Straight Line In Space

Given: line passes through point (3, 4, 5) and is parallel to

To find: equation of line in vector and Cartesian forms

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and is a vector parallel to the line.

Explanation:

Here, and

Therefore,

Vector form:

Cartesian form:

Given: line passes through (2, 1, -3) and is parallel to

To find: equation of line in vector and Cartesian forms

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and is a vector parallel to the line.

Explanation:

Here, and

Therefore,

Vector form:

Cartesian form:

Given: line passes through and is parallel to

To find: equation of line in vector and Cartesian forms

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and is a vector parallel to the line.

Explanation:

Here, and

Therefore,

Vector form:

Cartesian form:

Given: line passes through and is drawn in the direction of

To find: equation of line in vector and Cartesian forms

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and is a vector parallel to the line.

Explanation:

Since line is drawn in the direction of , it is parallel to

Here, and

Therefore,

Vector form:

Cartesian form:

Given: Cartesian equation of line

To find: equation of line in vector form

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and is a vector parallel to the line.

Explanation:

From the Cartesian equation of the line, we can find and

Here, and

Therefore,

Vector form:

Given: Cartesian equation of line are 3x + 1 = 6y – 2 = 1 - z

To find: fixed point through which the line passes through, its direction ratios and the vector equation.

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and is a vector parallel to the line and also its direction ratio.

Explanation:

The Cartesian form of the line can be rewritten as:

Therefore, and

So, the line passes through and direction ratios of the line are (2, 1, -6) and vector form is:

Given: line passes through (1, 3, -2) and is parallel to the line

To find: equation of line in vector and Cartesian form

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and is a vector parallel to the line.

Explanation:

Since the line (say L₁) is parallel to another line (say L₂), L₁ has the same direction ratios as that of L₂

Here,

Since the equation of L₂ is

Therefore,

Vector form of the line is:

Cartesian form of the line is:

Given: line passes through (1, -2, 3) and is parallel to the line

To find: equation of line in vector and Cartesian form

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and is a vector parallel to the line.

Explanation:

Since the line (say L₁) is parallel to another line (say L₂), L₁ has the same direction ratios as that of L₂

Here,

Since the equation of L₂ is

Therefore,

Vector form of the line is:

Cartesian form of the line is:

Given: line passes through (1, 2, 3) and is parallel to the line

To find: equation of line in Vector and Cartesian form

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and is a vector parallel to the line.

Explanation:

Since the line (say L₁) is parallel to another line (say L₂), L₁ has the same direction ratios as that of L₂

Here,

Equation of L₂ can be rewritten as:

Therefore,

Vector form of the line is:

Cartesian form of the line is:

Given: line passes through (-1, 3, -2) and is perpendicular to each of the lines and

To find: equation of line in Vector and Cartesian form

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and is a vector parallel to the line.

If 2 lines of direction ratios a₁:a₂:a₃ and b₁:b₂:b₃ are perpendicular, then a₁b₁+a₂b₂+a₃b₃ = 0

Explanation:

Here,

Let the direction ratios of the line be b₁:b₂:b₃

Direction ratios of the other two lines are 1 : 2 : 3 and -3 : 2 : 5

Since the other two line are perpendicular to the given line, we have

b₁ + 2b₂ + 3b₃ = 0

-3b₁ + 2b₂ + 5b₃ = 0

Solving,

Therefore,

Vector form of the line is:

Cartesian form of the line is:

Given: line passes through (1, 2, -4) and is perpendicular to each of the lines and

To find: equation of line in Vector and Cartesian form

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and is a vector parallel to the line.

If 2 lines of direction ratios a₁:a₂:a₃ and b₁:b₂:b₃ are perpendicular, then a₁b₁+a₂b₂+a₃b₃ = 0

Explanation:

Here,

Let the direction ratios of the line be b₁:b₂:b₃

Direction ratios of other two lines are 8 : -16 : 7 and 3 : 8 : -5

Since the other two line are perpendicular to the given line, we have

8b₁ – 16b₂ + 7b₃ = 0

3b₁ + 8b₂ – 5b₃ = 0

Solving,

Therefore,

Vector form of the line is:

Cartesian form of the line is:

Given: The equations of the two lines are

and

_{To Prove:The two lines intersect and to find their point of intersection.}

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and b₁ : b₂ : b₃ is the direction ratios of the line.

_Proof:

_Let

So a point on the first line is (λ₁ + 4, 4λ₁ – 3, 7λ₁ – 1)

A point on the second line is (2λ₂ + 1, -3λ₂ – 1, 8λ₂ – 10)

If they intersect they should have a common point.

λ₁ + 4 = 2λ₂ + 1 ⇒ λ₁ – 2λ₂ = -3 … (1)

4λ₁ – 3 = -3λ₂ – 1 ⇒ 4λ₁ + 3λ₂ = 2 … (2)

Solving (1) and (2),

11λ₂ = 14

Therefore,

Substituting for the z coordinate, we get

and

So, the lines do not intersect.

Given: The equations of the two lines are

and

_{To Prove:The two lines intersect and to find their point of intersection.}

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and b₁ : b₂ : b₃ is the direction ratios of the line.

_Proof:

_Let

So a point on the first line is (2λ₁ + 1, 3λ₁ + 2, 4λ₁ + 3)

A point on the second line is (5λ₂ + 4, 2λ₂ + 1, λ₂)

If they intersect they should have a common point.

2λ₁ + 1 = 5λ₂ + 4 ⇒ 2λ₁ – 5λ₂ = 3 … (1)

3λ₁ + 2 = 2λ₂ + 1 ⇒ 3λ₁ - 2λ₂ = -1 … (2)

Solving (1) and (2),

-11λ₂ = 11

λ₂ = -1

Therefore, λ₁ = -1

Substituting for the z coordinate, we get

4λ₁ + 3 = -1 and λ₂ = -1

So, the lines intersect and their point of intersection is (-1, -1, -1)

Given: The equations of the two lines are

and

_{To Prove:the lines do not intersect each other.}

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and b₁ : b₂ : b₃ is the direction ratios of the line.

_Proof:

_Let

So a point on the first line is (2λ₁ + 1, 3λ₁ – 1, λ₁)

A point on the second line is (5λ₂ - 1, λ₂ + 1, 2)

If they intersect they should have a common point.

2λ₁ + 1 = 5λ₂ - 1 ⇒ 2λ₁ – 5λ₂ = -2 … (1)

3λ₁ – 1 = λ₂ + 1 ⇒ 3λ₁ - λ₂ = 2 … (2)

Solving (1) and (2),

-13λ₂ = -10

Therefore,

Substituting for the z coordinate, we get

and z = 2

So, the lines do not intersect.

Given: Equation of line is

To find: coordinates of foot of the perpendicular from (1, 2, 3) to the line. And find the length of the perpendicular.

Formula Used:

1. Equation of a line is

Cartesian form:

where is a point on the line and b₁ : b₂ : b₃ is the direction ratios of the line.

2. Distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is

Explanation:

Let

So the foot of the perpendicular is (3λ + 6, 2λ + 7, -2λ + 7)

Direction ratio of the line is 3 : 2 : -2

Direction ratio of the perpendicular is

⇒ (3λ + 6 - 1) : (2λ + 7 - 2) : (-2λ + 7 - 3)

⇒ (3λ + 5) : (2λ + 5) : (-2λ + 4)

Since this is perpendicular to the line,

3(3λ + 5) + 2(2λ + 5) – 2(-2λ + 4) = 0

⇒ 9λ + 15 + 4λ + 10 + 4λ – 8 = 0

⇒ 17λ = -17

⇒ λ = -1

So the foot of the perpendicular is (3, 5, 9)

Distance

= 7 units

Therefore, the foot of the perpendicular is (3, 5, 9) and length of perpendicular is 7 units.

Given: Equation of line is

To find: coordinates of foot of the perpendicular from (2, -1, 5) to the line. And find the length of the perpendicular.

Formula Used:

1. Equation of a line is

Cartesian form:

where is a point on the line and b₁ : b₂ : b₃ is the direction ratios of the line.

2. Distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is

Explanation:

Let

So the foot of the perpendicular is (10λ + 11, -4λ - 2, -11λ - 8)

Direction ratio of the line is 10 : -4 : -11

Direction ratio of the perpendicular is

⇒ (10λ + 11 - 2) : (-4λ - 2 + 1) : (-11λ - 8 - 5)

⇒ (10λ + 9) : (-4λ - 1) : (-11λ - 13)

Since this is perpendicular to the line,

10(10λ + 9) - 4(-4λ - 1) - 11(-11λ - 13) = 0

⇒ 100λ + 90 + 16λ + 4 + 121λ + 143 = 0

⇒ 237λ = -237

⇒ λ = -1

So the foot of the perpendicular is (1, 2, 3)

Distance

= √14 units

Therefore, the foot of the perpendicular is (1, 2, 3) and length of perpendicular is √14 units.

Given: line passes through the points (3, 4, -6) and (5, -2, 7)

To find: equation of line in vector and Cartesian forms

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and with b₁ : b₂ : b₃ being the direction ratios of the line.

Explanation:

Here,

The direction ratios of the line are (3 - 5) : (4 + 2) : (-6 - 7)

⇒ -2 : 6 : -13

⇒ 2 : -6 : 13

So,

Therefore,

Vector form:

Cartesian form:

Given: line passes through the points (2, -3, 0) and (-2, 4, 3)

To find: equation of line in vector and Cartesian forms

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and with b₁ : b₂ : b₃ being the direction ratios of the line.

Explanation:

Here,

The direction ratios of the line are (2 + 2) : (-3 - 4) : (0 - 3)

⇒ 4 : -7 : -3

⇒ -4 : 7 : 3

So,

Therefore,

Vector form:

Cartesian form:

Given: line passes through the points whose position vectors are and

To find: equation of line in vector and Cartesian forms

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and with b₁ : b₂ : b₃ being the direction ratios of the line.

Explanation:

Here,

The direction ratios of the line are (1 - 1) : (-2 - 3) : (1 + 2)

⇒ 0 : -5 : 3

⇒ 0 : 5 : -3

So,

Therefore,

Vector form:

Cartesian form:

Given: line passes through the point (3, -2, 1) and is parallel to the line joining points B(-2, 4, 2) and C(2, 3, 3).

To find: equation of line in vector and Cartesian forms

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and with b₁ : b₂ : b₃ being the direction ratios of the line.

Explanation:

Here,

The direction ratios of the line are (-2 - 2) : (4 - 3) : (2 - 3)

⇒ -4 : 1 : -1

⇒ 4 : -1 : 1

So,

Therefore,

Vector form:

Cartesian form:

Given: line passes through the point with position vector and parallel to the line joining the points with position vectors and .

To find: equation of line in vector and Cartesian forms

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and with b₁ : b₂ : b₃ being the direction ratios of the line.

Explanation:

Here,

The direction ratios of the line are (1 - 2) : (-1 - 3) : (5 + 4)

⇒ -1 : -4 : 9

⇒ 1 : 4 : -9

So,

Therefore,

Vector form:

Cartesian form:

Given: perpendicular drawn from point A (1, 2, 1) to line joining points B (1, 4, 6) and C (5, 4, 4)

To find: foot of perpendicular

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and with b₁ : b₂ : b₃ being the direction ratios of the line.

If 2 lines of direction ratios a₁:a₂:a₃ and b₁:b₂:b₃ are perpendicular, then a₁b₁+a₂b₂+a₃b₃ = 0

Explanation:

B (1, 4, 6) is a point on the line.

Therefore,

Also direction ratios of the line are (1 - 5) : (4 – 4) : (6 – 4)

⇒ -4 : 0 : 2

⇒ -2 : 0 : 1

So, equation of the line in Cartesian form is

Any point on the line will be of the form (-2λ + 1, 4, λ + 6)

So the foot of the perpendicular is of the form (-2λ + 1, 4, λ + 6)

The direction ratios of the perpendicular is

(-2λ + 1 – 1) : (4 - 2) : (λ + 6 - 1)

⇒ (-2λ) : 2 : (λ + 5)

From the direction ratio of the line and the direction ratio of its perpendicular, we have

-2(-2λ) + 0 + λ + 5 = 0

⇒ 4λ + λ = -5

⇒ λ = -1

So, the foot of the perpendicular is (3, 4, 5)

Given: perpendicular drawn from point A (1, 8, 4) to line joining points B (0, -1, 3) and C (2, -3, -1)

To find: foot of perpendicular

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and with b₁ : b₂ : b₃ being the direction ratios of the line.

If 2 lines of direction ratios a₁:a₂:a₃ and b₁:b₂:b₃ are perpendicular, then a₁b₁+a₂b₂+a₃b₃ = 0

Explanation:

B (0, -1, 3) is a point on the line.

Therefore,

Also direction ratios of the line are (0 - 2) : (-1 + 3) : (3 + 1)

⇒ -2 : 2 : 4

⇒ -1 : 1 : 2

So, equation of the line in Cartesian form is

Any point on the line will be of the form (-λ, λ - 1, 2λ + 3)

So the foot of the perpendicular is of the form (-λ, λ - 1, 2λ + 3)

The direction ratios of the perpendicular is

(-λ - 1) : (λ – 1 - 8) : (2λ + 3 - 4)

⇒ (-λ - 1) : (λ – 9) : (2λ – 1)

From the direction ratio of the line and the direction ratio of its perpendicular, we have

-1(-λ - 1) + λ – 9 + 2(2λ – 1) = 0

⇒ λ + 1 + λ – 9 + 4λ – 2 = 0

⇒ 6λ = 10

So, the foot of the perpendicular is

Given: Equation of line is

_{To find:image of point (0, 2, 3)}

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and with b₁ : b₂ : b₃ being the direction ratios of the line.

If 2 lines of direction ratios a₁:a₂:a₃ and b₁:b₂:b₃ are perpendicular, then a₁b₁+a₂b₂+a₃b₃ = 0

_{Mid-point of line segment joining (}x₁, y₁, z₁) and (x₂, y₂, z₂) is

_Explanation:

Let

So the foot of the perpendicular is (5λ – 3, 2λ + 1, 3λ – 4)

The direction ratios of the perpendicular is

(5λ – 3 - 0) : (2λ + 1 - 2) : (3λ - 4 - 3)

⇒ (5λ – 3) : (2λ – 1) : (3λ – 7)

Direction ratio of the line is 5 : 2 : 3

From the direction ratio of the line and the direction ratio of its perpendicular, we have

5(5λ - 3) + 2(2λ – 1) + 3(3λ – 7) = 0

⇒ 25λ – 15 + 4λ – 2 + 9λ – 21 = 0

⇒ 38λ = 38

⇒ λ = 1

So, the foot of the perpendicular is (2, 3, -1)

The foot of the perpendicular is the mid-point of the line joining (0, 2, 3) and (α, β, γ)

So, we have

So, the image is (4, 4, -5)

Given: Equation of line is

_{To find:image of point (5, 9, 3)}

Formula Used: Equation of a line is

Vector form:

Cartesian form:

where is a point on the line and with b₁ : b₂ : b₃ being the direction ratios of the line.

If 2 lines of direction ratios a₁:a₂:a₃ and b₁:b₂:b₃ are perpendicular, then a₁b₁+a₂b₂+a₃b₃ = 0

_{Mid-point of line segment joining (}x₁, y₁, z₁) and (x₂, y₂, z₂) is

_Explanation:

Let

So the foot of the perpendicular is (2λ + 1, 3λ + 2, 4λ + 3)

The direction ratios of the perpendicular is

(2λ + 1 - 5) : (3λ + 2 - 9) : (4λ + 3 - 3)

⇒ (2λ – 4) : (3λ – 7) : (4λ)

Direction ratio of the line is 2 : 3 : 4

From the direction ratio of the line and the direction ratio of its perpendicular, we have

2(2λ – 4) + 3(3λ – 7) + 4(4λ) = 0

⇒ 4λ – 8 + 9λ – 21 + 16λ = 0

⇒ 29λ = 29

⇒ λ = 1

So, the foot of the perpendicular is (3, 5, 7)

The foot of the perpendicular is the mid-point of the line joining (5, 9, 3) and (α, β, γ)

So, we have

So, the image is (1, 1, 11)

Given: Point (2, -1, 5)

Equation of line

The equation of line can be re-arranged as

The general point on this line is

(10r + 11, -4r – 2, -11r – 8)

Let N be the foot of the perpendicular drawn from the point P(2, 1, -5) on the given line.

Then, this point is N(10r + 11, -4r – 2, -11r – 8) for some fixed value of r.

D.r.’s of PN are (10r + 9, -4r - 3, -11r - 3)

D.r.’s of the given line is 10, -4, -11.

Since, PN is perpendicular to the given line, we have,

10(10r + 9) – 4(-4r – 3) – 11(-11r – 3) = 0

100r + 90 + 16r + 12 + 121r + 33 = 0

237r = 135

r

Then, the image of the point is

Therefore, the image is (0, 5, 1).

Given -

A = (2,1,3)

B = (5,0,5)

C = (-4,3,-1)

To prove – A, B and C are collinear

Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line PQ is given by ((a’-a),(b’-b),(c’-c))

The direction ratios of the line AB can be given by

((5-2),(0-1),(5-3))

=(3,-1,-2)

Similarly, the direction ratios of the line BC can be given by

((-4-5),(3-0),(-1-5))

=(-9,3,-6)

Tip – If it is shown that direction ratios of AB=λ times that of BC , where λ is any arbitrary constant, then the condition is sufficient to conclude that points A, B and C will be collinear.

So, d.r. of AB

=(3,-1,-2)

=(-1/3)Χ(-9,3,-6)

=(-1/3)Хd.r. of BC

Hence, A, B and C are collinear

Given -

A = (2,3,-4)

B = (1,-2,3)

C = (3,8,-11)

To prove – A, B and C are collinear

Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line PQ is given by ((a’-a),(b’-b),(c’-c))

The direction ratios of the line AB can be given by

((1-2),(-2-3),(3+4))

=(-1,-5,7)

Similarly, the direction ratios of the line BC can be given by

((3-1),(8+2),(-11-3))

=(2,10,-14)

Tip – If it is shown that direction ratios of AB=λ times that of BC , where λ is any arbitrary constant, then the condition is sufficient to conclude that points A, B and C will be collinear.

So, d.r. of AB

=(-1,-5,7)

=(-1/2)Χ(2,10,-14)

=(-1/2)Хd.r. of BC

Hence, A, B and C are collinear

Given -

A = (2,5,1)

B = (1,2,-1)

C = (3,λ,3)

To find – The value of λ so that A, B and C are collinear

Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line PQ is given by ((a’-a),(b’-b),(c’-c))

The direction ratios of the line AB can be given by

((1-2),(2-5),(-1-1))

=(-1,-3,-2)

Similarly, the direction ratios of the line BC can be given by

((3-1),(λ-2),(3+1))

=(2,λ-2,4)

Tip – If it is shown that direction ratios of AB=α times that of BC , where λ is any arbitrary constant, then the condition is sufficient to conclude that points A, B and C will be collinear.

So, d.r. of AB

=(-1,-3,-2)

=(-1/2)Χ(2,λ-2,4)

=(-1/2)Хd.r. of BC

Since, A, B and C are collinear,

Given -

A = (3,2,-4)

B = (9,8,-10)

C = (λ,μ,-6)

To find – The value of λ and μ so that A, B and C are collinear

Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line PQ is given by ((a’-a),(b’-b),(c’-c))

The direction ratios of the line AB can be given by

((9-3),(8-2),(-10+4))

=(6,6,-6)

Similarly, the direction ratios of the line BC can be given by

((λ-9),(μ-8),(-6+10))

=(λ-9,μ-8,4)

Tip – If it is shown that direction ratios of AB=α times that of BC , where λ is any arbitrary constant, then the condition is sufficient to conclude that points A, B and C will be collinear.

So, d.r. of AB

=(6,6,-6)

=(-6/4)Χ(-4,-4,4)

=(-3/2)Хd.r. of BC

Since, A, B and C are collinear,

And,

Given -

A = (-1,4,-2)

B = (λ,μ,1)

C = (0,2,-1)

To find – The value of λ and μ so that A, B and C are collinear

Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line PQ is given by ((a’-a),(b’-b),(c’-c))

The direction ratios of the line AB can be given by

((λ+1),(μ-4),(1+2))

=(λ+1,μ-4,3)

Similarly, the direction ratios of the line BC can be given by

((0-λ),(2-μ),(-1-1))

=(-λ,2-μ,-2)

Tip – If it is shown that direction ratios of AB=α times that of BC , where λ is any arbitrary constant, then the condition is sufficient to conclude that points A, B and C will be collinear.

So, d.r. of AB

=(λ+1,μ-4,3)

Say, α be an arbitrary constant such that d.r. of AB = α Х d.r. of BC

So, 3 = α Х (-2)

i.e. α = -3/2

Since, A, B and C are collinear,

And,

Given -

It can thus be written as:

A = (-4,2,-3)

B = (1,3,-2)

C = (-9,1,-4)

To prove – A, B and C are collinear

Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line PQ is given by ((a’-a),(b’-b),(c’-c))

The direction ratios of the line AB can be given by

((1+4),(3-2),(-2+3))

=(5,1,1)

Similarly, the direction ratios of the line BC can be given by

((-9-1),(1-3),(-4+2))

=(-10,-2,-2)

Tip – If it is shown that direction ratios of AB=λ times that of BC , where λ is any arbitrary constant, then the condition is sufficient to conclude that points A, B and C will be collinear.

So, d.r. of AB

=(5,1,1)

=(-1/2)Χ(-10,-2,-2)

=(-1/2)Хd.r. of BC

Hence, A, B and C are collinear

Given –

&

To find – Angle between the two pair of lines

Direction ratios of L₁ = (1,-1,-2)

Direction ratios of L₂ = (3,-5,-4)

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

The angle between the lines

Given –

&

To find – Angle between the two pair of lines

Direction ratios of L₁ = (1,0,3)

Direction ratios of L₂ = (-1,1,1)

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

The angle between the lines

Given –

&

To find – Angle between the two pair of lines

Direction ratios of L₁ = (2,-2,1)

Direction ratios of L₂ = (1,2,-2)

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

The angle between the lines

Given –

&

To find – Angle between the two pair of lines

Direction ratios of L₁ = (1,1,2)

Direction ratios of L₂ = (3,5,4)

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

The angle between the lines

Given –

&

To find – Angle between the two pair of lines

Direction ratios of L₁ = (4,3,5)

Direction ratios of L₂ = (1,-1,1)

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

The angle between the lines

Given –

&

To find – Angle between the two pair of lines

Direction ratios of L₁ = (2,1,-3)

Direction ratios of L₂ = (3,2,-1)

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

The angle between the lines

Given –

&

To find – Angle between the two pair of lines

Direction ratios of L₁ = (1,0,-1)

Direction ratios of L₂ = (3,4,5)

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

The angle between the lines

Given –

&

To find – Angle between the two pair of lines

Direction ratios of L₁ = (-3,-2,0)

Direction ratios of L₂ = (1,-3,2)

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

The angle between the lines

Given –

&

To prove – The lines are perpendicular to each other

Direction ratios of L₁ = (2,-3,4)

Direction ratios of L₂ = (2,4,2)

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

The angle between the lines

Hence, the lines are perpendicular to each other.

Given –

&

To find – The value of λ

Direction ratios of L₁ = (-3,2λ,2)

Direction ratios of L₂ = (3λ,1,-5)

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

Since the lines are perpendicular to each other,

The angle between the lines

Given –

&

To prove – The lines are perpendicular to each other

Direction ratios of L₁ = (2,-2,1)

Direction ratios of L₂ = (2,1,-2)

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

The angle between the lines

Hence, the lines are perpendicular to each other.

(i): Given – Direction ratios of L₁ = (2,1,2) & Direction ratios of L₂ = (4,8,1)

To find – Angle between the two pair of lines

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

The angle between the lines

(ii): Given – Direction ratios of L₁ = (5,-12,13) & Direction ratios of L₂ = (-3,4,5)

To find – Angle between the two pair of lines

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

The angle between the lines

(iii) Given – Direction ratios of L₁ = (1,1,2) & Direction ratios of L₂ = (√3-1,-√3-1,4)

To find – Angle between the two pair of lines

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

The angle between the lines

(iv) Given – Direction ratios of L₁ = (a,b,c) & Direction ratios of L₂ = ((b-c),(c-a),(a-b))

To find – Angle between the two pair of lines

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

The angle between the lines

Given -

A = (1,2,3)

B = (4,5,7)

C = (-4,3,-6)

D = (2,9,2)

Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line PQ is given by ((a’-a),(b’-b),(c’-c))

The direction ratios of the line AB can be given by

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

=(3,3,4)

Similarly, the direction ratios of the line CD can be given by

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

=(6,6,8)

To find – Angle between the two pair of lines AB and CD

Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by

The angle between the lines

Given equations:

To Find: d

Formula:

1. Cross Product :

If are two vectors

then,

2. Dot Product :

If are two vectors

then,

3. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

For given lines,

Here,

Therefore,

Now,

= 3 + 0 + 7

= 10

Therefore, the shortest distance between the given lines is

Given equations:

To Find: d

Formula:

1. Cross Product :

If are two vectors

then,

2. Dot Product :

If are two vectors

then,

3. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

For given lines,

Here,

Therefore,

Now,

= 6 + 60 – 4

= 62

Therefore, the shortest distance between the given lines is

Given equations:

To Find: d

Formula:

1. Cross Product :

If are two vectors

then,

2. Dot Product :

If are two vectors

then,

3. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

For given lines,

Here,

Therefore,

Now,

= - 27 + 9 + 27

= 9

Therefore, the shortest distance between the given lines is

Given equations:

To Find: d

Formula:

1. Cross Product :

If are two vectors

then,

2. Dot Product :

If are two vectors

then,

3. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

For given lines,

Here,

Therefore,

= 3√2

Now,

= - 3 + 0 – 6

= - 9

Therefore, the shortest distance between the given lines is

Given equations:

To Find: d

Formula:

1. Cross Product :

If are two vectors

then,

2. Dot Product :

If are two vectors

then,

3. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

For given lines,

Here,

Therefore,

Now,

= 12 – 28 + 0

= - 16

Therefore, the shortest distance between the given lines is

Given equations:

To Find: d

Formula:

1. Cross Product :

If are two vectors

then,

2. Dot Product :

If are two vectors

then,

3. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

For given lines,

Here,

Therefore,

Now,

= 150 + 4 + 18

= 172

Therefore, the shortest distance between the given lines is

Given equations:

To Find: d

Formula:

1. Cross Product :

If are two vectors

then,

2. Dot Product :

If are two vectors

then,

3. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

Given lines,

Above equations can be written as

Here,

Therefore,

Now,

= - 2 – 33 + 0

= - 35

Therefore, the shortest distance between the given lines is

Given equations:

To Find: d

Formula:

1. Cross Product :

If are two vectors

then,

2. Dot Product :

If are two vectors

then,

3. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

Given lines,

Above equations can be written as

Here,

Therefore,

Now,

= 6 + 0 + 9

= 15

Therefore, the shortest distance between the given lines is

Given equations:

To Find: d

Formula:

1. Cross Product :

If are two vectors

then,

2. Dot Product :

If are two vectors

then,

3. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

For given lines,

Here,

Therefore,

Now,

= - 1 + 0 + 0

= - 1

Therefore, the shortest distance between the given lines is

As d ≠ 0

Hence, the given lines do not intersect.

Given equations:

To Find: d

Formula:

1. Cross Product :

If are two vectors

then,

2. Dot Product :

If are two vectors

then,

3. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

For given lines,

Here,

Therefore,

Now,

= - 68 + 256 + 0

= 188

Therefore, the shortest distance between the given lines is

As d ≠ 0

Hence, the given lines do not intersect.

Given equations:

To Find: d

Formula:

1. Cross Product :

If are two vectors

then,

2. Dot Product :

If are two vectors

then,

3. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

For given lines,

Here,

Therefore,

Now,

= 0 + 12 – 6

= 6

Therefore, the shortest distance between the given lines is

As d ≠ 0

Hence, the given lines do not intersect.

Given equations:

To Find: d

Formula:

1. Cross Product :

If are two vectors

then,

2. Dot Product :

If are two vectors

then,

3. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

For given lines,

Here,

Therefore,

Now,

= - 15 – 18 + 33

= 0

Therefore, the shortest distance between the given lines is

As d = 0

Hence, the given lines not intersect each other.

Now, to find point of intersection, let us convert given vector equations into Cartesian equations.

For that substituting in given equations,

General point on L1 is

x₁ = 2λ+1 , y₁ = 3λ+2 , z₁ = 4λ+3

let, P(x₁, y₁, z₁) be point of intersection of two given lines.

Therefore, point P satisfies equation of line L2.

⇒ 4λ – 6 = 15λ + 5

⇒ 11λ = -11

⇒ λ = -1

Therefore, x₁ = 2(-1)+1 , y₁ = 3(-1)+2 , z₁ = 4(-1)+3

⇒ x₁ = -1 , y₁ = -1 , z₁ = -1

Hence point of intersection of given lines is (-1, -1, -1).

Given equations:

To Find: d

Formula:

1. Cross Product :

If are two vectors

then,

2. Dot Product :

If are two vectors

then,

3. Shortest distance between two parallel lines :

The shortest distance between the parallel lines and

is given by,

Answer:

For given lines,

Here,

As , given lines are parallel to each other.

Therefore,

= 7

Therefore, the shortest distance between the given lines is

Given equations:

To Find: d

Formula:

1. Cross Product :

If are two vectors

then,

2. Dot Product :

If are two vectors

then,

3. Shortest distance between two parallel lines :

The shortest distance between the parallel lines and

is given by,

Answer:

For given lines,

Here,

Therefore, the shortest distance between the given lines is

Given: point A ≡ (2, 3, 2)

Equation of line :

To Find: i) equation of line

ii) distance d

Formulae:

1. Equation of line :

Equation of line passing through point A (a₁, a₂, a₃) and parallel to vector is given by

Where,

2. Cross Product :

If are two vectors

then,

3. Dot Product :

If are two vectors

then,

4. Shortest distance between two parallel lines :

The shortest distance between the parallel lines and

is given by,

Answer:

As the required line is parallel to the line

Therefore, the vector parallel to the required line is

Given point A ≡ (2, 3, 2)

Therefore, equation of line passing through A and parallel to is

Now, to calculate distance between above line and given line,

Here,

= 7

Therefore, the shortest distance between the given lines is

Given: Cartesian equations of lines

To Find: i) vector equations of given lines

ii) distance d

Formulae:

1. Equation of line :

Equation of line passing through point A (a₁, a₂, a₃) and having direction ratios (b₁, b₂, b₃) is

Where,

And

2. Cross Product :

If are two vectors

then,

3. Dot Product :

If are two vectors

then,

4. Shortest distance between two parallel lines :

The shortest distance between the parallel lines and

is given by,

Answer:

Given Cartesian equations of lines

Line L1 is passing through point (1, 2, -4) and has direction ratios (2, 3, 6)

Therefore, vector equation of line L1 is

And

Line L2 is passing through point (3, 3, -5) and has direction ratios (4, 6, 12)

Therefore, vector equation of line L2 is

Now, to calculate distance between the lines,

Here,

As , given lines are parallel to each other.

Therefore,

= 7

Therefore, the shortest distance between the given lines is

Given: Cartesian equations of lines

To Find: i) vector equations of given lines

ii) distance d

Formulae:

1. Equation of line :

Equation of line passing through point A (a₁, a₂, a₃) and having direction ratios (b₁, b₂, b₃) is

Where,

And

2. Cross Product :

If are two vectors

then,

3. Dot Product :

If are two vectors

then,

4. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

Given Cartesian equations of lines

Line L1 is passing through point (1, 2, 3) and has direction ratios (2, 3, 4)

Therefore, vector equation of line L1 is

And

Line L2 is passing through point (2, 3, 5) and has direction ratios (3, 4, 5)

Therefore, vector equation of line L2 is

Now, to calculate distance between the lines,

Here,

Therefore,

Now,

= - 2 + 2 - 2

= -2

Therefore, the shortest distance between the given lines is

Given: Cartesian equations of lines

To Find: distance d

Formulae:

1. Equation of line :

Equation of line passing through point A (a₁, a₂, a₃) and having direction ratios (b₁, b₂, b₃) is

Where,

And

2. Cross Product :

If are two vectors

then,

3. Dot Product :

If are two vectors

then,

4. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

Given Cartesian equations of lines

Line L1 is passing through point (1, -2, 3) and has direction ratios (-1, 1, -2)

Therefore, vector equation of line L1 is

And

Line L2 is passing through point (1, -1, -1) and has direction ratios (2, 2, -2)

Therefore, vector equation of line L2 is

Now, to calculate distance between the lines,

Here,

Therefore,

Now,

= 0 - 6 + 16

= 10

Therefore, the shortest distance between the given lines is

Given: Cartesian equations of lines

To Find: distance d

Formulae:

1. Equation of line :

Equation of line passing through point A (a₁, a₂, a₃) and having direction ratios (b₁, b₂, b₃) is

Where,

And

2. Cross Product :

If are two vectors

then,

3. Dot Product :

If are two vectors

then,

4. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

Given Cartesian equations of lines

Line L1 is passing through point (12, 1, 5) and has direction ratios (-9, 4, 2)

Therefore, vector equation of line L1 is

And

Line L2 is passing through point (23, 10, 23) and has direction ratios (-6, -4, 3)

Therefore, vector equation of line L2 is

Now, to calculate distance between the lines,

Here,

Therefore,

= 65

Now,

= 220 + 135 + 1080

= 1435

Therefore, the shortest distance between the given lines is

Given: Cartesian equations of lines

Formulae:

1. Condition for perpendicularity :

If line L1 has direction ratios (a₁, a₂, a₃) and that of line L2 are (b₁, b₂, b₃) then lines L1 and L2 will be perpendicular to each other if

2. Distance formula :

Distance between two points A≡(a₁, a₂, a₃) and B≡(b₁, b₂, b₃) is given by,

3. Equation of line :

Equation of line passing through points A≡(x₁, y₁, z₁) and B≡(x₂, y₂, z₂) is given by,

Answer:

Given equations of lines

Direction ratios of L1 and L2 are (3, -1, 1) and (-3, 2, 4) respectively.

Let, general point on line L1 is P≡(x₁, y₁, z₁)

x₁ = 3s+3 , y₁ = -s+8 , z₁ = s+3

and let, general point on line L2 is Q≡(x₂, y₂, z₂)

x₂ = -3t – 3 , y₂ = 2t – 7 , z₂ = 4t + 6

Direction ratios of are ((-3t – 3s - 6), (2t + s - 15), (4t – s + 3))

PQ will be the shortest distance if it perpendicular to both the given lines

Therefore, by the condition of perpendicularity,

3(-3t – 3s - 6) – 1(2t + s - 15) + 1(4t – s + 3) = 0 and

-3(-3t – 3s - 6) + 2(2t + s - 15) + 4(4t – s + 3) = 0

⇒ -9t – 9s – 18 – 2t – s + 15 + 4t – s + 3 = 0 and

9t + 9s + 18 + 4t + 2s – 30 + 16t – 4s + 12 = 0

⇒ -7t – 11s = 0 and

29t + 7s = 0

Solving above two equations, we get,

t = 0 and s = 0

therefore,

P ≡ (3, 8, 3) and Q ≡ (-3, -7, 6)

Now, distance between points P and Q is

Therefore, the shortest distance between two given lines is

Now, equation of line passing through points P and Q is,

Therefore, equation of line of shortest distance between two given lines is

Given: Cartesian equations of lines

Formulae:

1. Condition for perpendicularity :

If line L1 has direction ratios (a₁, a₂, a₃) and that of line L2 are (b₁, b₂, b₃) then lines L1 and L2 will be perpendicular to each other if

2. Distance formula :

Distance between two points A≡(a₁, a₂, a₃) and B≡(b₁, b₂, b₃) is given by,

3. Equation of line :

Equation of line passing through points A≡(x₁, y₁, z₁) and B≡(x₂, y₂, z₂) is given by,

Answer:

Given equations of lines

Direction ratios of L1 and L2 are (-1, 2, 1) and (1, 3, 2) respectively.

Let, general point on line L1 is P≡(x₁, y₁, z₁)

x₁ = -s+3 , y₁ = 2s+4 , z₁ = s-2

and let, general point on line L2 is Q≡(x₂, y₂, z₂)

x₂ = t+1 , y₂ = 3t – 7 , z₂ = 2t - 2

Direction ratios of are ((t + s – 2), (3t – 2s – 11), (2t – s))

PQ will be the shortest distance if it perpendicular to both the given lines

Therefore, by the condition of perpendicularity,

-1(t + s – 2) + 2(3t – 2s – 11) + 1(2t – s) = 0 and

1(t + s – 2) + 3(3t – 2s – 11) + 2(2t – s) = 0

⇒ - t – s + 2 + 6t – 4s – 22 + 2t – s = 0 and

t + s – 2 + 9t – 6s – 33 + 4t – 2s = 0

⇒ 7t – 6s = 20 and

14t - 7s = 35

Solving above two equations, we get,

t = 2 and s = -1

therefore,

P ≡ (4, 2, -3) and Q ≡ (3, -1, 2)

Now, distance between points P and Q is

Therefore, the shortest distance between two given lines is

Now, equation of line passing through points P and Q is,

Therefore, equation of line of shortest distance between two given lines is

Given: Cartesian equations of lines

Formulae:

1. Condition for perpendicularity :

If line L1 has direction ratios (a₁, a₂, a₃) and that of line L2 are (b₁, b₂, b₃) then lines L1 and L2 will be perpendicular to each other if

2. Distance formula :

Distance between two points A≡(a₁, a₂, a₃) and B≡(b₁, b₂, b₃) is given by,

3. Equation of line :

Equation of line passing through points A≡(x₁, y₁, z₁) and B≡(x₂, y₂, z₂) is given by,

Answer:

Given equations of lines

Direction ratios of L1 and L2 are (2, 1, -3) and (2, -7, 5) respectively.

Let, general point on line L1 is P≡(x₁, y₁, z₁)

x₁ = 2s-1 , y₁ = s+1 , z₁ = -3s+9

and let, general point on line L2 is Q≡(x₂, y₂, z₂)

x₂ = 2t+3 , y₂ = -7t – 15 , z₂ = 5t + 9

Direction ratios of are ((5t - 2s + 10), (-7t – s – 16), (5t + 3s))

PQ will be the shortest distance if it perpendicular to both the given lines

Therefore, by the condition of perpendicularity,

2(5t - 2s + 10) + 1(-7t – s – 16) - 3(5t + 3s) = 0 and

2(5t - 2s + 10) – 7(-7t – s – 16) + 5(5t + 3s) = 0

⇒ 10t – 4s + 20 - 7t – s - 16 - 15t – 9s = 0 and

10t - 4s + 20 + 49t + 7s + 112 + 25t + 15s = 0

⇒ -12t – 14s = -4 and

84t + 18s = -132

Solving above two equations, we get,

t = -2 and s = 2

therefore,

P ≡ (3, 3, 3) and Q ≡ (-1, -1, -1)

Now, distance between points P and Q is

Therefore, the shortest distance between two given lines is

Now, equation of line passing through points P and Q is,

⇒

Therefore, equation of line of shortest distance between two given lines is

x = y = z

Given: Cartesian equations of lines

Formulae:

1. Condition for perpendicularity :

If line L1 has direction ratios (a₁, a₂, a₃) and that of line L2 are (b₁, b₂, b₃) then lines L1 and L2 will be perpendicular to each other if

2. Distance formula :

Distance between two points A≡(a₁, a₂, a₃) and B≡(b₁, b₂, b₃) is given by,

3. Equation of line :

Equation of line passing through points A≡(x₁, y₁, z₁) and B≡(x₂, y₂, z₂) is given by,

Answer:

Given equations of lines

Direction ratios of L1 and L2 are (3, -1, 1) and (-3, 2, 4) respectively.

Let, general point on line L1 is P≡(x₁, y₁, z₁)

x₁ = 3s+6 , y₁ = -s+7 , z₁ = s+4

and let, general point on line L2 is Q≡(x₂, y₂, z₂)

x₂ = -3t , y₂ = 2t – 9 , z₂ = 4t + 2

Direction ratios of are ((-3t - 3s - 6), (2t + s – 16), (4t – s – 2))

PQ will be the shortest distance if it perpendicular to both the given lines

Therefore, by the condition of perpendicularity,

3(-3t - 3s - 6) - 1(2t + s – 16) + 1(4t – s – 2) = 0 and

-3(-3t - 3s - 6) + 2(2t + s – 16) + 4(4t – s – 2) = 0

⇒ -9t - 9s - 18 - 2t – s + 16 + 4t – s – 2 = 0 and

9t + 9s + 18 + 4t + 2s – 32 + 16t – 4s – 8 = 0

⇒ -7t – 11s = 4 and

29t + 7s = -22

Solving above two equations, we get,

t = 1 and s = -1

therefore,

P ≡ (3, 8, 3) and Q ≡ (-3, -7, 6)

Now, distance between points P and Q is

Therefore, the shortest distance between two given lines is

Now, equation of line passing through points P and Q is,

Therefore, equation of line of shortest distance between two given lines is

Given: Cartesian equations of lines

To Find: distance d

Formulae:

1. Equation of line :

Equation of line passing through point A (a₁, a₂, a₃) and having direction ratios (b₁, b₂, b₃) is

Where,

And

2. Cross Product :

If are two vectors

then,

3. Dot Product :

If are two vectors

then,

4. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

Given Cartesian equations of lines

Line L1 is passing through point (0, 2, -3) and has direction ratios (1, 2, 3)

Therefore, vector equation of line L1 is

And

Line L2 is passing through point (2, 6, 3) and has direction ratios (2, 3, 4)

Therefore, vector equation of line L2 is

Now, to calculate distance between the lines,

Here,

Therefore,

Now,

= - 2 + 8 – 6

= 0

Therefore, the shortest distance between the given lines is

As d = 0

Hence, given lines intersect each other.

Now, general point on L1 is

x₁ = λ , y₁ = 2λ+2 , z₁ = 3λ-3

let, P(x₁, y₁, z₁) be point of intersection of two given lines.

Therefore, point P satisfies equation of line L2.

⇒ 3λ – 6 = 4λ – 8

⇒ λ = 2

Therefore, x₁ = 2 , y₁ = 2(2)+2 , z₁ = 3(2)-3

⇒ x₁ = 2 , y₁ = 6 , z₁ = 3

Hence point of intersection of given lines is (2, 6, 3).

Given: Cartesian equations of lines

To Find: distance d

Formulae:

1. Equation of line :

Equation of line passing through point A (a₁, a₂, a₃) and having direction ratios (b₁, b₂, b₃) is

Where,

And

2. Cross Product :

If are two vectors

then,

3. Dot Product :

If are two vectors

then,

4. Shortest distance between two lines :

The shortest distance between the skew lines and

is given by,

Answer:

Given Cartesian equations of lines

Line L1 is passing through point (1, -1, 1) and has direction ratios (3, 2, 5)

Therefore, vector equation of line L1 is

And

Line L2 is passing through point (2, 1, -1) and has direction ratios (2, 3, -2)

Therefore, vector equation of line L2 is

Now, to calculate distance between the lines,

Here,

Therefore,

Now,

= - 19 + 32 – 10

= 3

Therefore, the shortest distance between the given lines is

As d ≠ 0

Hence, given lines do not intersect each other.

Given : A line has direction ratios 2, -1, -2

To find : Direction cosines of the line

Formula used : If (l,m,n) are the direction ratios of a given line then direction cosines are given by , ,

Here l = 2 , m = -1 , n = -2

Direction cosines of the line with direction ratios 2, -1, -2 is

, ,

= , , = , ,

= , ,

Direction cosines of the line with direction ratios 2, -1, -2 is , ,

Given : A line

To find : Direction cosines of the line

Formula used : If a line is given by = = then direction cosines are given by , ,

The line is = =

Here l = -2 , m = 6 , n = -3

Direction cosines of the line = = is

, ,

= , , = , ,

= , ,

Direction cosines of the line = = is , ,

Given : A line

To find : Direction cosines of the line parallel to

Formula used : If a line is given by = = then direction cosines are given by , ,

The line is = =

Parallel lines have same direction ratios and direction cosines

Here l = 3 , m = -2 , n = 6

Direction cosines of the line = = is

, ,

= , , = , ,

= , ,

Direction cosines of the line parallel to the line = = is

, ,