Straight Line In Space

vector equation of a line is

here &

so the vector equation of the line is

the Cartesian equation of the line is

where are the coordinates of the fixed-point ‘a’, and are coordinates of .

so, the cartesian equation becomes as follows

the direction ratios of the required line are

(3–(–1), 4–0, 6–2) = (4, 4, 4)

since the line passes through (–1, 0, 2)

the vector equation of the line,

the vector equation of a line that passes through a fixed point and is parallel to the vector is

the Cartesian equation of the line is

where are the coordinates of the fixed-point ‘a’, and are coordinates of .

so, the cartesian equation becomes as follows

The vector equation of a line passing through a fixed-point ‘a’ and having directions parallel to is given by

Here

the Cartesian equation of the line is

where are the coordinates of the fixed-point ‘a’, and are coordinates of .

so, the cartesian equation becomes as follows

Given: the vectors of point A = , B = and

C =

this means that the vector equation of line AB is given by

where and

so the vector equation of AB is

now the vector equation of BC is given as

where and

therefore the vector equation of BC is given as

The conceptof the question:

the vector equation of the diagonal BD is given by

and the vector equation of diagonal AC is given by

……(1)

can be written as

…..(2)

Comparing 1&2

x = –2–, y = –8 + 13, z = 14–17

Therefore, a cartesian equation of the required line is

the vector of point A can be written as and that of point B can be written as

vector equation of line AB is given by

Here and

⇒

Here x = 1 +

⇒

Hence the cartesian equation of the line is

here the vector for point A(1, 2, 3) is and the vector parallel to which the line is required is

vector equation of a line is

here &

so the vector equation of the line is

Here x = 1 + , y = 2–2, z = 3 + 3

Hence the cartesian equation of the above line is

the direction ratios of the line

Are‹

So in this question the direction ratios of the given line are ‹2, 7, –3›

So the vector equation of

Here

vector equation of a line is

so the vector equation of the line is

Here x = 2 + 2, y = –1 + 7, z = 1–3

Hence the cartesian equation of the above line is

Cartesian equations of a line is

Let this be equal to

….(1)

….(2)

Hence comparing 1&2

cartesian equation of the given line is

Hence its direction ratios are‹1, 2, –2›

The cartesian equation of the line is given by

the point is (1, –1, 2) and the direction ratios are ‹1, 2, –2›

The cartesian equation of the line is

Let this be equal to

….(1)

….(2)

Hence comparing 1&2

The vector equation of the line is given as

the equation of the line is

But to make it cartesian equation the coefficient of x, y, z must be one so the above equation becomes as

Now the direction ratios of this line is ‹–2, 6, –3›

Concept of the question

Direction cosines of line

Are given as

Here

Hence direction cosines of the above line are

cartesian equation of the line is

Let this be equal to

Hence ….(1)

….(2)

Hence comparing 1&2

the cartesian equation of the line can be written as

We have written the above equation ion this way because the coefficients of x, y, z must be 1

Therefore the direction ratios of the above line are ‹a, 1, c›

Let

Therefore x = b + a, y = , z = d + c….(1)

….(2)

Hence comparing 1&2

The vector equation of the line is given as

direction ratios of the line joining the points with position vector and is ‹(1–1), (–1–1), (4–2)›

i. e. ‹0, –2, 2›

the vector equation of the line passing through the point with position vector and having direction ratio ‹0, –2, 2› is given as

Here x = 1, y = –2–2, z = –3 + 2

⇒

hence the cartesian equation of the line is

let Q be the point on the given line

It is of the form (3–2, 2–1, 2 + 3)

By letting

distance of point P from PQ =

So the points on the line are ((3(0)–2), (2(0)–1), (2(0) + 3)) and (3(2)–2, 2(2)–1, 2(2) + 3)

i.e. (–2, –1, 3) and (4, 3, 7)

let the given points are A, B, C with position vectors respectively so

We know that equation of line passing through is

….(1)

If A, B, C are collinear then C must satisfy eq. (1)

On comparing the coefficient of on both sides of the equation for all

Hence the points A, B, C are collinear.

Firstly, we need to write the cartesian form of the given line i.e. of

Cartesian equation is

So the direction ratios of the given line are ‹–1, 7, 3/2›

Vector Equation of line passing through the point (1, 2, 3) i.e. position vector and having direction ratios ‹–1, 7, 3/2› is

And the cartesian equation is

The equation of line is

3x + 1 = 6y – 2 = 1 – z

This can be written as

To make it a cartesian equation we need to make the coefficient of x, y, z to be 1

Therefore, the cartesian equation is

….(1)

the Cartesian equation of the line is

….(2)

Therefore on comparing (1)&(2)

The fixed point through which it passes is

And the direction ratios are

Let

…(3)

….(4)

Hence, comparing 3&4 we get,

firstly we need to resolve the given line

The given line can be written as

We need to make the coefficients of x, y, z equal to 1

Therefore the line becomes as follows

Therefore the of the given line are

Vector equation of line passing through the point A(1, 2, –1) and having

l, m, n are generally written as direction cosines or the direction ratios of unit vector-*,

Let l₁ = , m₁ = , n₁ =

l₂ = , m₂ = , n₂ =

l₃ = , m₃ = , n₃ =

For the lines or vectors to be perpendicular their dot product or scalar product should be zero and for the lines or vectors to be parallel their cross product or vector product should be zero.

So we will use scalar product to prove these lines perpendicular to each other.

l₁l₂ + m₁m₂ + n₁n₂ =

l₂l₃ + m₂m₃ + n₂n₃ =

l₁l₃ + m₁m₃ + n₁n₃ =

Therefore, all three lines or vectors are mutually perpendicular to each other.

The direction ratios of a line can be found by subtracting the corresponding coordinates of two points through which the line passes i.e. (subtract x coordinates, subtract y coordinates, subtract z coordinates), this is the direction ratio of the line. There can be no direction ratio of a line passing through only one point, there should be at least two points.

The direction ratios of a line passing through the points (1,–1,2) and (3,4,–2) are,

(3–1,4–{–1},–2–2) = (2,5,–4)

Or it can also be the other way you can choose the first and the second point of your own choice.

The direction ratios of a line passing through the points (0,3,2) and (3,5,6) are,

(3–0,5–3,6–2) = (3,2,4)

The direction ratios of lines are,

(a₁,b₁,c₁) = (2,5,–4)

(a₂,b₂,c₂) = (3,2,4)

By using dot product.

cos θ = 0

θ =

Therefore,, the lines are perpendicular.

The direction ratios of a line can be found by subtracting the corresponding coordinates of two points through which the line passes i.e. (subtract x coordinates, subtract y coordinates, subtract z coordinates), this is the direction ratio of the line. There can be no direction ratio of a line passing through only one point, there should be at least two points.

For the two lines to be parallel or anti parallel to each other the fraction of their corresponding direction ratios should be equal.

The direction ratios of a line passing through the points (4,7,8) and (2,3,4) are (4–2,7–3,8–4) = (2,4,4)

The direction ratios of a line passing through the points (–1,–2,1) and (1,2,5) are (–1–1,–2–2,1–5) = (–2,–4,–4)

The direction ratios are proportional.

Hence the lines are mutually parallel and even overlapping each other because of the constant –1 or 1.

The Cartesian equation or the symmetrical form of equation is the one which is of the form

Where (x₁,y₁,z₁) is the point through which the line passes and a,b,c are the directional ratios of the line or the directional ratios of the line are proportional to them.

The Cartesian equation of a line passing through the point (–2,4,–5) and parallel to the line

is .

The Cartesian equation of the lines are and , as we know their direction ratios we can find weather they are perpendicular or not, for proving the lines to be perpendicular we can only consider the numerator of the dot product when we use cosθ= numerator

cos θ = 7×1 + 2×(–5) + 1×3

cos θ = 7 – 10 + 3

cos θ = 0

θ =

Therefore, the lines are perpendicular.

The direction ratios of a line can be found by subtracting the corresponding coordinates of two points through which the line passes i.e. (subtract x coordinates, subtract y coordinates, subtract z coordinates), this is the direction ratio of the line. There can be no direction ratio of a line passing through only one point, there should be at least two points.

The direction ratios of a line joining the origin to the point (2,1,1) are (2–0,1–0,1–0) = (2,1,1)

The direction ratios of a line joining the points (3,5,–1) and (4,3,–1) are (4–3,3–5,–1–{–1}) = (1,–2,0)

By using the dot product we can find the angle between the two lines,

cos θ =

cos θ = 0

θ =

Therefore, the lines are mutually perpendicular.

Vector equation of a line is where is the position vector of the point a through which our line passes through and is the vector parallel to our line and is the general vector of a line satisfying these conditions and is a constant.

The direction cosines of the x-axis are (1,0,0), direction cosines are the direction ratios of a unit vector.

The equation of a line parallel to x-axis and passing through the origin is,

= (0+0+0) + λ(1+0+0)

= λ

Therefore, this is the family of lines satisfying the condition of the question.

We know that angle between two lines given with their vector equation is the angle between their parallel vectors always.

By using dot product to find the angle between the lines using their parallel vectors .

= (4) + λ(+2–2)

= (+2) – μ(2+4–4)

= 4 – , = +2–2

= – + 2, = 2 + 4 – 4

= = 3

= = 6

Let θ be the angle between given lines. So using dot product we can find the angle.

cos θ =

=

= = 1

cos θ = 1

θ = 0^o

We know that angle between two lines given with their vector equation is the angle between their parallel vectors always.

By using dot product to find the angle between the lines using their parallel vectors .

= (3+2–4)+λ(+2+2)

= (5–2)+μ(3+2+6)

= 3+2–4, = +2+2

= 5–2, = 3+2+6

= = 3

= = 7

Let θ be the angle between given lines, so using the dot product,

=

= = 19

θ =

We know that angle between two lines given with their vector equation is the angle between their parallel vectors always.

By using dot product to find the angle between the lines using their parallel vectors .

= λ(++2)

= 2+μ[(√3 – 1)–(√3 + 1)+4]

= ++2, = (√3 – 1)–(√3 + 1)+4

= = √6

= = √24 = 2√6

Let θ be the angle between given lines, so using the dot product equation,

=

cos θ =

θ =

In the Cartesian or symmetrical form of equation the angle between two lines can be found by dot product equation and in this equation we will use the direction ratios which are in the denominator of the equation. in this equation a,b,c are the direction ratios of this equation.

Equations of the given lines are,

and

a₁ = 3, b₁ = 5, c₁ = 4 ; a₂ = 1, b₂ = 1, c₂ = 2

now to find the angle between two lines we use cross product equation,

cos θ =

cos θ =

cos θ =

θ =

In the Cartesian or symmetrical form of equation the angle between two lines can be found by dot product equation and in this equation we will use the direction ratios which are in the denominator of the equation. in this equation a,b,c are the direction ratios of this equation.

Equations of the given lines are,

and

a₁ = 2, b₁ = 3, c₁ = –3 ; a₂ = –1, b₂ = 8, c₂ = 4

now to find the angle between two lines we use cross product equation,

cos θ =

cos θ =

θ =

In the Cartesian or symmetrical form of equation the angle between two lines can be found by dot product equation and in this equation we will use the direction ratios which are in the denominator of the equation. in this equation a,b,c are the direction ratios of this equation.

Equations of the given lines are,

and these are not in the standard form but after converting them, we get,

and

a₁ = 2, b₁ = 1, c₁ = –3 ; a₂ = 3, b₂ = 2, c₂ = –1

now to find the angle between two lines we use cross product equation,

cos θ =

cos θ =

θ =

In the Cartesian or symmetrical form of equation the angle between two lines can be found by dot product equation and in this equation we will use the direction ratios which are in the denominator of the equation. in this equation a,b,c are the direction ratios of this equation.

Equations of the given lines are,

, z = 5 and these are not in the standard form but after converting them, we get,

and

a₁ = 3, b₁ = –2, c₁ = 0 ; a₂ = 1, b₂ = , c₂ = 2

now to find the angle between two lines we use cross product equation,

cos θ =

cos θ =

cos θ = 0

θ =

In the Cartesian or symmetrical form of equation the angle between two lines can be found by dot product equation and in this equation we will use the direction ratios which are in the denominator of the equation. in this equation a,b,c are the direction ratios of this equation.

Equations of the given lines are,

and these are not in the standard form but after converting them, we get,

and

Given there are two vectors which are parallel to these lines,

= 1–1+1, = 3+4+5

They are parallel because they have same direction ratios.

So angle between the lines is the angle between these vectors which are parallel to the lines.

By using dot product equation,

cos θ =

cos θ =

cos θ =

cos θ =

cos θ =

θ =

In the Cartesian or symmetrical form of equation the angle between two lines can be found by dot product equation and in this equation we will use the direction ratios which are in the denominator of the equation. in this equation a,b,c are the direction ratios of this equation.

Equations of the given lines are,

and

Given there are two vectors which are parallel to these lines,

= 2+7–3, = –1+2+4

They are parallel because they have same direction ratios.

So angle between the lines is the angle between these vectors which are parallel to the lines.

By using dot product equation,

cos θ =

cos θ =

cos θ =

cos θ = 0

θ =

In the Cartesian or symmetrical form of equation the angle between two lines can be found by dot product equation and in this equation we will use the direction ratios which are in the denominator of the equation. in this equation a,b,c are the direction ratios of this equation.

Equations of the given lines are,

and

We are given with the direction ratios and we have to find the angle between the lines having these direction ratios.

a₁ = 5, b₁ = –12, c₁ = 13 ; a₂ = –3, b₂ = 4, c₂ = 5

By using dot product equation, we get

cos θ =

cos θ =

cos θ =

θ =

In the Cartesian or symmetrical form of equation the angle between two lines can be found by dot product equation and in this equation we will use the direction ratios which are in the denominator of the equation. in this equation a,b,c are the direction ratios of this equation.

Equations of the given lines are,

and

We are given with the direction ratios and we have to find the angle between the lines having these direction ratios.

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

By using dot product equation, we get

cos θ =

cos θ =

cos θ =

θ =

In the Cartesian or symmetrical form of equation the angle between two lines can be found by dot product equation and in this equation we will use the direction ratios which are in the denominator of the equation. in this equation a,b,c are the direction ratios of this equation.

Equations of the given lines are,

and

We are given with the direction ratios and we have to find the angle between the lines having these direction ratios.

a₁ = 1, b₁ = 2, c₁ = –2 ; a₂ = –2, b₂ = 2, c₂ = 1

By using dot product equation, we get

cos θ =

cos θ =

cos θ = 0

θ =

We are given with the direction ratios and the vector equations of two lines, we have to find the angle between the two lines, and weather they are parallel or perpendicular to each other.

The direction ratios are a,b,c and b–c, c–a, a–b .

The vectors with these direction ratios are,

= a+b+c and = (b–c)+(c–a)+(a–b)

By using dot product equation to find the angle between them, we get,

cos θ =

cos θ =

cos θ =

cos θ =

cos θ = 0

θ =

In the Cartesian or symmetrical form of equation the angle between two lines can be found by dot product equation and in this equation we will use the direction ratios which are in the denominator of the equation. in this equation a,b,c are the direction ratios of this equation.

Direction ratios of the first line are a₁ = 2, b₁ = 2, c₁ = 1 which corresponds to 2,2,1.

Direction ratios of the line joining (3,1,4) and (7,2,12) .

= (7–3,2–1,12–4) = (4,1,8) .

a₂ = 4, b₂ = 1, c₂ = 8

now to find the angle between two lines we use cross product equation,

cos θ =

θ =

The Cartesian equation of a line passing through a point (x₁, y₁, z₁) and having directional ratios proportional to a,b,c is given by,

Now the point (x₁, y₁, z₁) = (1,2,–4) and the required line is parallel to a given line , now as we know that if two lines are parallel and direction ratios of one line are a,b,c then the direction ratios of other lines will be ka,kb,kc where k is a constant and which gets cancelled when we put these direction ratios in the equation of the required line.

So the direction ratios of the required line are ;

a = 4λ, b = 2λ, c = 3λ

hence the equation of the required line is,

The Cartesian equation of a line passing through a point (x₁, y₁, z₁) and having directional ratios proportional to a,b,c is given by,

Now the point (x₁, y₁, z₁) = (–1,2,1) and the required line is parallel to a given line which is not in the general form of the Cartesian equation because the coefficients of x,y,z in the Cartesian equation are 1, so the equation will reduce to the form now as we know that if two lines are parallel and direction ratios of one line are a,b,c then the direction ratios of other lines will be ka,kb,kc where k is a constant and which gets cancelled when we put these direction ratios in the equation of the required line.

So the direction ratios of the required line are ;

a = 2λ, b = λ, c = –3λ

hence the equation of the required line is,

Vector equation of a line is where is the position vector of the point a through which our line passes through and is the vector parallel to our line and is the general vector of a line satisfying these conditions and is a constant.

Now the point vector through which the line passes is = 2+3 and the required line is parallel to a line having vector equation,

= (–2+) + λ(2+3–5)

The parallel vector is,

= (2+3–5)

So the vector equation of the required line is,

= (2+3) + μ(2+3–5)

Where μ is a constant or a scalar.

The Cartesian equation of a line passing through a point (x₁, y₁, z₁) and having directional ratios proportional to a,b,c is given by,

The required line passes through the point (2,1,3), now we need to find the direction ratios of the line which are a,b,c . this equation of the required line is,

We are given with the Cartesian equation of the two lines which are perpendicular to the given equation, as the lines are perpendicular with the required line so the dot product will result in zero.

The first line is , the dot product equation is,

a×1 + b×2 + c×3 = 0

a+2b+3c = 0 ………(i).

The second line is , the dot product equation is,

a×(–3) + b×2 + c×5 = 0

–3a+2b+5c = 0 ……….(ii).

Solving equations (i) and (ii), we get, by cross multiplication method,

= λ

= λ

= λ

a = 2λ, b = –7λ, c = 4λ

using a,b,c in the required equation we get,

This is the required equation.

The vector equation of a line passing through a point with position vector and perpendicular to two lines with vector equations = + λ and = + λ, is given by = + λ() because as we know that vector product between two vectors will give you a vector whose direction is perpendicular to both the vectors, so the required equation is parallel to this vector and hence this is the equation.

Now we are given with,

= (+–3) and the lines perpendicular to required line are,

= +λ(2+–3) and = (2+)+μ(++)

= 2+–3 and = ++

Now,

=

=

=

Therefore, the required equation is,

= +–3 + ρ(4–5+)

The Cartesian equation of a line passing through a point (x₁, y₁, z₁) and having directional ratios proportional to a,b,c is given by,

The required line passes through the point (1,–1,1), now we need to find the direction ratios of the line which are a,b,c . this equation of the required line is,

Direction ratios of the line joining A(4,3,2) and B(1,–1,0)

= (4–1,3+1,2–0) = (3,4,2)

Direction ratios are 3,4,2

Direction ratios of the line joining C(1,2,–1) and D(2,1,1)

= (2–1,1–2,1+1) = (1,–1,2)

Direction ratios are 1,–1,2

It is given that the line AB is perpendicular to the required line, so the dot product equation will be equal to zero .

a×3 + b×4 + c×2 = 0

3a+4b+2c = 0 ……….(i).

It is given that line CD is perpendicular to the required line, so the dot product will be equal to zero .

a×1 + b×(–1) + c×2 = 0

a–b+2c = 0 ………(ii).

Solving equations (i) and (ii) by cross multiplication method, we get

= λ

a = 10λ, b = –4λ, c = –7λ

Therefore, the cartesian or symmetry form of equation of the required line is,

The Cartesian equation of a line passing through a point (x₁, y₁, z₁) and having directional ratios proportional to a,b,c is given by,

The required line passes through the point (1,2,–4), now we need to find the direction ratios of the line which are a,b,c . this equation of the required line is,

It is given that a line having Cartesian equation is perpendicular to the required line, so the dot product equation will be equal to zero.

a×8 + b×(–16) + c×7 = 0

8a–16b+7c = 0 ……(i).

It is given that a line having Cartesian equation is perpendicular to the required line,

So the dot product equation will be equal to zero.

a×3 + b×8 + c×(–5) = 0

3a+8b–5c = 0 ……..(ii).

By solving equation (i) and (ii), we get, by using cross multiplication method,

= λ

a = 24λ, b = 61λ, c = 112λ

Put these values in the required equation of line,

Therefore, this is the required equation of line.

The Cartesian equation of the lines are and and we need to find that weather the lines are perpendicular or not, so we will use the dot product equation, as we know the direction ratios of both the lines.

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

Hence the given lines are perpendicular because,

cos θ = 0

θ =

The Cartesian equation of a line passing through a point (x₁, y₁, z₁) and having directional ratios proportional to a,b,c is given by,

The required line passes through the point (2,–1,–1), now we need to find the direction ratios of the line which are a,b,c . this equation of the required line is,

It is given that a line is parallel to the required line and has the Cartesian equation 6x–2 = 3y+1 = 2z–2, which can be further solved to it’s generalized form, which is ,

So we get the direction ratios as, = λ

a = 1λ, b = 2λ, c = 3λ

as we know that two parallel lines have their direction ratios, suppose a line has direction ratios a,b,c and the line parallel to this line will have direction ratios ka,kb,kc .

putting these values in the required line equation, we get,

To convert this Cartesian form to the vector equation form, first equate the Cartesian form to a scalar,

= λ

Now equate all parts to this scalar individually,

x–2 = λ, y+1 = 2λ, z+1 = 3λ

x = 2+λ, y = 2λ – 1, z = 3λ – 1

we know that = x+y+z = +λ

x+y+z = (2+λ) + (2λ – 1) + (3λ – 1)

x+y+z = (2–1–1) + λ(1+2+3)

We are given with the Cartesian equation of two lines and their direction ratios are in the form of some variable we need to find the value of this variable so that these lines are perpendicular to each other. We can use the dot product equation to solve this problem, which is,

a₁a₂+b₁b₂+c₁c₂ = 0

the equation of the lines are,

By using the direction ratios of these lines in the dot product equation, we get,

(–3)3k + 2k(1) + 2(–5) = 0

–9k + 2k – 10 = 0

k =

We need to find the angle between two lines AB and CD but we are not given with the equations this time and we are given with only the points through which these lines start and end.

The coordinates of A, B, C, D are (1,2,3), (4,5,7), (–4,3,–6), (2,9,2).

The direction ratios of line AB are (4–1,5–2,7–3) = (3,3,4)

The direction ratios of line CD are (2+4,9–3,2+6) = (6,6,8)

We can see that the fraction of the corresponding direction ratios will be a constant,

(constant)

Therefore, lines AB and CD are parallel to each other, so the angle between them can be 0⁰ or 180⁰ .

We are given with the Cartesian equations of two lines and their direction ratios are in the form of variables, we need to find the value of this variable and give the complete equation, it is also given that the lines are perpendicular to each other.

The equations of the lines are,

These are not in their standard form, so after converting them, we get

Now as we know the direction ratios of both the lines, so by using the dot product equation, we get,

a₁a₂ + b₁ b₂ + c₁c₂ = 0

(5λ + 2)1 + (–5)2λ + 1(3) = 0

5λ + 2 – 10λ + 3 = 0

–5λ + 5 = 0

λ = 1

In this question we have to convert Cartesian equation to vector equation, we are given with the Cartesian equation of the line which is as we can see that this line is not in the standard form, so after converting it we get,

Now the direction ratios of this line is 2,3,–6

The direction cosines of the line are,

l =

m =

n =

To convert this Cartesian form to the vector equation form, first equate the Cartesian form to a scalar,

= λ

Now equate all parts to this scalar individually,

x+2 = 2λ, y – = 3λ, z–5 = –6λ

x = 2λ–2, y = 3λ+, z = 5–6λ

we know that = x+y+z = +λ

x+y+z = (2λ–2) + (3λ+) + (–6λ+5)

x+y+z = (–2++5) + λ(2+3–6)

Therefore, the vector equation of the line is the mentioned above.

Given: - Two lines equation: and

To find: - Intersection point

We have,

⇒ x = λ, y = 2λ + 2 and z = 3λ – 3

So, the coordinates of a general point on this line are

(λ, 2λ + 2, 3λ – 3)

The equation of the 2^nd line is

⇒ x = 2μ + 2, y = 3μ + 6 and z = 4μ + 3

So, the coordinates of a general point on this line are

(2μ + 2, 3μ + 6, 4μ + 3)

If the lines intersect, then they must have a common point.

Therefore for some value of λ and μ, we have

⇒ λ = 2μ + 2 , 2λ + 2 = 3μ + 6, and 3λ – 3 = 4μ + 3

⇒ λ = 2μ + 2 ……(i)

⇒ 2λ – 3μ = 4 ……(ii)

and 3λ – 4μ = 6 ……(iii)

putting value of λ from eq i in eq ii, we get

⇒ 2(2μ + 2) – 3μ = 4

⇒ 4μ + 4 – 3μ = 4

⇒ μ = 0

Now putting value of μ in eq i, we get

⇒ λ = 2μ + 2

⇒ λ = 2(0) + 2

⇒ λ = 2

As we can see by putting value of λ and μ in eq iii, that it satisfy the equation.

Check

⇒ 3λ – 4μ = 6

⇒ 3(2) = 6 ;Hence intersection point exist or line do intersects

We can find intersecting point by putting values of μ or λ in any one general point equation

Thus,

Intersection point

λ, 2λ + 2, 3λ – 3

⇒ 2, 6, 3

Given: - Two lines equation: and

To find: - Intersection point

We have,

⇒ x = 3λ + 1, y = 2λ – 1 and z = 5λ + 1

So, the coordinates of a general point on this line are

(3λ + 1, 2λ – 1, 5λ + 1)

The equation of the 2^nd line is

⇒ x = 4μ – 2, y = 3μ + 1 and z = – 2μ – 1

So, the coordinates of a general point on this line are

(4μ – 2, 3μ + 1, – 2μ – 1)

If the lines intersect, then they must have a common point.

Therefore for some value of λ and μ, we have

⇒ 3λ + 1 = 4μ – 2 , 2λ – 1 = 3μ + 1, and 5λ + 1 = – 2μ – 1

⇒ ……(i)

⇒ 2λ – 3μ = 2 ……(ii)

and 5λ + 2μ = – 2 ……(iii)

putting value of λ from eq i in eq ii, we get

⇒ – μ = 12

⇒ μ = – 12

Now putting value of μ in eq i, we get

⇒

⇒

⇒ λ = – 17

As we can see by putting value of λ and μ in eq iii, that it does not satisfy the equation.

Check

LHS

= 5λ + 2μ

= 5( – 17) + 2( – 12)

= – 85 – 24

= – 109

≠ RHS

Hence intersection point does not exist or line do not intersects

Given: - Two lines equation: and

To find: - Intersection point

We have,

⇒ x = 3λ – 1, y = 5λ – 3 and z = 7λ – 5

So, the coordinates of a general point on this line are

(3λ – 1, 5λ – 3, 7λ – 5)

The equation of the 2^nd line is

⇒ x = μ + 2, y = 3μ + 4 and z = 5μ + 6

So, the coordinates of a general point on this line are

(μ + 2, 3μ + 4, 5μ + 6)

If the lines intersect, then they must have a common point.

Therefore for some value of λ and μ, we have

⇒ 3λ – 1 = μ + 2 , 5λ – 3 = 3μ + 4, and 7λ – 5 = 5μ + 6

⇒ ……(i)

⇒ 5λ – 3μ = 7 ……(ii)

and 7λ – 5μ = 11 ……(iii)

putting value of λ from eq i in eq ii, we get

Now putting value of μ in eq i, we get

⇒

⇒

As we can see by putting the value of λ and μ in eq iii, that it satisfy the equation.

Check

⇒ 7λ – 5μ = 11

⇒

⇒

⇒ 11 = 11

⇒ LHS = RHS ; Hence intersection point exists or line do intersects

We can find an intersecting point by putting values of μ or λ in any one general point equation

Thus,

Intersection point

3λ – 1, 5λ – 3, 7λ – 5

Given: - Line joining A(0, – 1, – 1) and B(4, 5, 1).

Line joining C(3, 9, 4) and D( – 4, 4, 4).

To Prove: - Both lines intersects

Proof: - Equation of a line joined by two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is given by

Now equation of line joining A(0, – 1, – 1) and B(4, 5, 1)

=

⇒ x = 4λ, y = 6λ – 1 and z = 2λ – 1

So, the coordinates of a general point on this line are

(4λ, 6λ – 1, 2λ – 1)

And equation of line joining C(3, 9, 4) and D( – 4, 4, 4)

=

⇒ x = – 7μ + 3, y = – 5μ + 9 and z = 4

So, the coordinates of a general point on this line are

( – 7μ + 3, – 5μ + 9, 4)

If the lines intersect, then they must have a common point.

Therefore for some value of λ and μ, we have

⇒ 4λ = – 7μ + 3 , 6λ – 1 = – 5μ + 9, and 2λ – 1 = 4

⇒ ……(i)

⇒ 6λ + 5μ = 10 ……(ii)

and 2λ = 5 ……(iii)

from eq iii, we get

⇒

Now putting the value of λ in eq i, we get

⇒

⇒ – 1

As we can see by putting the value of λ and μ in eq ii, that it satisfy the equation.

Check

⇒ 6λ + 5μ = 10

⇒

⇒

⇒ 10 = 10

⇒ LHS = RHS ;Hence intersection point exist or line do intersects

We can find intersecting point by putting values of μ or λ in any one general point equation

Thus,

Intersection point

– 7μ + 3, – 5μ + 9, 4

– 7( – 1) + 3, – 5( – 1) + 9, 4

10, 14, 4

Given: – Two lines having vector notion and

The position vectors of arbitrary points on the given lines are

1^st line

2^nd line

If the lines intersect, then they must have a common point.

Therefore for some value of λ and μ, we have

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

⇒ 3λ – 2μ = 3 ……(i)

⇒ λ = 1 ……(ii)

and μ = 0 ……(iii)

from eq ii and eq iii we get

⇒ λ = 1 and μ = 0

As we can see by putting the value of λ and μ in eq i, that it satisfy the equation.

Check

⇒ 3λ – 2μ = 3

⇒ 3(1) – 2(0) = 3

⇒ 3 = 3

⇒ LHS = RHS ; Hence intersection point exists or line do intersect

We can find an intersecting point by putting values of μ or λ in any one general point equation

Thus,

Intersection point

⇒

⇒

⇒

Hence, Intersection point is (4,0, – 1)

Given: - Two lines having vector notion and

The position vectors of arbitrary points on the given lines are

1^st line

2^nd line

If the lines intersect, then they must have a common point.

Therefore for some value of λ and μ, we have

⇒ (2λ + 1) = μ + 2, – 1 = μ – 1, λ = – μ

⇒ 2λ – μ = 1 ……(i)

⇒ μ = 0 ……(ii)

and λ = μ ……(iii)

from eq ii and eq iii we get

⇒ λ = 0 and μ = 0

As we can see by putting the value of λ and μ in eq i, that it does not satisfy the equation.

Check

⇒ 2λ – μ = 1

⇒ 2(0) – (0) = 1

⇒ 0 = 1

⇒ LHS≠RHS ;Hence intersection point exist or line does not intersects

Given: - Two lines equation: and ; z = 3

We have,

⇒ x = 2λ + 1, y = 3λ – 1 and z = λ

So, the coordinates of a general point on this line are

(2λ + 1, 3λ – 1, λ)

The equation of the 2^nd line is

⇒ x = 5μ + 1, y = μ + 2 and z = 3

So, the coordinates of a general point on this line are

(5μ + 1, μ + 2, 3)

If the lines intersect, then they must have a common point.

Therefore for some value of λ and μ, we have

⇒ 2λ + 1 = 5μ + 1 , 3λ – 1 = μ + 2, and λ = 3

⇒ 2λ – 5μ = 0 ……(i)

⇒ 3λ – μ = 3 ……(ii)

and λ = 3 ……(iii)

putting the value of λ from eq iii in eq ii, we get

⇒ 3λ – μ = 3

⇒ 3(3) – μ = 3

⇒ μ = 6

As we can see by putting the value of λ and μ in eq i, that it does not satisfy the equation.

Check

⇒ 2λ – 5μ = 0

⇒ 2(3) – 5(6) = 0

⇒ – 24 = 0

⇒ LHS≠RHS; Hence intersection point exists or line does not intersects.

Given: - Two lines equation: and

We have,

⇒ x = 3λ + 1, y = – λ + 1 and z = – 1

So, the coordinates of a general point on this line are

(3λ + 1, – λ + 1, – 1)

The equation of the 2^nd line is

⇒ x = 2μ + 4, y = 0 and z = 3μ – 1

So, the coordinates of a general point on this line are

(2μ + 4, 0, 3μ – 1)

If the lines intersect, then they must have a common point.

Therefore for some value of λ and μ, we have

⇒ 3λ + 1 = 2μ + 4 , – λ + 1 = 0, and – 1 = 3μ – 1

⇒ 3λ – 2μ = 3 ……(i)

⇒ λ = 1 ……(ii)

and μ = 0 ……(iii)

from eq ii and eq iii, we get

⇒ λ = 1

and μ = 0

As we can see by putting the value of λ and μ in eq i, that it satisfy the equation.

Check

⇒ 3λ – 2μ = 3

⇒ 3(1) = 3

⇒ 3 = 3

⇒ LHS = RHS ;Hence intersection point exist or line do intersects

We can find intersecting point by putting values of μ or λ in any one general point equation

Thus,

Intersection point

2μ + 4, 0, 3μ – 1

4, 0, – 1

Given: - Two lines equation: and

To find: - Intersection point

We have,

⇒ x = 4λ + 5, y = 4λ + 7 and z = – 5λ – 3

So, the coordinates of a general point on this line are

(4λ + 5, 4λ + 7, – 5λ – 3)

The equation of the 2^nd line is

⇒ x = 7μ + 8, y = μ + 4 and z = 3μ + 5

So, the coordinates of a general point on this line are

(7μ + 8, μ + 4, 3μ + 5)

If the lines intersect, then they must have a common point.

Therefore for some value of λ and μ, we have

⇒ 4λ + 5 = 7μ + 8 , 4λ + 7 = μ + 4, and – 5λ – 3 = 3μ + 5

⇒ 4λ – 7μ = 3 ……(i)

⇒ μ = 4λ + 3 ……(ii)

and – 5λ – 3μ = 8 ……(iii)

putting the value of μ from eq ii in eq i, we get

⇒ 4λ – 7μ = 3

⇒ 4λ – 7(4λ + 3) = 3

⇒ 4λ – 28λ – 21 = 3

⇒ – 24λ = 24

⇒ λ = – 1

Now putting the value of λ in eq ii, we get

⇒ μ = 4λ + 3

⇒ μ = 4( – 1) + 3

⇒ μ = – 1

As we can see by putting the value of λ and μ in eq iii, that it satisfy the equation.

Check

⇒ – 5λ – 3μ = 8

⇒ – 5( – 1) – 3( – 1) = 8

⇒ 5 + 3 = 8

⇒ 8 = 8

⇒ LHS = RHS ;Hence intersection point exist or line do intersects

We can find intersecting point by putting values of μ or λ in any one general point equation

Thus,

Intersection point

4λ + 5, 4λ + 7, – 5λ – 3

4( – 1) + 5, 4( – 1) + 7, – 5( – 1) – 3

1, 3, 2

Given: – Two lines having vector notion and

To show: - Lines are intersecting

The position vectors of arbitrary points on the given lines are

1^st line

2^nd line

If the lines intersect, then they must have a common point.

Therefore for some value of λ and μ, we have

⇒ (λ + 3) = 3μ + 5, 2λ + 2 = 2μ – 2, 2λ – 4 = 6μ

⇒ λ = 3μ + 2 …… (i)

⇒ λ – μ = – 2 ……(ii)

and λ – 3μ = 2 ……(iii)

putting value of λ from eq i in eq ii, we get

⇒ λ – μ = – 2

⇒ 3μ + 2 – μ = – 2

⇒ 2μ = – 4

⇒ μ = – 2

Now putting value of μ in eq i, we get

⇒ λ = 3μ + 2

⇒ λ = 3( – 2) + 2

⇒ λ = – 6 + 2

⇒ λ = – 4

As we can see by putting value of λ and μ in eq iii, that it satisfy the equation.

Check

⇒ λ – 3μ = 2

⇒ – 4 – 3( – 2) = 2

⇒ – 4 + 6 = 2

⇒ 2 = 2

⇒ LHS = RHS ; Hence intersection point exists or line do intersect

We can find an intersecting point by putting values of μ or λ in any one general point equation

Thus,

Intersection point

⇒

⇒

⇒

Hence, Intersection point is ( – 1, – 6, – 12)

Given: - Point P(3, – 1, 11) and the equation of the line

Let, PQ be the perpendicular drawn from P to given line whose endpoint/ foot is Q point.

Thus to find Distance PQ we have to first find coordinates of Q

⇒ x = 2λ, y = – 3λ + 2, z = 4λ + 3

Therefore, coordinates of Q(2λ, – 3λ + 2,4λ + 3)

Now as we know (TIP) ‘if two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) on a line, then its direction ratios are proportional to (x₂ – x₁,y₂ – y₁,z₂ – z₁)’

Hence

Direction ratio of PQ is

= (2λ – 3), ( – 3λ + 2 + 1), (4λ + 3 – 11)

= (2λ – 3), ( – 3λ + 1), (4λ – 8)

and by comparing with given line equation, direction ratios of the given line are

(hint: denominator terms of line equation)

= (2, – 3,4)

Since PQ is perpendicular to given line, therefore by “condition of perpendicularity.”

a₁a₂ + b₁b₂ + c₁c₂ = 0 ; where a terms and b terms are direction ratio of lines which are perpendicular to each other.

⇒ 2(2λ – 3) + ( – 3)( – 3λ + 3) + 4(4λ – 8) = 0

⇒ 4λ – 6 + 9λ – 9 + 16λ – 32 = 0

⇒ 29λ – 47 = 0

Therefore coordinates of Q

i.e. Foot of perpendicular

By putting the value of λ in Q coordinate equation, we get

Now,

Distance between PQ

Tip: - Distance between two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is given by

unit

Given: - Point P(1, 0, 0) and equation of line

Let, PQ be the perpendicular drawn from P to given line whose endpoint/ foot is Q point.

Thus to find Distance PQ we have to first find coordinates of Q

⇒ x = 2λ + 1, y = – 3λ – 1, z = 8λ – 10

Therefore, coordinates of Q(2λ + 1, – 3λ – 1,8λ – 10)

Now as we know (TIP) ‘if two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) on a line, then its direction ratios are proportional to (x₂ – x₁,y₂ – y₁,z₂ – z₁)’

Hence

Direction ratio of PQ is

= (2λ + 1 – 1), ( – 3λ – 1 – 0), (8λ – 10 – 0)

= (2λ), ( – 3λ – 1), (8λ – 10)

and by comparing with given line equation, direction ratios of the given line are

(hint: denominator terms of line equation)

= (2, – 3,8)

Since PQ is perpendicular to given line, therefore by “condition of perpendicularity.”

a₁a₂ + b₁b₂ + c₁c₂ = 0 ; where a terms and b terms are direction ratio of lines which are perpendicular to each other.

⇒ 2(2λ) + ( – 3)( – 3λ – 1) + 8(8λ – 10) = 0

⇒ 4λ + 9λ + 3 + 64λ – 80 = 0

⇒ 77λ – 77 = 0

⇒ λ = 1

Therefore coordinates of Q

i.e. Foot of perpendicular

By putting the value of λ in Q coordinate equation, we get

Now,

Distance between PQ

Tip: – Distance between two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is given by

= 2√6 unit

Given: - Perpendicular from A(1, 0, 3) drawn at line joining points B(4, 7, 1) and C(3, 5, 3)

Let D be the foot of the perpendicular drawn from A(1, 0, 3) to line joining points B(4, 7, 1) and C(3, 5, 3).

Now let's find the equation of the line which is formed by joining points B(4, 7, 1) and C(3, 5, 3)

Tip: - Equation of a line joined by two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is given by

Now

Therefore,

⇒ x = - λ + 4, y = – 2λ + 7, z = 2λ + 1

Therefore, coordinates of D( – λ + 4, – 2λ + 7, 2λ + 1)

Now as we know (TIP) ‘if two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) on a line, then its direction ratios are proportional to (x₂ – x₁,y₂ – y₁,z₂ – z₁)’

Hence

Direction Ratios of AD

= ( – λ + 4 – 1), ( – 2λ + 7 – 0), (2λ – 2)

= ( – λ + 3), ( – 2λ + 7), (2λ – 2)

and by comparing with given line equation, direction ratios of the given line are

(hint: denominator terms of line equation)

= ( – 1, – 2,2)

Since AD is perpendicular to given line, therefore by “condition of perpendicularity”

a₁a₂ + b₁b₂ + c₁c₂ = 0 ; where a terms and b terms are direction ratio of lines which are perpendicular to each other.

⇒ – 1( – λ + 3) + ( – 2)( – 2λ + 7) + 2(2λ – 2) = 0

⇒ λ – 3 + 4λ – 14 + 4λ – 4 = 0

⇒ 9λ – 21 = 0

Therefore coordinates of D

i.e Foot of perpendicular

By putting value of λ in D coordinate equation, we get

Given: - Perpendicular from A(1, 0, 4) drawn at line joining points B(0, – 11, 3) and C(2, – 3, 1)

and D be the foot of the perpendicular drawn from A(1, 0, 4) to line joining points B(0, – 11, 3) and C(2, – 3, 1).

Now let's find the equation of the line which is formed by joining points B(0, – 11, 3) and C(2, – 3, 1)

Tip: - Equation of a line joined by two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is given by

Now

Therefore,

⇒ x = 2λ, y = 8λ – 11, z = – 2λ + 3

Therefore, coordinates of D(2λ, 8λ – 11, – 2λ + 3)

Now as we know (TIP) ‘if two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) on a line, then its direction ratios are proportional to (x₂ – x₁,y₂ – y₁,z₂ – z₁)’

Hence

Direction Ratios of AD

= (2λ – 1), (8λ – 11 – 0), ( – 2λ + 3 – 4)

= (2λ – 1), (8λ – 11), ( – 2λ – 1)

and by comparing with given line equation, direction ratios of the given line are

(hint: denominator terms of line equation)

= (2,8, – 2)

Since the AD is perpendicular to given line, therefore by “condition of perpendicularity.”

a₁a₂ + b₁b₂ + c₁c₂ = 0 ; where a terms and b terms are direction ratio of lines which are perpendicular to each other.

⇒ 2(2λ – 1) + (8)(8λ – 11) – 2( – 2λ – 1) = 0

⇒ 4λ – 2 + 64λ – 88 + 4λ + 2 = 0

⇒ 72λ – 88 = 0

Therefore coordinates of D

i.e. Foot of perpendicular

By putting the value of λ in D coordinate equation, we get

Given: - Point P(2, 3, 4) and the equation of the line

Let, PQ be the perpendicular drawn from P to given line whose endpoint/ foot is Q point.

Thus to find Distance PQ we have to first find coordinates of Q

⇒ x = 4 – 2λ, y = 6λ, z = 1 – 3λ

Therefore, coordinates of Q( – 2λ + 4, 6λ, – 3λ + 1)

Now as we know (TIP) ‘if two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) on a line, then its direction ratios are proportional to (x₂ – x₁,y₂ – y₁,z₂ – z₁)’

Hence

Direction ratio of PQ is

= ( – 2λ + 4 – 2), (6λ – 3), ( – 3λ + 1 – 4)

= ( – 2λ + 2), (6λ – 3), ( – 3λ – 3)

and by comparing with given line equation, direction ratios of the given line are

(hint: denominator terms of line equation)

= ( – 2,6, – 3)

Since PQ is perpendicular to given line, therefore by “condition of perpendicularity.”

a₁a₂ + b₁b₂ + c₁c₂ = 0 ; where a terms and b terms are direction ratio of lines which are perpendicular to each other.

⇒ – 2( – 2λ + 2) + (6)(6λ – 3) – 3( – 3λ – 3) = 0

⇒ 4λ – 4 + 36λ – 18 + 9λ + 9 = 0

⇒ 49λ – 13 = 0

Therefore coordinates of Q

i.e. Foot of perpendicular

By putting the value of λ in Q coordinate equation, we get

Now,

Distance between PQ

Tip: - Distance between two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is given by

units

Given: - Point P(2, 4, – 1) and equation of line

Let, PQ be the perpendicular drawn from P to given line whose endpoint/ foot is Q point.

Thus to find Distance PQ we have to first find coordinates of Q

⇒ x = λ – 5, y = 4λ – 3, z = – 9λ + 6

Therefore, coordinates of Q(λ – 5,4λ – 3, – 9λ + 6)

Now as we know (TIP) ‘if two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) on a line, then its direction ratios are proportional to (x₂ – x₁,y₂ – y₁,z₂ – z₁)’

Hence

Direction ratio of PQ is

= (λ – 5 – 2), (4λ – 3 – 4), ( – 9λ + 6 + 1)

= (λ – 7), (4λ – 7), ( – 9λ + 7)

and by comparing with given line equation, direction ratios of the given line are

(hint: denominator terms of line equation)

= (1,4, – 9)

Since PQ is perpendicular to given line, therefore by “condition of perpendicularity.”

a₁a₂ + b₁b₂ + c₁c₂ = 0 ; where a terms and b terms are direction ratio of lines which are perpendicular to each other.

⇒ 1(λ – 7) + (4)(4λ – 7) – 9( – 9λ + 7) = 0

⇒ λ – 7 + 16λ – 28 + 81λ – 63 = 0

⇒ 98λ – 98 = 0

⇒ λ = 1

Therefore coordinates of Q

i.e. Foot of perpendicular

By putting the value of λ in Q coordinate equation, we get

Now,

So, Equation of perpendicular PQ is

Tip: - Equation of a line joined by two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is given by

Given: - Point (5, 4, – 1) and equation of line

Let, PQ be the perpendicular drawn from P to given line whose endpoint/ foot is Q point.

As we know position vector is given by

Therefore,

Position vector of point P is

and, from a given line, we get

⇒

⇒

⇒

On comparing both sides we get,

⇒ x = 1 + 2λ, y = 9λ, z = 5λ

⇒ ; Equation of line

Thus, coordinates of Q i.e. General point on the given line

⇒ Q((1 + 2λ), 9λ, 5λ)

Now as we know (TIP) ‘if two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) on a line, then its direction ratios are proportional to (x₂ – x₁,y₂ – y₁,z₂ – z₁)’

Hence

Direction ratio of PQ is

= (2λ + 1 – 5), (9λ – 4), (5λ + 1)

= (2λ – 4), (9λ – 4), (5λ + 1)

and by comparing with line equation, direction ratios of the given line are

(hint: denominator terms of line equation)

= (2,9,5)

Since PQ is perpendicular to given line, therefore by “condition of perpendicularity.”

a₁a₂ + b₁b₂ + c₁c₂ = 0 ; where a terms and b terms are direction ratio of lines which are perpendicular to each other.

⇒ 2(2λ – 4) + (9)(9λ – 4) + 5(5λ + 1) = 0

⇒ 4λ – 8 + 81λ – 36 + 25λ + 5 = 0

⇒ 110λ – 39 = 0

Therefore coordinates of Q

i.e. Foot of perpendicular

By putting the value of λ in Q coordinate equation, we get

Now,

Distance between PQ

Tip: - Distance between two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is given by

units

Given: - Point with position vector and equation of line

Let, PQ be the perpendicular drawn from P () to given line whose endpoint/ foot is Q point.

Q is on line

⇒ is the position vector of Q

Hence,

⇒

⇒

⇒

⇒

Since, PQ is perpendicular on line

Therefore, their Dot product is zero

Compare given line equation with

⇒

⇒

⇒

⇒

⇒ 14λ – 14 = 0

⇒ λ = 1

Hence Position vector of Q by putting the value of λ

⇒

⇒ ; Foot of perpendicular

and Distance PQ, putting the value of λ in PQ vector equation, we get

⇒

⇒

⇒

Now, Magnitude of PQ

We know that,

; where x,y,z are coefficient of vector

Hence

⇒

⇒

⇒ units

Given: -

Point P( – 1, 3, 2) and equation of line

Let, PQ be the perpendicular drawn from P to given line whose endpoint/ foot is Q point.

As we know position vector is given by

Therefore,

Position vector of point P is

and, from a given line, we get

⇒

⇒

⇒

On comparing both sides we get,

⇒ x = 2λ, y = λ + 2, z = 3λ + 3

⇒ ; Equation of line

Thus, coordinates of Q i.e. General point on the given line

⇒ Q(2λ, (λ + 2), (3λ + 3))

Now as we know (TIP) ‘if two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) on a line, then its direction ratios are proportional to (x₂ – x₁,y₂ – y₁,z₂ – z₁)’

Hence

Direction ratio of PQ is

= (2λ + 1), (λ + 2 – 3), (3λ + 3 – 2)

= (2λ + 1), (λ – 1), (3λ + 1)

and by comparing with line equation, direction ratios of the given line are

(hint: denominator terms of line equation)

= (2,1,3)

Since PQ is perpendicular to given line, therefore by “condition of perpendicularity.”

a₁a₂ + b₁b₂ + c₁c₂ = 0 ; where a terms and b terms are direction ratio of lines which are perpendicular to each other.

⇒ 2(2λ + 1) + (λ – 1) + 3(3λ + 1) = 0

⇒ 4λ + 2 + λ – 1 + 9λ + 3 = 0

⇒ 14λ + 4 = 0

Therefore coordinates of Q

i.e. Foot of perpendicular

By putting the value of λ in Q coordinate equation, we get

2λ, (λ + 2), (3λ + 3)

Position Vector of Q

Now,

Equation of line passing through two points with position vectoris given by

Here,

⇒

Given: - Point P(0, 2, 7) and equation of line

Let, PQ be the perpendicular drawn from P to given line whose endpoint/ foot is Q point.

Thus to find Distance PQ we have to first find coordinates of Q

⇒ x = – λ – 2, y = 3λ + 1, z = – 2λ + 3

Therefore, coordinates of Q( – λ – 2, 3λ + 1, – 2λ + 3)

Now as we know (TIP) ‘if two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) on a line, then its direction ratios are proportional to (x₂ – x₁,y₂ – y₁,z₂ – z₁)’

Hence

Direction ratio of PQ is

= ( – λ – 2 – 0), (3λ + 1 – 2), ( – 2λ + 3 – 7)

= ( – λ – 2), (3λ – 1), ( – 2λ – 4)

and by comparing with given line equation, direction ratios of the given line are

(hint: denominator terms of line equation)

= ( – 1,3, – 2)

Since PQ is perpendicular to given line, therefore by “condition of perpendicularity.”

a₁a₂ + b₁b₂ + c₁c₂ = 0 ; where ‘a’ terms and ‘b’ terms are direction ratio of lines which are perpendicular to each other.

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

⇒ λ + 2 + 9λ – 3 + 4λ + 8 = 0

⇒ 14λ + 7 = 0

⇒

Therefore coordinates of Q

i.e. Foot of perpendicular

By putting the value of λ in Q coordinate equation, we get

Given: – Point P(1, 2, – 3) and equation of line

Let, PQ be the perpendicular drawn from P to given line whose endpoint/ foot is Q point.

Thus to find Distance PQ we have to first find coordinates of Q

⇒ x = 2λ – 1, y = – 2λ + 3, z = – λ

Therefore, coordinates of Q(2λ – 1, – 2λ + 3, – λ)

Now as we know (TIP) ‘if two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) on a line, then its direction ratios are proportional to (x₂ – x₁,y₂ – y₁,z₂ – z₁)’

Hence

Direction ratio of PQ is

= (2λ – 1 – 1), ( – 2λ + 3 – 2), ( – λ + 3)

= (2λ – 2), ( – 2λ + 1), ( – λ + 3)

and by comparing with given line equation, direction ratios of the given line are

(hint: denominator terms of line equation)

= (2, – 2, – 1)

Since PQ is perpendicular to given line, therefore by “condition of perpendicularity.”

a₁a₂ + b₁b₂ + c₁c₂ = 0 ; where a terms and b terms are direction ratio of lines which are perpendicular to each other.

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

⇒ 4λ – 4 + 4λ – 2 + λ – 3 = 0

⇒ 9λ – 9 = 0

⇒ λ = 1

Therefore coordinates of Q

i.e. Foot of perpendicular

By putting the value of λ in Q coordinate equation, we get

Given: – Line passing through the points A(0, 6, – 9) and B( – 3, – 6, 3). Point C(7, 4, – 1).

We know that

Tip: - Equation of a line joined by two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is given by

Hence equation of line AB

⇒ x = – 3λ, y = – 12λ + 6, z = 12λ – 9

Now coordinates of point D

D( – 3λ, ( – 12λ + 6),( 12λ – 9))

Now as we know (TIP) ‘if two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) on a line, then its direction ratios are proportional to (x₂ – x₁,y₂ – y₁,z₂ – z₁)’

Hence

Direction ratio of CD is

= ( – 3λ – 7), ( – 12λ + 6 – 4), (12λ – 9 + 1)

= ( – 3λ – 7), ( – 12λ + 2), (12λ – 8)

and by comparing with given line equation, direction ratios of the given line are

(hint: denominator terms of line equation)

= ( – 3, – 12,12)

Since CD is perpendicular to given line, therefore by “condition of perpendicularity.”

a₁a₂ + b₁b₂ + c₁c₂ = 0 ; where a terms and b terms are direction ratio of lines which are perpendicular to each other.

⇒ ( – 3)( – 3λ – 7) + ( – 12)( – 12λ + 2) + 12(12λ – 8) = 0

⇒ 9λ + 21 + 144λ – 24 + 144λ – 96 = 0

⇒ 297λ – 99 = 0

Therefore coordinates of D

i.e. Foot of perpendicular

By putting the value of λ in D coordinate equation, we get

– 3λ, ( – 12λ + 6),( 12λ – 9)

Hence equation of line CD

Given: - Point P(2, 4, – 1) and equation of line

Let, Q be a point through which line passes

Thus from given equation of line coordinates of Q is

Q( – 5, – 3,6)

As we know line equation with direction ratio of given line is parallel to given line.

Hence Line is parallel to

Now,

⇒

⇒

Now let's find cross product of this two vectors

⇒

⇒

The magnitude of this cross product

⇒

⇒

And Magnitude of

⇒

⇒

Thus distance of point from line is

⇒

⇒

⇒ units

Given: – Perpendicular from A(1, 8, 4) drawn at line joining points B(0, – 1, 3) and C(2, – 3, – 1)

and D be the foot of the perpendicular drawn from A(1, 8, 4) to line joining points B(0, – 1, 3) and C(2, – 3, – 1).

Now let's find the equation of the line which is formed by joining points B(0, – 1, 3) and C(2, – 3, – 1)

Tip: - Equation of a line joined by two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is given by

Now

Therefore,

⇒ x = 2λ, y = – 2λ – 1, z = – 4λ + 3

Therefore, coordinates of D(2λ, – 2λ – 1, – 4λ + 3)

Now as we know (TIP) ‘if two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) on a line, then its direction ratios are proportional to (x₂ – x₁,y₂ – y₁,z₂ – z₁)’

Hence

Direction Ratios of AD

= (2λ – 1), ( – 2λ – 1 – 8), ( – 4λ + 3 – 4)

= (2λ – 1), ( – 2λ – 9), ( – 4λ – 1)

and by comparing with given line equation, direction ratios of the given line are

(hint: denominator terms of line equation)

= (2, – 2, – 4)

Since AD is perpendicular to given line, therefore by “condition of perpendicularity”

a₁a₂ + b₁b₂ + c₁c₂ = 0 ; where a terms and b terms are direction ratio of lines which are perpendicular to each other.

⇒ 2(2λ – 1) + ( – 2)( – 2λ – 9) – 4( – 4λ – 1) = 0

⇒ 4λ – 2 + 4λ + 18 + 16λ + 4 = 0

⇒ 24λ + 20 = 0

⇒

Therefore coordinates of D

i.e Foot of perpendicular

By putting value of λ in D coordinate equation, we get

Equation of line in vector form

Line I:

Line II:

Here

The shortest distance between lines is

Putting these values in the expression,

Equation of line in vector form

Line I:

Line II:

Here,

The shortest distance between lines is

Putting these values in the expression,

Equation of line in vector form

Line I:

Line II:

Here,

The shortest distance between lines is

Putting these values in the expression,

Equation of line in vector form

Line I:

Line II:

Here,

The shortest distance between lines is

Putting these values in the expression,

V. Equation of line in vector form

Line I:

Line II:

Here,

The shortest distance between lines is

Putting these values in the expression,

VI. Equation of line in vector form

Line I:

Line II:

Here,

The shortest distance between lines is

Putting these values in the expression,

VII. Equation of line in vector form

Line I:

Line II:

Here,

The shortest distance between lines is

Putting these values in the expression,

VIII. Equation of line in vector form

Line I:

Line II:

Here,

The shortest distance between lines is

Putting these values in the expression,

Given data:

Pair of lines:

To find the shortest distance between these two lines

Solution: We need to write this in vector form

Line I:

Line II:

Here

The shortest distance between two skew lines is

Putting these values in the expression,

Given data:

Pair of lines:

To find the shortest distance between these two lines

Solution: We need to write this in vector form

Line I:

Line II:

Here

The shortest distance between two skew lines is

Putting these values in the expression,

Given data:

Pair of lines:

To find the shortest distance between these two lines

Solution: We need to write this in vector form

Line I:

Line II:

Here

The shortest distance between two skew lines is

Putting these values in the expression,

Given data:

Pair of lines:

To find the shortest distance between these two lines

Solution: We need to write this in vector form

Line I:

Line II:

Here

The shortest distance between two skew lines is

Putting these values in the expression,

Equation of line in vector form

Line I:

Line II:

Here,

The shortest distance between lines is

Putting these values in the expression,

Shortest distance d between the lines is not 0. So lines are not intersecting.

Equation of line in vector form

Line I:

Line II:

Here,

The shortest distance between lines is

Putting these values in the expression,

d = 0

Shortest distance d between the lines is 0. So lines are intersecting.

Given data:

Pair of lines:

To find the shortest distance between these two lines

Solution: We need to write this in vector form

Line I:

Line II:

Here

The shortest distance between two skew lines is

Putting these values in the expression,

Shortest distance d between the lines is not 0. So lines are not intersecting.

Given data:

Pair of lines:

To find the shortest distance between these two lines

Solution: We need to write this in vector form

Line I:

Line II:

Here

The shortest distance between two skew lines is

Putting these values in the expression,

Shortest distance d between the lines is not 0. So lines are not intersecting.

Equation of a line in vector form

Line I:

Line II:

Here,

Since lines are parallel therefore they have common normal

The shortest distance between lines is

Putting these values in the expression,

Equation of a line in vector form

Line I:

Line II:

Here,

Since lines are parallel therefore they have common normal

The shortest distance between lines is

Putting these values in the expression,

We need to write this in vector form

Line I:

Equation of line passing through a (0, 0, 0) and b (1, 0, 2)

Line II:

Equation of line passing through a (1, 3, 0) and b (0, 3, 0)

Here

The shortest distance between two skew lines is

Putting these values in the expression,

d = 3 units

Given data:

Pair of lines:

To find the shortest distance between these two lines

We need to write this in vector form

Line I:

Line II:

Here the lines are parallel as can be seen from above vector equation

Since lines are parallel therefore they have common normal

The shortest distance between lines is

Putting these values in the expression,

Equation of a line in vector form

Line I:

Line II:

Here,

The shortest distance between lines is

Putting these values in the expression,

Given data:

Pair of lines:

To find the shortest distance between these two lines

Solution: We need to write this in vector form

Line I:

Line II:

Here

Shortest distance between two skew lines is

Putting these values in the expression,

We need to write this in vector form

Line I:

Line II:

Here

Shortest distance between two skew lines is

Putting these values in the expression,

We need to write this in vector form

Line I:

Line II:

Here

Shortest distance between two skew lines is

Putting these values in the expression,

d = 9 units

We need to write this in vector form

Line I:

Line II:

Since lines are parallel therefore they have common normal

The shortest distance between lines is

Putting these values in the expression,

X-axis passes through the point (0, 0, 0).

Position Vector →

Since, it is also parallel to the vector, having direction ratios proportional to 1, 0, 0, the Cartesian equation of x-axis is,

Also, its vector equation is

Y-axis passes through the point (0, 0, 0).

Position Vector →

Since, it is also parallel to the vector, having direction ratios proportional to 0, 1, 0, the Cartesian equation of y-axis is,

Also, its vector equation is

Z-axis passes through the point (0, 0, 0).

Position Vector →

Since, it is also parallel to the vector, having direction ratios proportional to 0, 0, 1, the Cartesian equation of z-axis is,

Also, its vector equation is

The vector equation of the line passing through the point having position vector

And parallel to vector is,

We have

Now, the equation of the line AB can be re-written as,

The direction ratios of the line parallel to AB proportional to → 1, -7, 2

We have

6x – 2 = 3y + 1 = 2z – 4

The equation of the given line can be re-written as,

The direction ratios of the line parallel to AB are proportional to → 1,2,3

The direction cosines of the line parallel to AB are proportional to →

, ,

, ,

We have

, z = 2

The equation of the given line can be re-written as,

The direction ratios of the line parallel to AB are proportional to → 4, -3, 0

The direction cosines of the line parallel to AB are proportional to →

, ,

, , 0

We have,

The given line is parallel to the vector,

Now let,

be perpendicular to the given line.

Now,

3x+4y+0z = 0

It is satisfied by the coordinates of z-axis, i.e. (0,0,1).

Hence, the given line is perpendicular to z-axis.

We have,

The given lines are parallel to the vectors

→ and

Let θ be the angle between the given lines.

Now,

cos θ

= 0

→ θ

We have

2x = 3y = -z

The equation of the given line can be re-written as,

The direction ratios of the line parallel to AB are proportional to → 3, 2, -6

The direction cosines of the line parallel to AB are proportional to →

, ,

, ,

We have,

2x = 3y = -z

6x = -y = -4z

The given lines can be re-written as

These lines are parallel to vectors,

and

Let θ be the angle between these lines.

Now,

cosθ

= 0

→ θ

We have,

The given lines are parallel to vectors,

and

Now, for we must have,

-7λ -10 = 0

λ

The shortest distance d between the parallel lines and is given by –

d

The shortest distance between the lines and is given by –

d

For the lines to be intersecting, d = 0

We have,

The equation of the given line can be re-written as,

The direction ratios of the line parallel to AB are proportional to → , 4, 6

The direction cosines of the line parallel to AB are proportional to →

, ,

, ,

We have,

The equation of the line AB can be re-written as,