4z = 5 is

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

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

Direction ratio of the plane (2λ + 1, 3λ + 1, 4λ + 1)

and Direction ratio of another plane (1, -1, 1)

∵ Both are ⊥ hence

(2λ + 1 × 1) + (3λ + 1 × (-1)) + (4λ + 1 × 1) = 0

2λ + 1 - 3λ - 1 + 4λ + 1 = 0

Put the value of λ in (i)eq., we get

x – z + 2 = 0

This is the required eq. of the plane.

The eq. of the given planes are

If n₁ and n₂ are normal to the planes,

Angle between two planes

The eq. of the given planes are

7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

Two planes are ⊥ if the direction ratio of the normal to the plane is

a₁a₂ + b₁b₂ + c₁c₂ = 0

21 – 5 – 60

-44 ≠ 0

∴ Both the planes are not ⊥ to each other.

Two planes are || to each other if the direction ratio of the normal to the plane is

∴ Both the planes are not || to each other.

The angle between them is given by

The eq. of the given planes are

2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

Two planes are ⊥ if the direction ratio of the normal to the plane is

a₁a₂ + b₁b₂ + c₁c₂ = 0

2 × 1 + 1 × (-2) + 3 × 0

= 0

Thus, the given planes are ⊥ to each other.

The eq. of the given planes are

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

Two planes are ⊥ if the direction ratio of the normal to the plane is

a₁a₂ + b₁b₂ + c₁c₂ = 0

6 + 6 + 24

36 ≠ 0

∴ Both the planes are not ⊥ to each other.

Two planes are || to each other if the direction ratio of the normal to the plane is

Thus, the given planes are || to each other.

The eq. of the given planes are

2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0

Two planes are ⊥ if the direction ratio of the normal to the plane is

a₁a₂ + b₁b₂ + c₁c₂ = 0

2 × 2 + (-1) × (-1) + 3 × 3

14 ≠ 0

∴ Both the planes are not ⊥ to each other.

Two planes are || to each other if the direction ratio of the normal to the plane is

Thus, the given planes are || to each other.

The eq. of the given planes are

4x + 8y + z – 8 = 0 and y + z – 4 = 0

Two planes are ⊥ if the direction ratio of the normal to the plane is

a₁a₂ + b₁b₂ + c₁c₂ = 0

0 + 8 + 1

9 ≠ 0

∴ Both the planes are not ⊥ to each other.

Two planes are || to each other if the direction ratio of the normal to the plane is

∴ Both the planes are not || to each other.

The angle between them is given by

= 45⁰

Distance of point P(x₁,y₁,z₁) from the plane Ax + By + Cz – D = 0 is

Given point is (0,0,0) and the plane is 3x – 4y + 12z = 3

Given point is (3,-2,1) and the plane is 2x – y + 2z + 3 = 0

Given point is (2,3,-5) and the plane is x + 2y – 2z = 9

= 3

Given point is (-6,0,0) and the plane is 2x – 3y + 6z – 2 = 0

= 2

Let OA be the line joining the origin (0,0,0) and the point A(2,1,1).

Let BC be the line joining the points B(3,5,−1) and C(4,3,−1)

Direction ratios of OA = (a₁, b₁, c₁) ≡ [(2 - 0), (1 - 0), (1 - 0)] ≡ (2,1,1)

Direction ratios of BC = (a₂, b₂, c₂) ≡ [(4 - 3), (3 - 5), (-1 + 1)]

≡ (1, -2, 0)

Given-

OA is ⊥ to BC

∴ we have to prove that -

a₁a₂ + b₁b₂ + c₁c₂ = 0

L.H.S = a₁a₂ + b₁b₂ + c₁c₂ = 2 × 1 + 1 × (−2) + 1 × 0 = 2 - 2 = 0

R.H.S = 0

Thus, L.H.S = R.H.S ….PROVED

Hence OA is ⊥ to BC.

Let l, m, n be the direction cosines of the line perpendicular to each of the given lines. Then,

ll₁ + mm₁ + nn₁ = 0 …(1)

and, ll₂ + mm₂ + nn₂ = 0 …(2)

On solving (1) and (2) by cross - multiplication, we get -

Thus, the direction cosines of the given line are proportional to

(m₁n₂ - m₂n₁), (n₁l₂ - n₂l₁), (l₁m₂ - l₂m₁)

So, its direction cosines are

where .

we know that -

(l₁² + m₁² + n₁²) (l₂² + m₂² + n₂²) - (l₁l₂ + m₁m₂ + n₁n₂)²

= (m₁n₂ - m₂n₁)² + (n₁l₂ - n₂l₁)² + (l₁m₂ - l₂m₁)² …(3)

It is given that the given lines are perpendicular to each other. Therefore,

l₁l₂ + m₁m₂ + n₁n₂ = 0

Also, we have

l₁² + m₁² + n₁² = 1

and, l₂² + m₂² + n₂² = 1

Putting these values in (3), we get -

(m₁n₂ - m₂n₁)² + (n₁l₂ - n₂l₁)² + (l₁m₂ - l₂m₁)² = 1

⇒ λ = 1

Hence, the direction cosines of the given line are (m₁n₂ - m₂n₁), (n₁l₂ - n₂l₁), (l₁m₂ - l₂m₁)

Angle between the lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ is given by

Given -

a₁ = a, b₁ = b, c₁ = c

a₂ = b - c, b₂ = c - a, c₂ = a - b

So,

= 0

∴ cosθ = 0

So, θ = 90°

Hence, Angle between the given pair of Lines is 90°.

Equation of a line passing through (x₁, y₁, z₁) and parallel to a line with direction ratios a,b,c is

Since the line passes through origin i.e. (0,0,0)

x₁ = 0, y₁ = 0, z₁ = 0

Since line is parallel to x - axis,

a = 1, b = 0, c = 0

Equation of Line is given by –

Angle between the lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ is given by

A line passing through A(x₁, y₁, z₁) and B(x₂, y₂, z₂) has direction ratios (x₁ - x₂), (y₁ - y₂), (z₁ - z₂)

Direction ratios of line joining the points A(1,2,3) and B(4,5,7)

= (4 - 1), (5 - 2), (7 - 3)

= (3,3,4)

∴ a₁ = 3, b₁ = 3, c₁ = 4

Direction ratios of line joining the points C(-4, 3, -6) and B(2,9,2)

= (2 - (-4)), (9 - 3), (2-(-6))

= (6,6,8)

∴ a₂ = 6, b₂ = 6, c₂ = 8

Now,

∴ cosθ = 1

So, θ = 0°

Hence, Angle between the lines AB and CD is 0°.

Two lines and

are perpendicular to each other if

a₁a₂ + b₁b₂ + c₁c₂ = 0

Given -

comparing with

we get -

x₁ =1, y₁ = 2, z₁ = 3

& a₁ = - 3, b₁ = 2k, c₁ = 2

Similarly,

comparing with

we get -

x₂ = 1, y₂ = 2, z₂ = 3

& a₂ = 3k, b₂ = 1, c₂ = -5

Since the two lines are perpendicular,

a₁a₂ + b₁b₂ + c₁c₂ = 0

⇒ (-3) × 3k + 2k × 1 + 2 × (-5) = 0

⇒ -9k + 2k - 10 = 0

⇒ -7k = 10

∴ k = -10/7

Hence, the value of k is -10/7.

The vector equation of a line passing through a point with position vector and parallel to vector is

⇒

Given, the line passes through (1,2,3)

So,

Finding normal of plane

Comparing with ,

Since line is perpendicular to plane, the line will be parallel of the plane

∴

Hence,

This is the required vector equation of line.

The equation of a plane passing through (x₁,y₁,z₁) and perpendicular to a line with direction ratios A, B, C is

A(x - x₁) + B(y - y₁) + C(z - z₁) = 0

The plane passes through (a,b,c)

So, x₁ = a, y₁ = b, z₁ = c

Since both planes are parallel to each other, their normal will be parallel

∴ Direction ratios of normal

= Direction ratios of normal of

Direction ratios of normal = (1,1,1)

∴ A = 1, B =1, C = 1

Thus,

Equation of plane in cartesian form is

A(x - x₁) + B(y - y₁) + C(z - z₁) = 0

⇒ 1(x - a) + 1(y - b) + 1(z - c) = 0

⇒ x + y + z - (a + b + c) = 0

Thus, x + y + z = a + b + c is the required equation of plane.

Shortest distance between lines with vector equations

and is

⇒

Given -

Comparing with , we get -

&

Similarly,

Comparing with , we get -

&

Now,

⇒

⇒

⇒

⇒

⇒

Magnitude of =

= √144

= 12

Also,

⇒

= -80 + (-16) + (-12)

= -108

Shortest Distance = 9

Hence, the shortest distance between the given two lines is 9.

The vector equation of a line passing through two points with position vectors & is

⇒

The position vector of point A(5,1,6) is given as -

⇒

The position vector of point B(3,4,1) is given as -

⇒

⇒

∴ ...(1)

Let the coordinates of the point where the line crosses the YZ plane be (0,y,z)

So, …(2)

Since point lies in line, it will satisfy its equation,

Putting (2) in (1)

⇒

Two vectors are equal if their corresponding components are equal

So,

0 = 5 - 2λ…(3)

y = 1 + 3λ…(4)

and, z = 6 - 5λ…(5)

From equation (3), we get -

λ = 5/2

Substitute the value of λ in equation (4) and (5), we get -

y = 1 + 3λ = 1 + 3 × (5/2) = 1 + (15/2) = 17/2

and

z = 6 - 5λ = 6 - 5 × (5/2) = 6 - (25/2) = - 13/2

Therefore, the coordinates of the required point is

(0, 17/2, -13/2).

The vector equation of a line passing through two points with position vectors & is

⇒

The position vector of point A(5,1,6) is given as -

.

The position vector of point B(3,4,1) is given as -

⇒

⇒

∴ ...(1)

Let the coordinates of the point where the line crosses the ZX plane be (0,y)

So, …(2)

Since point lies in line, it will satisfy its equation,

Putting (2) in (1)

Two vectors are equal if their corresponding components are equal

So,

x = 5 - 2λ …(3)

0 = 1 + 3λ …(4)

and, z = 6 - 5λ …(5)

From equation (4), we get -

λ = -1/3

Substitute the value of λ in equation (3) and (5), we get -

x = 5 - 2λ = 5 - 2 × (-1/3) = 5 + (2/3) = 17/3

and

z = 6 - 5λ = 6 - 5 × (-1/3) = 6 + (5/3) = 23/3

Therefore, the coordinates of the required point is

(17/3, 0, 23/3).

The equation of a line passing through two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is

⇒

Given the line passes through the points A(3, –4, –5) and

B(2, –3, 1)

∴ x₁ = 3, y₁ = -4, z₁ = -5

and, x₂ = 2, y₂ = -3, z₂ = 1

So, the equation of line is

⇒

⇒

So,

x = -k + 3 | y = k - 4 | z = 6k - 5 …(1)

Let (x, y, z) be the coordinates of the point where the line crosses the plane 2x + y + z + 7 = 0

Putting the value of x,y,z from (1) in the equation of plane,

2x + y + z + 7 = 0

⇒ 2(-k + 3) + (k - 4) + (6k - 5) = 7

⇒ 5k - 3 = 7

⇒ 5k = 10

∴ k = 2

Putting the value of k in x, y, z

x = - k + 3 = - 2 + 3 = 1

y = k - 4 = 2 - 4 = - 2

z = 6k - 5 = 12 - 5 = 7

Hence, the coordinates of the required point are (1, -2,7).

The equation of a plane passing through (x₁,y₁,z₁) is given by

A(x - x₁) + B(y - y₁) + C(z - z₁) = 0

where, A, B, C are the direction ratios of normal to the plane.

Now the plane passes through (-1,3,2)

So, equation of plane is

A(x + 1) + B(y - 3) + C(z - 2) = 0 …(1)

Since this plane is perpendicular to the given two planes.

So, their normal to the plane would be perpendicular to normals of both planes.

we know that -

is perpendicular to both &

So, required normal is cross product of normals of planes

x + 2y + 3z = 5 and 3x + 3y + z = 0

Required Normal

Hence, direction ratios = -7, 8, -3

∴ A = -7, B = 8, C = -3

Putting above values in (1), we get -

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

⇒ -7(x + 1) + 8(y - 3) + (-3)(z - 2) = 0

⇒ -7x - 7 + 8y - 24 - 3z + 6 = 0

⇒ -7x + 8y - 3z - 25 = 0

∴ 7x - 8y + 3z + 25 = 0

Therefore, equation of the required plane is 7x - 8y + 3z + 25 = 0.

The distance of a point with position vector from the plane is .

The position vector of point (1,1,p) is given as -

⇒

The position vector of point (-3,0,1) is given as -

⇒

It is given that the points (1,1,p) and (-3,0,1) are equidistant from the plane

∴

⇒ 20 - 12p = 8

⇒ 20 - 12p = 8 or, 20 - 12p = -8

⇒ 12p = 12 or, 12p = 28

∴ p = 1 or, p = 7/3

us, the possible values of p are 1 and 7/3.

The equation of any plane through the line of intersection of the planes and is given by -

.

So, the equation of any plane through the line of intersection of the given planes is

.

.

∴ . …(1)

Since this plane is parallel to x-axis.

So, the normal vector of the plane (1) will be perpendicular to x-axis.

Direction ratios of Normal (a_1, b_1, c₁)≡ [(1 - 2λ), (1 - 3λ), (1 +)]

Direction ratios of x–axis (a_2, b_2, c₂)≡ (1,0,0)

Since the two lines are perpendicular,

a₁a₂ + b₁b₂ + c₁c₂ = 0

(1 - 2λ) × 1 + (1 - 3λ) × 0 + (1 + λ) × 0 = 0

⇒ (1 - 2λ) = 0

∴ λ = 1/2

Putting the value of λ in (1), we get -

⇒

⇒

Hence, the equation of the required plane is

The equation of a plane passing through (x₁,y₁,z₁) and perpendicular to a line with direction ratios A, B, C is

A(x - x₁) + B(y - y₁) + C(z - z₁) = 0

The plane passes through P(1,2,3)

So, x₁ = 1, y₁ = 2, z₁ = - 3

Normal vector to plane =

where O(0,0,0), P(1,2, - 3)

Direction ratios of = (1 - 0), (2 - 0), (-3 - 0)

= (1,2, - 3)

∴ A = 1, B = 2, C = -3

Equation of plane in cartesian form is

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

⇒ x - 1 + 2y - 4 - 3z - 9 = 0

⇒ x + 2y - 3z - 14 = 0

The equation of any plane through the line of intersection of the planes and is given by -

⇒

So, the equation of any plane through the line of intersection of the given planes is

⇒

⇒

∴ …(1)

Since this plane is perpendicular to the plane

…(2)

So, the normal vector of the plane (1) will be perpendicular to the normal vector of plane (2).

Direction ratios of Normal of plane (1) = (a_1, b_1, c₁)

≡ [(1 - 2λ), (2 - λ), (3 + λ)]

Direction ratios of Normal of plane (2) = (a_2, b_2, c₂)

≡ (-5, -3,6)

Since the two lines are perpendicular,

a₁a₂ + b₁b₂ + c₁c₂ = 0

⇒ (1 - 2λ) × (-5) + (2 - λ) × (-3) + (3 + λ) × 6 = 0

⇒ -5 + 10λ - 6 + 3λ + 18 + 6λ = 0

⇒ 19λ + 7 = 0

∴ λ = -7/19

Putting the value of λ in (1), we get -

⇒

⇒

⇒

Hence, the equation of the required plane is

Given -

The equation of line is

and the equation of the plane is

To find the intersection of line and plane, putting value of from equation of line into equation of plane, we get -

⇒

⇒

⇒ (2 + 3λ) × 1 + (-1 + 4λ) × (-1) + (2 + 2λ) × 1 = 5

⇒ 2 + 3λ + 1 - 4λ + 2 + 2λ = 5

⇒ λ = 0

So, the equation of line is

Let the point of intersection be (x,y,z)

So,

∴

Hence, x = 2, y = -1, z = 2

Therefore, the point of intersection is (2, -1, 2).

Now, the distance between points (x₁, y₁, z₁) and (x₂, y₂, z₂) is given by -

units

Distance between the points A(2, -1, 2) and B(-1, -5, -10) is given by -

= 13 units

The vector equation of a line passing through a point with position vector and parallel to a vector is

⇒

Given, the line passes through (1,2,3)

So,

Given, line is parallel to both planes

∴ Line is perpendicular to normal of both planes.

i.e is perpendicular to normal of both planes.

we know that -

is perpendicular to both &

So, is cross product of normals of planes

and

Required Normal

Thus,

Now, putting the value of & in formula

Therefore, the equation of the line is

The vector equation of a line passing through a point with position vector and parallel to a vector is

⇒

Given, the line passes through (1, 2, -4)

So,

Given, line is parallel to both planes

∴ Line is perpendicular to normal of both planes.

i.e is perpendicular to normal of both planes.

we know that -

is perpendicular to both &

So, is cross product of normals of planes

and

Required Normal

Thus,

Now, putting the value of & in formula

Therefore, the equation of the line is

Distance of the point (x₁,y₁,z₁) from the plane Ax + By + Cz = D is

⇒

The equation of a plane having intercepts a, b, c on the x-, y-, z- axis respectively is

⇒

Comparing with Ax + By + Cz = D, we get -

A = 1/a, B = 1/b, C = 1/c, D = 1

Given, the plane is at a distance of 'p' units from the origin.

So, The point is O(0,0,0)

∴ x₁ = 0, y₁ = 0, z₁ = 0

Now,

Distance

Substituting all values, we get -

⇒

squaring both sides, we get -

⇒

Hence Proved.

Distance between two parallel planes Ax + By + Cz = d₁ and Ax + By + Cz = d₂ is

⇒

Given -

First Plane is

2x + 3y + 4z = 4

Comparing with Ax + By + Cz = d₁, we get -

A = 2, B = 3, C = 4, d₁ = 4

Second Plane is

4x + 6y + 8z = 12

After Dividing by 2,

2x + 3y + 4z = 6

Comparing with Ax + By + Cz = d₁, we get -

A = 2, B = 3, C = 4, d₂ = 6

So,

Distance between two planes

= 2/√29

Hence, (D) is the correct option.

Given -

First Plane is

2x – y + 4z = 5

Multiply both sides by 2.5, we get -

5x - 2.5y + 10z = 12.5 …(1)

Second Plane is

5x – 2.5y + 10z = 6 …(2)

Clearly, the direction ratios of normals of both the plane (1) and (2) are same.

Hence, Both the given planes are parallel.