`color{red} ♦` Equation of a Line in Space

`color{red} ♦` Angle between Two Lines

`color{red} ♦` Shortest Distance between Two Lines

`color{red} ♦` Angle between Two Lines

`color{red} ♦` Shortest Distance between Two Lines

● We have studied equation of lines in two dimensions, A line is uniquely determined if

(i) it passes through a given point and has given direction, or

(ii) it passes through two given points.

(i) it passes through a given point and has given direction, or

(ii) it passes through two given points.

● Let `vec a` be the position vector of the given point

`=>` Let `l` be the line which passes through the point A and is parallel to a given vector `vec b` . Let `vec r` be the position vector of an arbitrary point P on the line (Fig).

`=>` Then `bar (AP)` is parallel to the vector ` vec B` , I.E.,

`BAR (AP)= lambda vec b` , where `lambda` is some real number.

But `bar (AP) = bar (OP) - bar (OA)`

i.e., `color{red}{lambda bar b = vec r - vec a}`

Conversely, for each value of the parameter λ, this equation gives the position vector of a point P on the line. Hence, the vector equation of the line is given by

`color{green}{vec r = vec a + lambda vec b}` .......(1)

`color{blue} "Key Concept :"`

`=>` If `vec b = a hat i + b hat j + c hat k` , then `a,b,c` are direction ratios of the line and conversely, if a, b, c are direction ratios of a line, then `vec b = a hat i + b hat j + c hat k` will be the parallel to the line. Here, b should not be confused with `| vec b |` .

`color{brown} "Derivation of cartesian form from vector form"`

`=>` Let the coordinates of the given point A be `(x_1, y_1, z_1)` and the direction ratios of the line be `a, b, c.` Consider the coordinates of any point P be `(x, y, z).` Then

`vec r = xhati + yhatj + zhatk ; vec a = x_1 hat i + y_1 hatj + z_1 hatk`

and `vec b = a hat i + b hatj + c hatk`

`=>` Substituting these values in (1) and equating the coefficients of `hat i , hat j ` and `hat k`, we get

`x = x_1 + λ a; y = y_1 + λ b; z = z_1+ λ c` .......................(2)

`=>` These are parametric equations of the line. Eliminating the parameter `λ` from (2),

we get

`color{red}{( x - x_1 )/a = (y - y_1 )/b = (z -z_1 )/c}` .....................(3)

`=>` This is the Cartesian equation of the line.

`=>` Let `l` be the line which passes through the point A and is parallel to a given vector `vec b` . Let `vec r` be the position vector of an arbitrary point P on the line (Fig).

`=>` Then `bar (AP)` is parallel to the vector ` vec B` , I.E.,

`BAR (AP)= lambda vec b` , where `lambda` is some real number.

But `bar (AP) = bar (OP) - bar (OA)`

i.e., `color{red}{lambda bar b = vec r - vec a}`

Conversely, for each value of the parameter λ, this equation gives the position vector of a point P on the line. Hence, the vector equation of the line is given by

`color{green}{vec r = vec a + lambda vec b}` .......(1)

`color{blue} "Key Concept :"`

`=>` If `vec b = a hat i + b hat j + c hat k` , then `a,b,c` are direction ratios of the line and conversely, if a, b, c are direction ratios of a line, then `vec b = a hat i + b hat j + c hat k` will be the parallel to the line. Here, b should not be confused with `| vec b |` .

`color{brown} "Derivation of cartesian form from vector form"`

`=>` Let the coordinates of the given point A be `(x_1, y_1, z_1)` and the direction ratios of the line be `a, b, c.` Consider the coordinates of any point P be `(x, y, z).` Then

`vec r = xhati + yhatj + zhatk ; vec a = x_1 hat i + y_1 hatj + z_1 hatk`

and `vec b = a hat i + b hatj + c hatk`

`=>` Substituting these values in (1) and equating the coefficients of `hat i , hat j ` and `hat k`, we get

`x = x_1 + λ a; y = y_1 + λ b; z = z_1+ λ c` .......................(2)

`=>` These are parametric equations of the line. Eliminating the parameter `λ` from (2),

we get

`color{red}{( x - x_1 )/a = (y - y_1 )/b = (z -z_1 )/c}` .....................(3)

`=>` This is the Cartesian equation of the line.

`color{red} "Key Piont :"` If l, m, n are the direction cosines of the line, the equation of the line is

`(x -x_1)/l = (y - y_1)/m = (z -z_1 )/n`

`(x -x_1)/l = (y - y_1)/m = (z -z_1 )/n`

Q 3177578486

Find the vector and the Cartesian equations of the line through the point

(5, 2, – 4) and which is parallel to the vector `3 hat i + 2 hat j − 8 hat k` .

Class 12 Chapter 11 Example 6

(5, 2, – 4) and which is parallel to the vector `3 hat i + 2 hat j − 8 hat k` .

Class 12 Chapter 11 Example 6

We have

`vec a = 5 hat i +2 hat j -4 hat k` and ` vec b = 3 hat i +2 hat j - 8 hat k`

Therefore, the vector equation of the line is

`vec r = 5 hat i + 2 hat j - 4 hat k + lambda ( 3 hat i +2 hat j - 8 hat k )`

Now, `vec r` is the position vector of any point P(x, y, z) on the line.

Therefore, ` x hat i + y hat j + z hat k = 5 hat i +2 hat j -4 hat k + lambda ( 3 hat i +2 hat j - 8 hat k )`

`= (5+ 3 lambda) hat i + (2+2 2 lambda) hat j + (-4 -8 lambda) hat k`

Eliminating λ , we get

`(x-5)/3 = (y-2)/2 = (z+4)/(-8)`

which is the equation of the line in Cartesian form.

`=>` Let `vec a` and `vec b` be the position vectors of two points `A (x_1, y_1, z_1)` and `B(x_2, y_2, z_2)`, respectively that are lying on a line (Fig 11.5).

`=>` Let ` vec r` be the position vector of an arbitrary point `P(x, y, z)`, then P is a point on the line if and only if `bar (AP) = vec r - vec a` and `bar (AB) = vec b - vec a` are collinear vectors. Therefore, P is on the line if and only if

`vec r - vec a= lambda (vec b - vec a)`

or `color{blue}{vec r = vec a + lambda (vec b - vec a) , lambda ϵ R}` ....(1)

`=>` This is the vector equation of the line.

`color{brown} "Derivation of cartesian form from vector form"`

`vec r = x hat i + y hat j + z hat k , vec a = x_1 hat i + y_1 hat j + z_1 hat k` and `vec b = x_2 hat i + y_2 hat j + z_2 hat k` ,

Substituting these values in (1), we get

`x hat i + y hat j + z hat k = x_1 hat i+ y_1 hat j + z_1 hat k +lambda [ (x_2-x_1) hat i + (y_2-y_1) hat j + (z_2 -z_1) hat k ]`

Equating the like coefficients of `hat i, hat j, hat k` , we get

`x =x_1+ lambda (x_2-x_1) ; y = y_1 + lambda (y_2-y_1) ;z = z_1 + lambda (z_2 - z_1)`

On eliminating λ, we obtain

`=>` Let ` vec r` be the position vector of an arbitrary point `P(x, y, z)`, then P is a point on the line if and only if `bar (AP) = vec r - vec a` and `bar (AB) = vec b - vec a` are collinear vectors. Therefore, P is on the line if and only if

`vec r - vec a= lambda (vec b - vec a)`

or `color{blue}{vec r = vec a + lambda (vec b - vec a) , lambda ϵ R}` ....(1)

`=>` This is the vector equation of the line.

`color{brown} "Derivation of cartesian form from vector form"`

`vec r = x hat i + y hat j + z hat k , vec a = x_1 hat i + y_1 hat j + z_1 hat k` and `vec b = x_2 hat i + y_2 hat j + z_2 hat k` ,

Substituting these values in (1), we get

`x hat i + y hat j + z hat k = x_1 hat i+ y_1 hat j + z_1 hat k +lambda [ (x_2-x_1) hat i + (y_2-y_1) hat j + (z_2 -z_1) hat k ]`

Equating the like coefficients of `hat i, hat j, hat k` , we get

`x =x_1+ lambda (x_2-x_1) ; y = y_1 + lambda (y_2-y_1) ;z = z_1 + lambda (z_2 - z_1)`

On eliminating λ, we obtain

` "Equation of line " color{red} {(x -x_1)/( x_2 - x_1) = (y-y_1)/(y_2 - y_1) = (z-z_1)/(z_2- z_1)}`

Q 3107578488

Find the vector equation for the line passing through the points (–1, 0, 2)

and (3, 4, 6).

Class 12 Chapter 11 Example 7

and (3, 4, 6).

Class 12 Chapter 11 Example 7

Let `vec a ` and ` vec b` be the position vectors of the point A(– 1, 0, 2) and B(3, 4, 6).

Then `vec a = - hat i + 2 hat k`

and `vec b = 3 hat i +4 hat j + 6 hat k`

Therefore `vec b - vec a= 4 hat i +4 hat j + 4 hat k`

Let `vec r` be the position vector of any point on the line. Then the vector equation of

the line is

`vec r = - hat i + 2 hat k + lambda (4 hat i + 4 hat j +4 hat k )`

Q 3117578489

The Cartesian equation of a line is

`(x+3)/2 = (y-5)/4 = (z+6)/2`

Find the vector equation for the line.

Class 12 Chapter 11 Example 8

`(x+3)/2 = (y-5)/4 = (z+6)/2`

Find the vector equation for the line.

Class 12 Chapter 11 Example 8

Comparing the given equation with the standard form

`(x-x_1)/a = (y-y_1)/b = (z-z_1)/c`

We observe that `x_1 = -3, y_1 = 5, z_1 = -6 ; a= 2 , b =4, c=2`

Thus, the required line passes through the point (– 3, 5, – 6) and is parallel to the

vector `2 hat i + 4 hat j + 2 hat k` . Let `vec r` be the position vector of any point on the line, then the

vector equation of the line is given by

`vec r = (-3 hat i + 5 hat j - 6 hat k ) + lambda (2 hat i + 4 hat j +2 hat k)`

`=>` Let `L_1` and `L_2` be two lines passing through the origin and with direction ratios `a_1, b_1, c_1` and `a_2, b_2, c_2`, respectively.

`=>` Let `P` be a point on `L_1` and Q be a point on `L_2`. Consider the directed lines `OP` and `OQ` as given in Fig .

`=>` Let `θ` be the acute angle between `OP` and `OQ.` Now recall that the directed line segments `OP` and `OQ` are vectors with components `a_1, b_1, c_1` and `a_2, b_2, c_2`, respectively. Therefore, the angle `θ` between them is given by

`color{red}{cos theta= | (a_1 a_2 + b_1 b_2 + c_1 c_2 )/(sqrt (a_(1)^2 + b_(1)^2 + c_(1)^2) sqrt ( a_(2)^2 + b_(2)^2 + c_(2)^2) ) |}`

`=>` The angle between the lines in terms of sin θ is given by

`sin theta = sqrt(1- cos^2 theta)`

`= sqrt (1- ( (a_1 a_2 + b_1 b_2 + c_1 c_2 )^2)/( ( a_(1)^2 + b_(1)^2 + c_(1)^2) ( a_(2)^2 + b_(2)^2 + c_(2)^2) ) )`

`= ( sqrt ( ( a_(1)^2 + b_(1)^2 + c_(1)^2) ( a_(2)^2 + b_(2)^2 + c_(2)^2) - (a_(1) a_2 + b_1 b_2 + c_1 c_2)^2 ) )/(sqrt ( (a_(1)^2 + b_(1)^2 + c_(1)^2) ) sqrt (a_(2)^2 + b_(2)^2 + c_(2)^2) )`

`color{red}{sintheta= ( sqrt ( (a_1 b_2 - a_2 b_1)^2 + ( b_1 c_2 - b_2 c_1)^2 + ( c_1 a_2 - c_2 a_1)^2 ) )/( sqrt (a_(1)^2 + b_(1)^2 + c_(1)^2) sqrt ( a_(2)^2 + b_(2)^2 + c_(2)^2) )}` ................(2)

`\color{green} ✍️` If instead of direction ratios for the lines `L_1` and `L_2`, direction cosines, namely, `l_1, m_1, n_1` for `L_1` and `l_2, m_2, n_2` for `L_2` are given, then (1) and (2) takes the following form:

`color{red}{cos theta = | l_1 l_2 + m_1 m_2 + n_1 n_2 |}`

(as `l_(1)^2 + m_(1)^2 + n_(1)^2 =1 = l_(2)^2 + m_(2)^2 + n_(2)^2)` .....(3)

and `sin theta = sqrt ( (l_1 m_2 - l_2 m_1 )^2 - (m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2 )` ....(4)

`=>` Two lines with direction ratios `a_1, b_1, c_1` and `a_2, b_2, c_2` are

(i) perpendicular i.e. if `θ = 90° `by (1)

`color{blue}{a_1 a_2 + b_1 b_2 + c_1 c_2 = 0}`

(ii) parallel i.e. if `θ = 0` by (2)

`color{blue}{a_1/a_2 =b_1/b_2 =c_1/c_2}`

`\color{green} ✍️` Now, we find the angle between two lines when their equations are given. If θ is acute the angle between the lines

`vec r = bar (a_1) + lambda bar (b_1)` and `vec r = bar (a_2) + mu bar ( b_2)`

then `cos theta= | ( bar (b_1) * bar (b_2) )/( | bar (b_1) | | bar (b_2) | ) |`

`\color{green} ✍️` So In Cartesian form, if θ is the angle between the lines

` (x-x_1)/(a_1) = (y-y_1)/(b_1) = (z-z_1)/(c_1)` ........(1)

and `(x-x_2)/(a_2) = (y-y_2)/(b_2) = (z-z_2)/(c_2)` ........(2)

where, `a_1, b_1, c_1` and `a_2, b_2, c_2` are the direction ratios of the lines (1) and (2), respectively, then

`=>` Let `P` be a point on `L_1` and Q be a point on `L_2`. Consider the directed lines `OP` and `OQ` as given in Fig .

`=>` Let `θ` be the acute angle between `OP` and `OQ.` Now recall that the directed line segments `OP` and `OQ` are vectors with components `a_1, b_1, c_1` and `a_2, b_2, c_2`, respectively. Therefore, the angle `θ` between them is given by

`color{red}{cos theta= | (a_1 a_2 + b_1 b_2 + c_1 c_2 )/(sqrt (a_(1)^2 + b_(1)^2 + c_(1)^2) sqrt ( a_(2)^2 + b_(2)^2 + c_(2)^2) ) |}`

`=>` The angle between the lines in terms of sin θ is given by

`sin theta = sqrt(1- cos^2 theta)`

`= sqrt (1- ( (a_1 a_2 + b_1 b_2 + c_1 c_2 )^2)/( ( a_(1)^2 + b_(1)^2 + c_(1)^2) ( a_(2)^2 + b_(2)^2 + c_(2)^2) ) )`

`= ( sqrt ( ( a_(1)^2 + b_(1)^2 + c_(1)^2) ( a_(2)^2 + b_(2)^2 + c_(2)^2) - (a_(1) a_2 + b_1 b_2 + c_1 c_2)^2 ) )/(sqrt ( (a_(1)^2 + b_(1)^2 + c_(1)^2) ) sqrt (a_(2)^2 + b_(2)^2 + c_(2)^2) )`

`color{red}{sintheta= ( sqrt ( (a_1 b_2 - a_2 b_1)^2 + ( b_1 c_2 - b_2 c_1)^2 + ( c_1 a_2 - c_2 a_1)^2 ) )/( sqrt (a_(1)^2 + b_(1)^2 + c_(1)^2) sqrt ( a_(2)^2 + b_(2)^2 + c_(2)^2) )}` ................(2)

`color{red} "Key Point"` - In case the lines `L_1` and `L_2` do not pass through the origin, we may take lines `L'_(1)` and `L′_(2)` which are parallel to `L_1` and `L_2` respectively and pass through the origin.

`\color{green} ✍️` If instead of direction ratios for the lines `L_1` and `L_2`, direction cosines, namely, `l_1, m_1, n_1` for `L_1` and `l_2, m_2, n_2` for `L_2` are given, then (1) and (2) takes the following form:

`color{red}{cos theta = | l_1 l_2 + m_1 m_2 + n_1 n_2 |}`

(as `l_(1)^2 + m_(1)^2 + n_(1)^2 =1 = l_(2)^2 + m_(2)^2 + n_(2)^2)` .....(3)

and `sin theta = sqrt ( (l_1 m_2 - l_2 m_1 )^2 - (m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2 )` ....(4)

`=>` Two lines with direction ratios `a_1, b_1, c_1` and `a_2, b_2, c_2` are

(i) perpendicular i.e. if `θ = 90° `by (1)

`color{blue}{a_1 a_2 + b_1 b_2 + c_1 c_2 = 0}`

(ii) parallel i.e. if `θ = 0` by (2)

`color{blue}{a_1/a_2 =b_1/b_2 =c_1/c_2}`

`\color{green} ✍️` Now, we find the angle between two lines when their equations are given. If θ is acute the angle between the lines

`vec r = bar (a_1) + lambda bar (b_1)` and `vec r = bar (a_2) + mu bar ( b_2)`

then `cos theta= | ( bar (b_1) * bar (b_2) )/( | bar (b_1) | | bar (b_2) | ) |`

`\color{green} ✍️` So In Cartesian form, if θ is the angle between the lines

` (x-x_1)/(a_1) = (y-y_1)/(b_1) = (z-z_1)/(c_1)` ........(1)

and `(x-x_2)/(a_2) = (y-y_2)/(b_2) = (z-z_2)/(c_2)` ........(2)

where, `a_1, b_1, c_1` and `a_2, b_2, c_2` are the direction ratios of the lines (1) and (2), respectively, then

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{red} {cos theta = | (a_1 a_2 + b_1 b_2 + c_1 c_2)/( sqrt (a_(1)^2 + b_(1)^2 + c_(1)^2) sqrt (a_(2)^2 + b_(2)^2 + c_(2)^2) ) |}`

Q 3127678581

Find the angle between the pair of lines given by

`vec r = 3 hat i + 2 hat j - 4 hat k + lambda (hat i +2 hat j + 2 hat k )`

and `vec r = 5 hat i -2 hat j + mu (3 hat i +2 hat j + 6 hat k)`

Class 12 Chapter 11 Example 9

`vec r = 3 hat i + 2 hat j - 4 hat k + lambda (hat i +2 hat j + 2 hat k )`

and `vec r = 5 hat i -2 hat j + mu (3 hat i +2 hat j + 6 hat k)`

Class 12 Chapter 11 Example 9

Here `vec (b_1) = hat i + 2 hat j + 2 hat k` and `vec (b_2) = 3 hat i + 2 hat j + 6 hat k`

The angle θ between the two lines is given by

`cos theta = | (vec (b_1) * vec ( b_2) )/( | vec b_1 | | vec b_2 | ) | = | ( (hat i +2 hat j +2 hat k)* (3 hat i + 2 hat j + 6 hat k) )/( sqrt (1+4+) sqrt (9+4+36) ) |`

`= | (3+4+12)/(3 xx 7 ) | = 19/21`

Hence `theta= cos^(-1) (19/21)`

Q 3147678583

Find the angle between the pair of lines

`(x+3)/3 = (y-1)/5 = (z+3)/4`

and `(x+1)/1 = ( y-4)/1 = (z-5)/2`

Class 12 Chapter 11 Example 10

`(x+3)/3 = (y-1)/5 = (z+3)/4`

and `(x+1)/1 = ( y-4)/1 = (z-5)/2`

Class 12 Chapter 11 Example 10

The direction ratios of the first line are 3, 5, 4 and the direction ratios of the

second line are 1, 1, 2. If θ is the angle between them, then

`cos theta = | (3 * 1 + 5* 1 + 4 *2)/( sqrt (362 + 5^2 + 4^2 ) sqrt (1^2 + 1^2 + 2^2) ) | = 16/(sqrt (50) sqrt 6 ) =16/(5 sqrt 2 sqrt 6) = (8 sqrt 3)/15`

Hence, the required angle is `cos^(-1) ( (8 sqrt 3)/15)`

`●` If two lines in space intersect at a point, then the shortest distance between them is zero.

`●` if two lines in space are parallel, then the shortest distance between them will be the perpendicular distance, i.e. the length of the perpendicular drawn from a point on one line onto the other line.

`●` Further, in a space, there are lines which are neither intersecting nor parallel. In fact, such pair of lines are non coplanar and are called skew lines.

`=>` For example, let us consider a room of size 1, 3, 2 units along x, y and z-axes respectively Fig.

`=>` The line GE that goes diagonally across the ceiling and the line DB passes through one corner of the ceiling directly above A and goes diagonally down the wall. These lines are skew because they are not parallel and also never meet.

`●` For skew lines, the line of the shortest distance will be perpendicular to both the lines.

`●` if two lines in space are parallel, then the shortest distance between them will be the perpendicular distance, i.e. the length of the perpendicular drawn from a point on one line onto the other line.

`●` Further, in a space, there are lines which are neither intersecting nor parallel. In fact, such pair of lines are non coplanar and are called skew lines.

`=>` For example, let us consider a room of size 1, 3, 2 units along x, y and z-axes respectively Fig.

`=>` The line GE that goes diagonally across the ceiling and the line DB passes through one corner of the ceiling directly above A and goes diagonally down the wall. These lines are skew because they are not parallel and also never meet.

`●` For skew lines, the line of the shortest distance will be perpendicular to both the lines.

`=>` We now determine the shortest distance between two skew lines in the following way:

`=>` Let `l_1` and `l_2` be two skew lines with equations (Fig.)

`vec r = vec (a_1) + lambda vec ( b_1)` .........(1)

and `vec r= vec (a_2) + mu vec (b_2)` .........(2)

`=>` Take any point S on `l_1` with position vector `vec (a_1)` and T on `l_2` , , with position vector `vec (a_2)`

`=>` Then the magnitude of the shortest distance vector will be equal to that of the projection of ST along the direction of the line of shortest distance.

`=>` If `bar (PQ)` is the shortest distance vector between `l_1` and `l_2` , then it being perpendicular to both `bar ( b_1)` and `bar(b_2)` , the unit vector `hat n` along `bar (PQ)` would therefore be

`color{orange}{hat n = (vec b_1 xx vec b_2 )/( | vec b_1 xx vec b_2 | )}` ...........(3)

Then `bar (PQ) = d hat n`

`=>` where, d is the magnitude of the shortest distance vector. Let θ be the angle between `bar (ST)` and `bar (PQ)` . Then

`PQ=ST | cos theta |`

But `cos theta = | ( bar (PQ) *bar (ST) )/( | bar(PQ) | | bar (ST) |) |`

`= | (d hat n * (vec (a_2) - vec (a_1) ) )/( d ST) |` ( since `bar(ST = vec a_2 - vec a_1 )`

`= | ( (vec b_1 xx vec b_2 )*( vec a_2 - vec a_1) )/(ST | vec b_1 xx vec b_2 ) |` [From (3)]

`\color{green} ✍️`Hence, the required shortest distance is

`d= PQ =ST |cos theta |`

or `color{blue}{d = | ( (vec b_1 xx vec b_2)* (vec a_2 - vec a_1) )/(| vec b_1 xx vec b_2 |) |}`

`color{brown} "Shortest Distance In Cartesian form"`

The shortest distance between the lines

`l_1 : (x-x_1)/(a_1) = (y-y_1)/(b_1) = (z-z_1)/(c_1)`

and `l_2 : (x-x_2)/(a_2) = (y-y_2)/(b_2) = (z-z_2)/(c_2)` is

`=>` Let `l_1` and `l_2` be two skew lines with equations (Fig.)

`vec r = vec (a_1) + lambda vec ( b_1)` .........(1)

and `vec r= vec (a_2) + mu vec (b_2)` .........(2)

`=>` Take any point S on `l_1` with position vector `vec (a_1)` and T on `l_2` , , with position vector `vec (a_2)`

`=>` Then the magnitude of the shortest distance vector will be equal to that of the projection of ST along the direction of the line of shortest distance.

`=>` If `bar (PQ)` is the shortest distance vector between `l_1` and `l_2` , then it being perpendicular to both `bar ( b_1)` and `bar(b_2)` , the unit vector `hat n` along `bar (PQ)` would therefore be

`color{orange}{hat n = (vec b_1 xx vec b_2 )/( | vec b_1 xx vec b_2 | )}` ...........(3)

Then `bar (PQ) = d hat n`

`=>` where, d is the magnitude of the shortest distance vector. Let θ be the angle between `bar (ST)` and `bar (PQ)` . Then

`PQ=ST | cos theta |`

But `cos theta = | ( bar (PQ) *bar (ST) )/( | bar(PQ) | | bar (ST) |) |`

`= | (d hat n * (vec (a_2) - vec (a_1) ) )/( d ST) |` ( since `bar(ST = vec a_2 - vec a_1 )`

`= | ( (vec b_1 xx vec b_2 )*( vec a_2 - vec a_1) )/(ST | vec b_1 xx vec b_2 ) |` [From (3)]

`\color{green} ✍️`Hence, the required shortest distance is

`d= PQ =ST |cos theta |`

or `color{blue}{d = | ( (vec b_1 xx vec b_2)* (vec a_2 - vec a_1) )/(| vec b_1 xx vec b_2 |) |}`

`color{brown} "Shortest Distance In Cartesian form"`

The shortest distance between the lines

`l_1 : (x-x_1)/(a_1) = (y-y_1)/(b_1) = (z-z_1)/(c_1)`

and `l_2 : (x-x_2)/(a_2) = (y-y_2)/(b_2) = (z-z_2)/(c_2)` is

` \ \ \ \ \"d" = color{red} {( | ( x_2- x_1 , y_2-y_1 , z_2 -z_1), ( a_1 , b_1 , c_1), ( a_2, b_2 , c_2) | )/( sqrt ( (b_1 c_2 - b_2 c_1)^2 + ( c_1 a_2 -c_2 a_1)^2 +(a_1 b_2 - a_2 b_1)^2) )}`

`=>` If two lines `l_1` and `l_2` are parallel, then they are coplanar. Let the lines be given by

`vec r = vec a_1 + λ vec b` ... (1)

and `vec r = vec a_2 + μ vec b` … (2)

`=>` where, `vec a_1` is the position vector of a point S on `l_1` and `vec a_2` is the position vector of a point T on `l_2` Fig.

`=>` As `l_1, l_2` are coplanar, if the foot of the perpendicular from `T` on the line `l_1` is P, then the distance between the lines `l_1` and `l_2 = |TP|`

`=>` Let θ be the angle between the vectors `bar (ST)` and `vec b`

Then, `vec b xx bar (ST) = ( | vec b | | bar (ST) | sin theta) hat n` ........(3)

`=>` where `hat n` is the unit vector perpendicular to the plane of the lines `l_1` and `l_2`.

But `bar (ST ) = vec a_2 - vec a_1`

Therefore, from (3), we get

`vec b xx (vec a_2 - vec a_1) = | bar b| PT hat n` (since PT = ST sin θ)

i.e., `| bar b xx ( vec a_2 - vec a_1 ) | = | vec b | PT * 1` (as `| hat n | = 1 |`

Hence, the distance between the given parallel lines is

`vec r = vec a_1 + λ vec b` ... (1)

and `vec r = vec a_2 + μ vec b` … (2)

`=>` where, `vec a_1` is the position vector of a point S on `l_1` and `vec a_2` is the position vector of a point T on `l_2` Fig.

`=>` As `l_1, l_2` are coplanar, if the foot of the perpendicular from `T` on the line `l_1` is P, then the distance between the lines `l_1` and `l_2 = |TP|`

`=>` Let θ be the angle between the vectors `bar (ST)` and `vec b`

Then, `vec b xx bar (ST) = ( | vec b | | bar (ST) | sin theta) hat n` ........(3)

`=>` where `hat n` is the unit vector perpendicular to the plane of the lines `l_1` and `l_2`.

But `bar (ST ) = vec a_2 - vec a_1`

Therefore, from (3), we get

`vec b xx (vec a_2 - vec a_1) = | bar b| PT hat n` (since PT = ST sin θ)

i.e., `| bar b xx ( vec a_2 - vec a_1 ) | = | vec b | PT * 1` (as `| hat n | = 1 |`

Hence, the distance between the given parallel lines is

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{red} {d= | bar (PT) | = | (vec b xx ( vec a_2 - vec a_1 ) ) /( | vec b | ) |}`

Q 3167678585

Find the shortest distance between the lines `l_1` and `l_2` whose vector

equations are

`vec r = hat i + hat j + lambda (2 hat i - hat j + hat k)` .......(1)

and `vec r = 2 hat i + hat j - hat k + mu (3 hat i -5 hat j + 2 hat k)` ..(2)

Class 12 Chapter 11 Example 11

equations are

`vec r = hat i + hat j + lambda (2 hat i - hat j + hat k)` .......(1)

and `vec r = 2 hat i + hat j - hat k + mu (3 hat i -5 hat j + 2 hat k)` ..(2)

Class 12 Chapter 11 Example 11

Comparing (1) and (2) with `vec = vec a_1 + λ vec b_1`

and `vec r = vec a_1 + mu vec b_2` respectively

we get ` vec a_1 = hat i + hat j , vec b_1 = 2hat i - hat j + hat k`

`vec a_1 = 2 hat i +hat j - hat k` and `vec b_2= 3 hat i - 5 hat j + 2 hat k`

Therefore `vec a_2 - vec a_2 = hat i - hat k`

and `vec b_1 xx vec b_2 = (2 hat i - hat j + hat k) xx (3 hat i - 5 hat j + 2 hat k)`

`= | ( hat i , hat j , hat k ), ( 2,-1,1 ), (3,-5,2) | = 3 hat i - hat j - 7 hat k`

So `| vec b_1 xx vec b_2 | = sqrt(9+1+49) = sqrt (59)`

Hence, the shortest distance between the given lines is given by

`d= | ( (vec b_1 xx vec b_2) * ( vec a_2 - vec a_1) )/( | vec b_1 xx vec b_2) | = ( | 3- 0 + 7 | )/(sqrt 59) =10/(sqrt 59)`

Q 3187678587

Find the distance between the lines `l_1` and `l_2` given by

`vec r = hat i + 2 hat j -4 hat k + lambda (2 hat i +3 hat j + 6 hat k)`

and `vec r = 3 hat i + 3 hat j - 5 hat k + mu (2 hat i +3 hat j +6 hat k )`

Class 12 Chapter 11 Example 12

`vec r = hat i + 2 hat j -4 hat k + lambda (2 hat i +3 hat j + 6 hat k)`

and `vec r = 3 hat i + 3 hat j - 5 hat k + mu (2 hat i +3 hat j +6 hat k )`

Class 12 Chapter 11 Example 12

The two lines are parallel (Why? ) We have

`vec a_1 = ht i + 2 hat j -4 hat k , vec a_2 = 3 hat i + 3 hat j - 5 hat k` and `vec b = 2 hat i + 3 hat j + 6 hat k`

Therefore, the distance between the lines is given by

`d = | (vec b xx (vec a_2 - vec a_1) )/( sqrt 49) | = (sqrt 293)/( sqrt 49) = (sqrt 293)/7`