Find the angle between the following pair of lines r → = 2 + s i ^ + s - 1 j ^ + 2 - 3 s k ^ r → = 1 - t ı ^ + 2 t + 3 ȷ ^ + k ^

Given: r → = 2 + s i ^ + s - 1 j ^ + 2 - 3 s k ^ = 2 i ^ - j ^ + 2 k ^ + s i ^ + j ^ - 3 k ^ r → = 1 - t ı ^ + 2 t + 3 ȷ ^ + k ^ = i ^ + 3 j ^ + k ^ + t - i ^ + 2 j ^ Let the angle between two given lines is θ ∴ cos θ = b 1 → · b 2 → b 1 → b 2 → = ( i ^ + j ^ - 3 k ^ ) ( - i ^ + 2 j ^ ) ( 1 ) 2 + ( 1 ) 2 + ( 3 ) 2 ( - 1 ) 2 + ( 2 ) 2 = ( 1 × - 1 ) + ( 1 × 2 ) + ( - 3 × 0 ) 11 5 = - 1 + 2 55 = 1 55 ⇒ θ = cos - 1 1 55 Therefore, angle between given lines will be cos - 1 1 55 or π - cos - 1 1 55 .

Find the shortest distance between the following pair of lines- r → = i ^ + ȷ ^ + λ 2 i ^ - ȷ ^ + k ^ and r → = 2 i ^ + j ^ - k ^ + μ 3 i ^ - 5 j ^ + 2 k ^ .

The shortest distance between two vector lines r → = a 1 → + λ b 1 → and r → = a 2 → + μ b 2 → is given by b 1 → × b 2 → . a 2 → - a 1 → b 1 → × b 2 → Given equations: ⇒ r → = i ^ + j ^ + λ 2 i ^ - j ^ + k ^ and r → = 2 i ^ + j ^ - k ^ + μ 3 i ^ - 5 j ^ + 2 k ^ Comparing it with standard forms, we get ⇒ a 1 → = i ^ + j ^ a 2 → = 2 i ^ + j ^ - k ^ ⇒ b 1 → = 2 i ^ - j ^ + k ^ b 2 → = 3 i ^ - 5 j ^ + 2 k ^ Now, ⇒ a 2 → - a 1 → = 2 - 1 i ^ + 1 - 1 j ^ + - 1 - 0 k ^ ⇒ i ^ + 0 j ^ - k ^ ⇒ b 1 → × b 2 → = i ^ j ^ k ^ 2 - 1 1 3 - 5 2 ⇒ i ^ - 2 - - 5 - j ^ 4 - 3 + k ^ - 10 - - 3 ⇒ 3 i ^ - j ^ - 7 k ^ Magnitude of b 1 → × b 2 → : ⇒ 3 2 + - 1 2 + - 7 2 ⇒ 9 + 1 + 49 = 59 Now, b 1 → × b 2 → . a 2 → - a 1 → = 3 i ^ - j ^ - 7 k ^ . i ^ - k ^ ⇒ 3 + 7 = 10 The shortest distance is given by: b 1 → × b 2 → . a 2 → - a 1 → b 1 → × b 2 → = 10 59

Find the shortest distance between the following pair of lines r → = 1 - t i ^ + t - 2 j ^ + 3 - 2 t k ^ and r → = s + 1 i ^ + 2 s - 1 j ^ - 2 s + 1 k ^ .

The shortest distance between two vector lines r → = a 1 → + λ b 1 → and r → = a 2 → + μ b 2 → is given by b 1 → × b 2 → . a 2 → - a 1 → b 1 → × b 2 → Given equations: r ⇀ = 1 - t i ^ + t - 2 j ^ + 3 - 2 t k ^ . . . i r → = s + 1 i ^ + 2 s - 1 j ^ - 2 s + 1 k ^ . . . ii ⇒ r → = i ^ - i ^ t + t j ^ - 2 j ^ + 3 k ^ - 2 t k ^ r → = s i ^ + i ^ + 2 s j ^ - j ^ - 2 s k ^ - k ^ ⇒ r → = i ^ - 2 j ^ + 3 k ^ + t - i ^ + j ^ - 2 k ^ r → = i ^ - j ^ - k ^ + s i ^ + 2 j ^ - 2 k ^ Comparing with standard forms: ⇒ a 1 → = i ^ - 2 j ^ + 3 k ^ a 2 → = i ^ - j ^ - k ^ ⇒ b 1 → = - i ^ + j ^ - 2 k ^ b 2 → = i ^ + 2 j ^ - 2 k ^ ⇒ a 2 → - a 1 → = i ^ - j ^ - k ^ - i ^ - 2 j ^ + 3 k ^ = i ^ - i ^ + - j ^ + 2 j ^ + - k ^ - 3 k ^ = j ^ - 4 k ^ Now, b 1 → × b 2 → = i ^ j ^ k ^ - 1 1 - 2 1 2 - 2 = i ^ - 2 - - 4 - j ^ 2 - - 2 + k ^ - 2 - 1 = 2 i ^ - 4 j ^ - 3 k ^ Magnitude of b 1 → × b 2 → = 2 2 + - 4 2 + - 3 2 = 4 + 16 + 9 = 29 Also, ⇒ a 2 → - a 1 → . b 1 → × b 2 → = j ^ - 4 k ^ 2 i ^ - 4 j ^ - 3 k ^ = - 4 + 12 = 8 The shortest distance is given by: ⇒ b 1 → × b 2 → . a 2 → - a 1 → b 1 → × b 2 → = 8 29 = 8 29

Find the shortest distance between the following pair of lines r → = i ^ + 2 j ^ + 3 k ^ + λ i ^ - 3 j ^ + 2 k ^ and r → = 4 i ^ + 5 j ^ + 6 k ^ + μ 2 i ^ + 3 j ^ + k ^ .

The shortest distance between two lines r → = a 1 → + λ b 1 → and r → = a 2 → + μ b 2 → is given by: b 1 → × b 2 → . a 2 → - a 1 → b 1 → × b 2 → Given vector equations: ⇒ r → = i ^ + 2 j ^ + 3 k ^ + λ i ^ - 3 j ^ + 2 k ^ , r → = 4 i ^ + 5 j ^ + 6 k ^ + μ 2 i ^ + 3 j ^ + k ^ Comparing with the standard form we get: ⇒ a 1 → = i ^ + 2 j ^ + 3 k ^ a 2 → = 4 i ^ + 5 j ^ + 6 k ^ ⇒ b 1 → = i ^ - 3 j ^ + 2 k ^ b 2 → = 2 i ^ + 3 j ^ + k ^ a 2 → - a 1 → = 4 i ^ + 5 j ^ + 6 k ^ - i ^ + 2 j ^ + 3 k ^ ⇒ 4 i ^ - i ^ + 5 j ^ - 2 j ^ + 6 k ^ - 3 k ^ = 3 i ^ + 3 j ^ + 3 k ^ b 1 → × b 2 → = i j k 1 - 3 2 2 3 1 ⇒ i - 3 - 6 - j 1 - 4 + k 3 - - 6 ⇒ - 9 i + 3 j + 9 k Magnitude of ( b 1 ⇀ × b 2 → ) = - 9 2 + 3 2 + 9 2 ⇒ 81 + 9 + 81 = 171 Now, a 2 → - a 1 → . b 1 → × b 2 → = 3 i ^ + 3 j ^ + 3 k ^ - 9 i ^ + 3 j ^ + 9 k ^ ⇒ - 27 + 9 + 27 = 9 The shortest distance is given by: ⇒ b 1 → × b 2 → . a 2 → - a 1 → b 1 → × b 2 → = 9 171 = 9 3 19 ⇒ 3 19

Find the shortest distance between the following pair of lines: x + 1 7 = y + 1 - 6 = z + 1 1 x - 3 1 = y - 5 - 2 = z - 7 1 .

Given: ⇒ x + 1 7 = y + 1 - 6 = z + 1 1 ⇒ x - 3 1 = y - 5 - 2 = z - 7 1 Comparing the above equations with the standard form, we get, ⇒ x - x 1 a 1 = y - y 1 b 1 = z - z 1 c 1 ⇒ x - x 2 a 2 = y - y 2 b 2 = z - z 2 b 2 We get, x 1 = - 1 , y 1 = - 1 , z 1 = - 1 , a 1 = 7 , b 1 = - 6 , c 1 = 1 And, x 2 = 3 , y 2 = 5 , z 2 = 7 , a 2 = 1 , b 2 = - 2 , c 2 = 1 The shortest distance between the two lines can be given as, x - x 1 a 1 = y - y 1 b 1 = z - z 1 c 1 x - x 2 a 2 = y - y 2 b 2 = z - z 2 b 2 ⇒ 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 Then, we get, ⇒ x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 = 3 - - 1 5 - - 1 7 - - 1 7 - 6 1 1 - 2 1 ⇒ 4 6 8 7 - 6 1 1 - 2 1 = 4 - 6 + 2 - 6 7 - 1 + 8 - 14 + 6 ⇒ - 16 - 36 - 64 = - 116 And, ⇒ 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 Now, we get, ⇒ d = - 116 2 29 d = 58 29 d = 2 29

Find the shortest distance between the lines x - 1 2 = y - 2 3 = z - 3 4 and x - 2 3 = y - 4 4 = z - 5 5 .

Given: ⇒ x - 1 2 = y - 2 3 = z - 3 4 x - 2 3 = y - 4 4 = z - 5 5 Comparing the equations with the standard cartesian form, we get, ⇒ x - x 1 a 1 = y - y 1 b 1 = z - z 1 c 1 x - x 2 a 2 = y - y 2 b 2 = z - z 2 c 2 We get, x 1 = 1 , y 1 = 2 , z 1 = 3 , a 1 = 2 , b 1 = 3 , c 1 = 4 And, x 2 = 2 , y 2 = 4 , z 2 = 5 , a 2 = 3 , b 2 = 4 , c 2 = 5 The shortest distance between the lines can be given as, ⇒ 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 Now, we get, x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 = 2 - 1 4 - 2 5 - 3 2 3 4 3 4 5 ⇒ 1 2 2 2 3 4 3 4 5 = 1 15 - 16 - 2 10 - 12 + 2 8 - 9 ⇒ - 1 + 4 - 2 = 1 Now, we get, 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 = ( 15 - 16 ) 2 + ( 12 - 10 ) 2 + ( 8 - 9 ) 2 = - 1 2 + 2 2 + - 1 2 = 6 The shortest distance between the lines can be given as, ⇒ d = 1 6 d = 1 6

E M B I B E

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>3-D Geometry>Chapter 7 - Straight Lines>EXERCISE

K. C. Sinha and Abhishek Chandra Solutions for Exercise 1: EXERCISE

Author:K. C. Sinha & Abhishek Chandra

K. C. Sinha Mathematics Solutions for Exercise - K. C. Sinha and Abhishek Chandra Solutions for Exercise 1: EXERCISE

Attempt the practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. VECTOR AND 3-D GEOMETRY for JEE Main and Advanced solutions are prepared by Experienced Embibe Experts.

Questions from K. C. Sinha and Abhishek Chandra Solutions for Exercise 1: EXERCISE with Hints & Solutions

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 7 - Straight Lines>EXERCISE>Q 1

Find the angle between the following pair of lines

$\vec{r} = (2 + s) \hat{i} + (s - 1) \hat{j} + (2 - 3 s) \hat{k}$

$\vec{r} = (1 - t) \hat{ı} + (2 t + 3) \hat{ȷ} + \hat{k}$

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 7 - Straight Lines>EXERCISE>Q 43

Find the length of the perpendicular from point $(3, 4, 5)$ on the line $\frac{x - 2}{2} = \frac{y - 3}{5} = \frac{z - 1}{3}$ .

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 7 - Straight Lines>EXERCISE>Q 43

Find the length and the foot of the perpendicular drawn from the point $(2, - 1, 5)$ to the line $\frac{x - 11}{10} = \frac{y + 2}{- 4} = \frac{z + 8}{- 11}$ .

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 7 - Straight Lines>EXERCISE>Q 44

Find the shortest distance between the following pair of lines- $\vec{r} = \hat{i} + \hat{ȷ} + λ (2 \hat{i} - \hat{ȷ} + \hat{k})$ and $\vec{r} = 2 \hat{i} + \hat{j} - \hat{k} + μ (3 \hat{i} - 5 \hat{j} + 2 \hat{k})$ .

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 7 - Straight Lines>EXERCISE>Q 44

Find the shortest distance between the following pair of lines $\vec{r} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - 2 t) \hat{k}$ and $\vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} - (2 s + 1) \hat{k}$ .

HARD

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 7 - Straight Lines>EXERCISE>Q 44

Find the shortest distance between the following pair of lines $\vec{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ (\hat{i} - 3 \hat{j} + 2 \hat{k})$ and $\vec{r} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + μ (2 \hat{i} + 3 \hat{j} + \hat{k})$ .

HARD

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 7 - Straight Lines>EXERCISE>Q 45

Find the shortest distance between the following pair of lines:

$\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}$

$\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}$ .

HARD

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 7 - Straight Lines>EXERCISE>Q 46

Find the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ .

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