Find the co-ordinates of foot of the perpendicular drawn from the point 2 i ^ - j ^ + 5 k ^ to the line r ¯ = ( 11 i ^ - 2 j ^ - 8 k ^ ) + λ ( 10 i ^ - 4 j ^ - 11 k ^ ) Also, find the length of the perpendicular.

Let L be the foot of the perpendicular drawn from P 2 i ^ - j ^ + 5 k ^ on the line r ¯ = ( 11 i ^ - 2 j ^ - 8 k ^ ) + λ ( 10 i ^ - 4 j ^ - 11 k ^ ) Let the position vector of L be 11 i ^ - 2 j ^ - 8 k ^ + λ ( 10 i ^ - 4 j ^ - 11 k ^ ) = ( 11 + 10 λ ) i ^ + ( - 2 - 4 λ ) j ^ + ( - 8 - 11 λ ) k ^ (As L lies on the line) Then P L → = O L → - O P → = [ ( 11 + 10 λ ) i ^ + ( - 2 - 4 λ ) j ^ + ( - 8 - 11 λ ) k ^ ] - ( 2 i ^ - j ^ + 5 k ^ ) ∴ P L → = ( 9 + 10 λ ) i ^ + ( - 1 - 4 λ ) j ^ + ( - 13 - 11 λ ) k ^ Since P L → is perpendicular to the given line and the given line is parallel to b → = 10 i ^ - 4 j ^ - 11 k ^ ∴ P L → · b → = 0 i.e, [ ( 9 + 10 λ ) i ^ + ( - 1 - 4 λ ) j ^ + ( - 13 - 11 λ ) k ^ ] · ( 10 i ^ - 4 j ^ - 11 k ^ ) = 0 ∴ 10 ( 9 + 10 λ ) - 4 ( - 1 - 4 λ ) - 11 ( - 13 - 11 λ ) = 0 ∴ 237 λ = - 237 ∴ λ = - 1 Putting the value of λ = - 1 , we obtain the position vector of L as i ^ + 2 j ^ + 3 k ^ Now, P L → = ( i ^ + 2 j ^ + 3 k ^ ) - ( 2 i ^ - j ^ + 5 k ^ ) = - i ^ + 3 j ^ - 2 k ^ Hence, the length of the perpendicular from P on the given line is | P L → | = 1 + 9 + 4 = 14 units. Hence, the length of the perpendicular from the point 2 i ^ - j ^ + 5 k ^ is 14 units.

Find the shortest distance between the lines r = 4 i ^ - j ^ + λ i ^ + 2 j ^ - 3 k ^ and r = i ^ - j ^ + 2 k ^ + μ i ^ + 4 j ^ - 5 k ^

We know that, The shortest distance between the lines r → = a → 1 + λ b → 1 and r → = a → 2 + λ b → 2 is given by d = a → 2 - a → 1 · b → 1 × b → 2 b → 1 × b → 2 Here, we have, a → 1 = 4 i ^ - j ^ , a → 2 = i ^ - j ^ + 2 k ^ , b → 1 = i ^ + 2 j ^ - 3 k ^ and b → 2 = i ^ + 4 j ^ - 5 k ^ Now, a → 2 - a → 1 = i ^ - j ^ + 2 k ^ - 4 i ^ + j ^ = - 3 i ^ + 0 j ^ + 2 k ^ And b → 1 × b → 2 = i ^ j ^ k ^ 1 2 - 3 1 4 - 5 = - 10 + 12 i ^ - - 5 + 3 j ^ + 4 - 2 k ^ = 2 i ^ + 2 j ^ + 2 k ^ Therefore, a → 2 - a → 1 · b → 1 × b → 2 = ( - 3 i ^ + 0 j ^ + 2 k ^ ) · ( 2 i ^ + 2 j ^ + 2 k ^ ) = - 6 + 0 + 4 = - 2 And b → 1 × b → 2 = 4 + 4 + 4 = 2 3 Therefore, d = a → 2 - a → 1 · b → 2 × b → 1 b → 2 × b → 1 = - 2 2 3 = 1 3 units. Hence, the shortest distance between the lines is 1 3 units.

Find the shortest distance between the lines x + 1 7 = y + 1 - 6 = z + 1 1 and x - 3 1 = y - 5 - 2 = z - 7 1 .

We know that the shortest distance between the two lines 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 is given by, d = 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 . . . . . . . i Given equations of line are x + 1 7 = y + 1 - 6 = z + 1 1 and x - 3 1 = y - 5 - 2 = z - 7 1 . Comparing with the given equations, we get x 1 = - 1 , y 1 = - 1 , z 1 = - 1 a 1 = 7 , b 1 = - 6 , c 1 = 1 x 2 = 3 , y 2 = 5 , z 2 = 7 a 2 = 1 , b 2 = - 2 , c 2 = 1 Therefore, x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 = 4 6 8 7 - 6 1 1 - 2 1 = 4 - 6 + 2 - 6 7 - 1 + 8 - 14 + 6 = - 16 - 36 - 64 = - 116 ⇒ ( 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 = ( - 6 + 2 ) 2 + ( 1 - 7 ) 2 + ( - 14 + 6 ) 2 = 16 + 36 + 64 = 116 = 2 29 Substituting all the values in equation i , we get d = - 116 2 29 = - 58 29 = - 2 × 29 29 = 2 29 units The distance between the lines x + 1 7 = y + 1 - 6 = z + 1 1 and x - 3 1 = y - 5 - 2 = z - 7 1 is 2 29 units.

Find the perpendicular distance of the point 1 , 0 , 0 from the line x - 1 2 = y + 1 - 3 = z + 10 8 . Also find the coordinates of the foot of the perpendicular.

Let the point 1 , 0 , 0 be P . Given equation: x - 1 2 = y + 1 - 3 = z + 10 8 So, the given line passes through Q 1 , - 1 , - 10 and the line is parallel to vector b → = 2 i ^ - 3 j ^ + 8 k ^ . So, P Q = 1 - 1 , - 1 - 0 , - 10 - 0 i.e P Q ≡ 0 , - 1 , - 10 ⇒ P Q → = 0 i ^ - j ^ - 10 k ^ Now, b → × P Q → = i ^ j ^ k ^ 2 - 3 8 0 - 1 - 10 = i ^ - 10 × - 3 - - 1 × 8 - j ^ - 10 × 2 - 8 × 0 + k ^ 2 × - 1 - - 3 × 0 = i ^ 30 + 8 - j ^ - 20 + k ^ - 2 + 0 = 38 i ^ + 20 j ^ - 2 k ^ ⇒ b → × P Q → = 38 2 + 20 2 + - 2 2 = 1444 + 400 + 4 = 1848 Hence, the perpendicular distance d = b → × P Q → b → = 1848 2 2 + - 3 2 + 8 2 = 1848 4 + 9 + 64 = 1848 77 = 24 = 2 6 Let F be the foot of the perpendicular from the point P 1 , 0 , 0 on the given line. The coordinates of any general point on the line x - 1 2 = y + 1 - 3 = z + 10 8 = k will be x = 2 k + 1 , y = - 3 k - 1 , z = 8 k - 10 Let the coordinates of L be 2 k + 1 , - 3 k - 1 , 8 k - 10 . Then, the direction ratios of P L is 2 k + 1 - 1 , - 3 k - 1 - 0 , 8 k - 10 - 0 i.e. 2 k , - 3 k - 1 , 8 k - 10 The direction ratios of the given line are 2 , - 3 , 8 . Also, it is perpendicular to P L . According to the perpendicular condition: 2 2 k - 3 - 3 k - 1 + 8 8 k - 10 = 0 ⇒ 4 k + 9 k + 3 + 64 k - 80 = 0 ⇒ 77 k = 77 ⇒ k = 77 77 = 1 Substituting k in coordinates of L , we get 2 1 + 1 , - 3 1 - 1 , 8 1 - 10 = 3 , - 4 , - 2 Hence, the coordinates of foot of perpendicular are ( 3 , - 4 , - 2 ) .

If lines x - 1 2 = y + 1 3 = z - 1 4 and x - 3 1 = y - k 2 = z 1 intersect each other then find k .

Given, x - 1 2 = y + 1 3 = z - 1 4 . . . . 1 x - 3 1 = y - k 2 = z 1 . . . . 2 Now let x - 1 2 = y + 1 3 = z - 1 4 = λ So, x = 2 λ + 1 , y = 3 λ - 1 & z = 4 λ + 1 And let x - 3 1 = y - k 2 = z 1 = μ So, x = μ + 3 , y = 2 μ + k & z = μ Now if the line intersect each other then, 2 λ + 1 = μ + 3 , 3 λ - 1 = 2 μ + k , 4 λ + 1 = μ ⇒ 2 λ - μ = 2 , 3 λ - 2 μ = k + 1 , 4 λ - μ = - 1 Now solving above equations we get, λ = - 3 2 & μ = - 5 and putting the value in 3 λ - 2 μ = k + 1 we get, k = 9 2

E M B I B E

Mathematics & Statistics (Arts & Science) Part 1 Standard 12>Chapter 6 - Line and Plane>Exercise 6.2

Maharashtra Board Solutions for Chapter: Line and Plane, Exercise 2: Exercise 6.2

Author:Maharashtra Board

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Line and Plane, Exercise 2: Exercise 6.2

Attempt the practice questions on Chapter 6: Line and Plane, Exercise 2: Exercise 6.2 with hints and solutions to strengthen your understanding. Mathematics & Statistics (Arts & Science) Part 1 Standard 12 solutions are prepared by Experienced Embibe Experts.

Questions from Maharashtra Board Solutions for Chapter: Line and Plane, Exercise 2: Exercise 6.2 with Hints & Solutions

HARD

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 1 Standard 12>Chapter 6 - Line and Plane>Exercise 6.2>Q 1

Find the length of the perpendicular from $A (2, - 3, 1)$ on the line $\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{- 1}$

HARD

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 1 Standard 12>Chapter 6 - Line and Plane>Exercise 6.2>Q 2

Find the co-ordinates of foot of the perpendicular drawn from the point $2 \hat{i} - \hat{j} + 5 \hat{k}$ to the line $\bar{r} = (11 \hat{i} - 2 \hat{j} - 8 \hat{k}) + λ (10 \hat{i} - 4 \hat{j} - 11 \hat{k})$
Also, find the length of the perpendicular.

MEDIUM

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 1 Standard 12>Chapter 6 - Line and Plane>Exercise 6.2>Q 3

Find the shortest distance between the lines $r = (4 \hat{i} - \hat{j}) + λ (\hat{i} + 2 \hat{j} - 3 \hat{k})$ and $r = (\hat{i} - \hat{j} + 2 \hat{k}) + μ (\hat{i} + 4 \hat{j} - 5 \hat{k})$

HARD

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 1 Standard 12>Chapter 6 - Line and Plane>Exercise 6.2>Q 4

Find the shortest distance between the lines $\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}$ and $\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}$ .

MEDIUM

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 1 Standard 12>Chapter 6 - Line and Plane>Exercise 6.2>Q 5

Find the perpendicular distance of the point $(1, 0, 0)$ from the line $\frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 10}{8}$ . Also find the coordinates of the foot of the perpendicular.

MEDIUM

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 1 Standard 12>Chapter 6 - Line and Plane>Exercise 6.2>Q 6

$A (1, 0, 4), B (0, - 11, 13), C (2, - 3, 1)$ are three points and $D$ is the foot of the perpendicular from $A$ to $B C$ . Find the co-ordinates of $D$ .

MEDIUM

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 1 Standard 12>Chapter 6 - Line and Plane>Exercise 6.2>Q 7

By computing the shortest distance determine whether the following pairs of lines intersect or not: $\frac{x - 5}{4} = \frac{y - 7}{- 5} = \frac{z + 3}{- 5}$ and $\frac{x - 8}{7} = \frac{y - 7}{1} = \frac{z - 5}{3}$

MEDIUM

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 1 Standard 12>Chapter 6 - Line and Plane>Exercise 6.2>Q 8

If lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect each other then find $k$ .

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