Find the shortest distance between the two given straight lines r → = 2 i ^ + 3 j ^ + 4 k ^ + t - 2 i ^ + j ^ - 2 k ^ and x - 3 2 = y - 1 = z + 2 2 .

The line r → = 2 i ^ + 3 j ^ + 4 k ^ + t - 2 i ^ + j ^ - 2 k ^ can be written in cartesian form as x - 2 - 2 = y - 3 1 = z - 4 - 2 and other line is x - 3 2 = y - 1 = z + 2 2 On comparing the given equations of lines with x - x 1 a 1 = y - y 1 b 1 = z - z 1 c 1 and x - x 2 a 2 = y - y 2 b 2 = z - z 2 c 2 , We get x 1 = 2 , y 1 = 3 , z 1 = 4 ; a 1 = - 2 , b 1 = 1 , c 1 = - 2 and x 2 = 3 , y 2 = 0 , z 2 = - 2 ; a 2 = 2 , b 2 = - 1 , c 2 = 2 . We need to find the shortest distance. Shortest distance = x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 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 On putting the values in x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 we get 3 - 2 0 - 3 - 2 - 4 - 2 1 - 2 2 - 1 2 = 1 - 3 - 6 - 2 1 - 2 2 - 1 2 = 1 1 × 2 - - 1 × - 2 + 3 - 2 × 2 - 2 × - 2 - 6 - 2 × - 1 - 1 × 2 = 1 2 - 2 + 3 - 4 + 4 - 6 2 - 2 = 0 Hence the lines are intersecting lines. Therefore, the shortest distance between the lines is 0 . .

Show that the lines x - 1 2 = y - 2 3 = z - 3 4 and x - 4 5 = y - 1 2 = z intersect. Also, find their point of intersection.

Given lines are x - 1 2 = y - 2 3 = z - 3 4 → 1 and x - 4 5 = y - 1 2 = z → 2 We know that the equation of a line through a point x 1 , y 1 , z 1 and having direction ratios a , b , and c is given by x - x 1 a = y - y 1 b = z - z 1 c → 3 ∴ Let line 1 passes through a point x 1 , y 1 , z 1 and have direction ratios a 1 , b 1 , c 1 We have, x 1 = 1 , y 1 = 2 , z 1 = 3 and a 1 = 2 , b 1 = 3 , c 1 = 4 Similarly, let the line 2 passes through a point x 2 , y 2 , z 2 and has direction ratios a 2 , b 2 , c 2 Also, x 2 = 4 , y 2 = 1 , z 2 = 0 and a 2 = 5 , b 2 = 2 , c 2 = 1 We know that the shortest distance between the two intersecting lines is zero. ∴ Shortest distance between two given lines = x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 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 = 4 - 1 1 - 2 0 - 3 2 3 4 5 2 1 ( 3 · 1 - 2 · 4 ) 2 + ( 4 · 5 - 1 · 2 ) 2 + ( 2 · 2 - 5 · 3 ) 2 = 3 - 1 - 3 2 3 4 5 2 1 25 + 324 + 121 = 3 ( 3 - 8 ) + 1 ( 2 - 20 ) - 3 ( 4 - 15 ) 470 = - 15 - 18 + 33 470 = 0 470 = 0 Therefore, the given two lines are intersecting. Now, to find the point of intersection of the given lines. Let x - 1 2 = y - 2 3 = z - 3 4 = λ ⇒ x = 2 λ + 1 , y = 3 λ + 2 and z = 4 λ + 3 Since the lines are intersecting. The above coordinates will satisfy the equation of the second line. We can put the above values in the equation of another line. Thus, 2 λ + 1 - 4 5 = 3 λ + 2 - 1 2 = 4 λ + 3 1 ⇒ 2 λ - 3 5 = 3 λ + 1 2 = 4 λ + 3 1 ⇒ 2 λ - 3 5 = 4 λ + 3 1 ⇒ 2 λ - 3 = 20 λ + 15 ⇒ 18 λ = - 18 ⇒ λ = - 1 So, the required point of intersection is x = 2 ( - 1 ) + 1 = - 1 y = 3 ( - 1 ) + 2 = - 1 z = 4 ( - 1 ) + 3 = - 1 Thus, the lines intersect at ( - 1 , - 1 , - 1 ) .

Find the shortest distance between the lines x - 3 3 = y - 8 - 1 = z - 3 1 and x + 3 - 3 = y + 7 2 = z - 6 4 . Find also the equation of the line of the shortest distance.

Given that: x - 3 3 = y - 8 - 1 = z - 3 1 = r 1 . . . . . 1 and x + 3 - 3 = y + 7 2 = z - 6 4 = r 2 . . . . . 2 Any point on line ( 1 ) is P ( 3 r 1 + 3 , 8 - r 1 , r 1 + 3 ) , and on line ( 2 ) is Q ( - 3 - 3 r 2 , 2 r 2 - 7 , 4 r 2 + 6 ) . If P Q is the line of the shortest distance, then direction ratios of P Q = ( 3 r 1 + 3 ) - ( - 3 - 3 r 2 ) , ( 8 - r 1 ) - ( 2 r 2 - 7 ) , ( r 1 + 3 ) - ( 4 r 2 + 6 ) . That is, 3 r 1 + 3 r 2 + 6 , - r 1 - 2 r 2 + 15 , r 1 - 4 r 2 - 3 As P Q is perpendicular to lines ( 1 ) and ( 2 ) , Therefore 3 ( 3 r 1 + 3 r 2 + 6 ) - 1 ( - r 1 - 2 r 2 + 15 ) + 1 ( r 1 - 4 r 2 + 3 ) = 0 ⇒ 11 r 1 + 7 r 2 = 0 . . . . . ( 3 ) And - 3 ( 3 r 1 + 3 r 2 + 6 ) + 2 ( - r 1 - 2 r 2 + 15 ) + 4 ( r 1 - 4 r 2 + 3 ) = 0 ⇒ 7 r 1 + 11 r 2 = 0 . . . . . ( 4 ) On solving equations ( 3 ) and ( 4 ) , we get r 1 = r 2 = 0 . So, points P ( 3 , 8 , 3 ) and Q ( - 3 , - 7 , 6 ) . Therefore, length of the shortest distance P Q = - 3 - 3 2 + - 7 - 8 2 + 6 - 3 2 P Q = 3 30 Direction ratios of shortest distance line are 2 , 5 and - 1 . Therefore, equation of the shortest distance line is x - 3 2 = y - 8 5 = z - 3 - 1

Find the vector equation of the line passing through the point ( - 1 , 2 , 1 ) and parallel to the line r → = 2 i ^ + 3 j ^ - k ^ + λ ( i ^ - 2 j ^ + k ^ ) .Also, find the distance between these lines.

Equation of line be r → = 2 i + 3 j - k + λ ( i - 2 j + k ) ... ( 1 ) Comparing with r → = a 1 → + λ b → where a 1 → = 2 i + 3 j - k ; b → = i - 2 j + k Now vector equation of line through the point ( - 1 , 2 , 1 ) and parallel to line ( 1 ) be given by r → = a 2 → + λ b → where a 2 → = - i + 2 j + k and b → = i - 2 j + k ∴ a 2 → - a 1 → = ( - i + 2 j + k ) - ( 2 i + 3 j - k ) = - 3 i - j + 2 k b → × ( a 2 → - a 1 → ) = i j k 1 - 2 1 - 3 - 1 2 = i ( - 4 + 1 ) - j ( 2 + 3 ) + k ( - 1 - 6 ) = - 3 i - 5 j - 7 k ∴ | b → × ( a 2 → - a 1 → ) | = ( - 3 ) 2 + ( - 5 ) 2 + ( - 7 ) 2 = 9 + 25 + 49 = 83 and | b → | = 1 2 + ( - 2 ) 2 + 1 2 = 6 Thus required shortest distance between given lines = | b → × ( a 2 → - a 1 → ) | | b → | = 83 6 = 83 6 units

Find the vector equation of the line parallel to the line x - 1 5 = 3 - y 2 = z + 1 4 and passing through ( 3 , 0 , - 4 ) . Also find the distance between these two lines.

Given equation of line be, x - 1 5 = y - 3 - 2 = z + 1 4 ... ( 1 ) Since line ( 1 ) passes through line point ( 1 , 3 , - 1 ) and having direction ratios < 5 , - 2 , 4 > So the vector equation of line ( 1 ) be r → = a 1 → + λ b → where a 1 → = i + 3 j - k ; b → = 5 i - 2 j + 4 k Now vector equation of line through the point ( 3 , 0 , - 4 ) and parallel to given line ( 1 ) is given by r → = 3 i + 0 j - 4 k + λ ( 5 i - 2 j + 4 k ) and it is of the form r → = a 2 → + λ b → where a 2 → = 3 i + 0 j - 4 k ∴ a 2 → - a 1 → = ( 3 i + 0 j - 4 k ) - ( i + 3 j - k ) = 2 i - 3 j - 3 k b → × ( a 2 → - a 1 → ) = i j k 5 - 2 4 2 - 3 - 3 = i ( 6 + 12 ) - j ( - 15 - 8 ) + k ( - 15 + 4 ) = 18 i + 23 j - 11 k ∴ | b → × ( a 2 → - a 1 → ) | = 18 2 + 23 2 + ( - 11 ) 2 = 324 + 529 + 121 = 974 and | b → | = 5 2 + ( - 2 ) 2 + 4 2 = 25 + 4 + 16 = 45 Required shortest distance between given lines = | b → × ( a 2 → - a 1 → ) | | b → | = 974 45 units

The vector equation of the line passing through the point ( - 1 , 5 , 4 ) and perpendicular to the plane z = 0 is

r → = - i ^ + 5 j ^ + ( 4 + λ ) k ^

r → = - i ^ + 5 j ^ + 4 k ^ + λ ( i ^ + j ^ )

r → = i ^ - 5 j ^ - 4 k ^ + λ k ^

An equation of the line passing through 3 i ^ - 5 j + 7 k ^ and perpendicular to the plane 3 x - 4 y + 5 z = 8 is

r = 3 i ^ + 5 j ^ + 7 k ^ + λ ( 3 i ^ + 4 j ^ - 5 k ^ )

r = 3 i ^ - 4 j ^ - 5 k ^ + μ ( 3 i ^ + 5 j ^ + 7 k ^ )

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Three Dimensional Geometry>Exercise>Q 71

HARD

12th Uttar Pradesh Board

IMPORTANT

Earn 100

Find the vector and Cartesian equations of the line that passes through origin and $(5, - 2, 3)$ .

Important Questions on Three Dimensional Geometry

HARD

12th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Three Dimensional Geometry>Exercise>Q 72

Prove that the line through

A (0, - 1, - 1) and B (4, 5, 1)

intersects the line through

C (3, 9, 4) and D (- 4, 4, 4)

.

HARD

12th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Three Dimensional Geometry>Exercise>Q 73

Find the shortest distance between the two given straight lines

\vec{r} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + t (- 2 \hat{i} + \hat{j} - 2 \hat{k})

and

\frac{x}{5} = \frac{y}{- 2} = \frac{z}{3}

HARD

12th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Three Dimensional Geometry>Exercise>Q 74

Show that the lines

$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = z$ intersect. Also, find their point of intersection.

HARD

12th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Three Dimensional Geometry>Exercise>Q 75

Show that the shortest distance between the lines

x + a = 2 y = - 12 z

and

x + y + 2 a = 6 z - 6 a

, is

2 a

.

HARD

12th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Three Dimensional Geometry>Exercise>Q 76

Find the shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} and \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ . Find also the equation of the line of the shortest distance.

EASY

12th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Three Dimensional Geometry>Exercise>Q 77

Find the vector equation of the line passing through the point

(- 1, 2, 1)

and parallel to the line

\vec{r} = 2 \hat{i} + 3 \hat{j} - \hat{k} + λ (\hat{i} - 2 \hat{j} + \hat{k})

.Also, find the distance between these lines.

MEDIUM

12th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Three Dimensional Geometry>Exercise>Q 78

Find the vector equation of the line parallel to the line

\frac{x - 1}{5} = \frac{3 - y}{2} = \frac{z + 1}{4}

and passing through

(3, 0, - 4)

. Also find the distance between these two lines.

MEDIUM

12th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 11 - Three Dimensional Geometry>Exercise>Q 79

Define the line of shortest distance between the two lines. Find the shortest distance and the vector equation of the line of shortest distance between the lines given by $\vec{r} = (- 4 \hat{i} + 4 \hat{j} + \hat{k}) + λ (\hat{i} + \hat{j} - \hat{k}) and \vec{r} = (- 3 \hat{i} - 8 \hat{j} - 3 \hat{k}) + μ (2 \hat{i} + 3 \hat{j} + 3 \hat{k})$ .