Find the shortest distance between the lines r → = i ^ + 2 j ^ + 3 k ^ + λ ( i ^ - j ^ + k ^ ) and r → = 2 i ^ - j ^ - k ^ + μ ( - i ^ + j ^ - k ^ ) .

Given lines are r → = ( i + 2 j + 3 k ) + λ ( i - j + k ) r → = ( 2 i - j - k ) + μ ( - i + j - k ) ⇒ r → = ( 2 i - j - k ) + μ ' ( i - j + k ) Where μ ' = - μ Hence, both given lines are parallel to b → = i - j + k Here a 1 → = i + 2 j + 3 k ; a 2 → = 2 i - j - k Now a 2 → - a 1 → = ( 2 i - j - k ) - ( i + 2 j + 3 k ) = i - 3 j - 4 k ∴ ( a 2 → - a 1 → ) × b → = i j k 1 - 3 - 4 1 - 1 1 = i ( - 3 - 4 ) - j ( 1 + 4 ) + k ( - 1 + 3 ) = - 7 i - 5 j + 2 k | ( a 2 → - a 1 → ) | × b → | = ( - 7 ) 2 + ( - 5 ) 2 + 2 2 = 49 + 25 + 4 = 78 | b → | = 1 2 + 1 2 + 1 2 = 3 ∴ Shortest distance = | ( a 2 → - a 1 → ) × b → | | b → | = 78 3 = 26

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

E M B I B E

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE

M L Aggarwal Solutions for Chapter: Three Dimensional Geometry, Exercise 4: EXERCISE

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Three Dimensional Geometry, Exercise 4: EXERCISE

Attempt the free practice questions on Chapter 2: Three Dimensional Geometry, Exercise 4: EXERCISE with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 12 Volume 2 solutions are prepared by Experienced Embibe Experts.

Questions from M L Aggarwal Solutions for Chapter: Three Dimensional Geometry, Exercise 4: EXERCISE with Hints & Solutions

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE>Q 1

Find the shortest distance between pair of lines: $\vec{r} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} + λ (\hat{i} - 2 \hat{j} + 2 \hat{k})$ and $\vec{r} = - 4 \hat{i} - \hat{k} + μ (3 \hat{i} - 2 \hat{j} - 2 \hat{k})$

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE>Q 1

Find the shortest distance between pair of lines: $\vec{r} = (8 + 3 λ) \hat{i} - (9 + 16 λ) \hat{j} + (10 + 7 λ) \hat{k}$ and $\vec{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + μ (3 \hat{i} + 8 \hat{j} - 5 \hat{k})$

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE>Q 2

Find the shortest distance between the lines $\frac{x - 8}{3} = \frac{y + 9}{- 16} = = \frac{10 - z}{- 7} and \frac{x - 15}{3} = \frac{58 - 2 y}{- 16} = \frac{z - 5}{- 5}$ .

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE>Q 7

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

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE>Q 8

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 ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE>Q 9

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.