Find the angle between the following pair of lines: r → = 5 i ^ - 7 j ^ + λ - i ^ + 4 j ^ + 2 k ^ r → = 2 ı ^ + k ^ + μ 3 ı ^ + 4 k ^

Angle between two lines r → = a 1 → + λ b 1 → r → = a 2 → + μ b 2 → is given by cos θ = b 1 → · b 2 → b 1 → b 2 → b 1 → · b 2 → = - i ^ + 4 j ^ + 2 k ^ · 3 i ^ + 4 k ^ = - 1 × 3 + 4 × 0 + 2 × 4 = - 3 + 0 + 8 = 5 ∵ b → 1 = - 1 2 + 4 2 + 2 2 ⇒ b 1 → = 1 + 16 + 4 = 21 ∵ b 2 → = 3 2 + 0 2 + 4 2 ⇒ b 2 → = 9 + 16 = 25 = 5 cos θ = b 1 → · b 2 → b 1 → b 2 → = 5 21 × 5 = 1 21 θ = cos - 1 1 21

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 angle between the following pair of lines: x - 1 2 = y - 2 3 = z - 4 6 ; x + 1 = y + 2 2 = z - 4 2

As equations are given in the form 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 The angle between two lines θ is given by cos θ = a 1 a 2 + b 1 b 2 + c 1 c 2 a 1 2 + b 1 2 + c 1 2 a 2 2 + b 2 2 + c 2 2 Comparing with the equations given x - 1 2 = y - 2 3 = z - 4 6 , x - - 1 1 = y - - 2 2 = z - - 4 2 Now, cos θ = a 1 a 2 + b 1 b 2 + c 1 c 2 a 1 2 + b 1 2 + c 2 1 a 2 2 + b 2 2 + c 2 2 = 2 × 1 + 3 × 2 + ( 6 × 2 ) ( 2 ) 2 + ( 3 ) 2 + ( 6 ) 2 1 2 + 2 2 + 2 2 = 2 + 6 + 12 4 + 9 + 36 9 = 20 49 9 = 20 21 ⇒ θ = cos - 1 20 21 Therefore, angle between given lines will be cos - 1 20 21 or π - cos - 1 20 21 .

Find the angle between the following pair of lines x + 1 3 = y - 1 5 = z + 3 4 ; x + 1 1 = y - 4 4 = z - 5 2 .

Given: x + 1 3 = y - 1 5 = z + 3 4 x + 1 1 = y - 4 4 = z - 5 2 Here the vectors parallel to the given lines are v 1 → = 3 i ^ + 5 j ^ + 4 k ^ v 2 → = i ^ + 4 j ^ + 2 k ^ Let θ is the angle between the given lines cos θ = v 1 → · v 2 → v 1 → v 2 → = 3 i + 5 j + 4 k · i + 4 j + 2 k 3 2 + 5 2 + 4 2 1 2 + 4 2 + 2 2 = 3 × 1 + 5 × 4 + 4 × 2 9 + 25 + 16 1 + 16 + 4 = 3 + 20 + 8 50 21 = 31 1050 θ = cos - 1 31 1050 Therefore, angle between given lines will be cos - 1 31 1050 or π - cos - 1 31 1050 .

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

Given: x + 4 3 = y - 1 5 = z + 3 4 x + 1 1 = y - 4 1 = z + 3 2 So, direction cosines of first line 3 , 5 , 4 and direction cosines of second line 1 , 1 , 2 . So, the vectors parallel to the given lines are 3 i ^ + 5 j ^ + 4 k ^ and i ^ + j ^ + 2 k ^ Let θ is the angle between the given lines cos θ = v 1 → · v 2 → v 1 → v 2 → = 3 i ^ + 5 j ^ + 4 k ^ i ^ + j ^ + 2 k ^ 3 2 + 5 2 + 4 2 1 2 + 1 2 + 2 2 = 3 × 1 + 5 × 1 + 4 × 2 9 + 25 + 16 1 + 1 + 4 = 3 + 5 + 8 50 6 = 16 300 = 16 10 3 = 8 5 3 = 8 3 15 θ = cos - 1 8 3 15 Therefore, angle between given lines will be cos - 1 8 3 15 or π - cos - 1 8 3 15 .

Find the angle between the lines x - 3 1 = y - 2 2 = z - 2 - 4 and x - 0 3 = y - 5 2 = z + 2 - 6 .

Given: x - 3 1 = y - 2 2 = z - 2 - 4 x - 0 3 = y - 5 2 = z + 2 - 6 Comparing it with 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 ⇒ x 1 = 3 , y 1 = 2 , z 1 = 2 , a 1 = 1 , b 1 = 2 , c 1 = - 4 ⇒ x 2 = 0 , y 2 = 5 , z 2 = - 2 , a 2 = 3 , b 2 = 2 , c 2 = - 6 By formula: cos θ = a 1 a 2 + b 1 b 2 + c 1 c 2 a 1 2 + b 1 2 + c 1 2 a 2 2 + b 2 2 + c 2 2 ⇒ 1 × 3 + 2 × 2 + - 4 × - 6 1 2 + 2 2 + - 4 2 3 2 + 2 2 + - 6 2 = 3 + 4 + 24 1 + 4 + 16 9 + 4 + 36 ⇒ 31 21 49 = 31 7 21 ⇒ θ = cos - 1 31 7 21

Find the angle between the lines r → = 4 i ^ - ȷ ^ + λ i ^ + 2 j ^ - 2 k ^ and r → = i ^ - ȷ ^ + 2 k ^ - μ 2 ı ^ + 4 ȷ ^ - 4 k ^ .

We know that if θ is the angle between two lines r → = a 1 → + λ b 1 → and r → = a 2 → + μ b 2 → , then cos θ = b 1 → · b 2 → b 1 → b 2 → r → = 4 i ^ - j ^ + λ i ^ + 2 j ^ - 2 k ^ Here: b 1 → = i ^ + 2 j ^ - 2 k ^ b 2 → = 2 i ^ + 4 j ^ - 4 k ^ By formula: cos θ = b 1 → · b 2 → b 1 → b 2 → = i ^ + 2 j ^ - 2 k ^ 2 i ^ + 4 j ^ - 4 k ^ 1 2 + 2 2 + - 2 2 2 2 + 4 2 + - 4 2 = 1 × 2 + 2 × 4 + - 2 × - 4 1 + 4 + 4 4 + 16 + 16 = 2 + 8 + 8 9 36 = 18 3 × 6 = 1 θ = cos - 1 1 ∴ θ = 0 °

Show that the line x - 3 2 = y + 1 - 3 = z - 2 4 is perpendicular to the line x + 2 2 = y - 4 4 = z + 5 2 .

Equation given: x - 3 2 = y + 1 - 3 = z - 2 4 x + 2 2 = y - 4 4 = z + 5 2 Comparing with the general form: 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 a 1 = 2 , a 2 = 2 , b 1 = - 3 , b 2 = 4 , c 1 = 4 , c 2 = 2 According to the condition: ⇒ ( a 1 ) · ( a 2 ) + ( b 1 ) · ( b 2 ) + ( c 1 ) · ( c 2 ) = 2 × 2 + - 3 × 4 + 4 × 2 ⇒ ( a 1 ) · ( a 2 ) + ( b 1 ) · ( b 2 ) + ( c 1 ) · ( c 2 ) = 4 - 12 + 8 = 0 The lines are perpendicular.

E M B I B E

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 7 - Straight Lines>EXERCISE>Q 20

MEDIUM

JEE Advanced

IMPORTANT

Earn 100

The acute angle between the lines $\vec{r} = (4 \hat{i} - \hat{j}) + 5 (2 \hat{i} + \hat{j} - 3 \hat{k})$ and $\vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) + t (\hat{i} - 3 \hat{j} + 2 \hat{k})$ is

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Important Questions on Straight Lines

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:

$\vec{r} = 5 \hat{i} - 7 \hat{j} + λ (- \hat{i} + 4 \hat{j} + 2 \hat{k})$

$\vec{r} = 2 \hat{ı} + \hat{k} + μ (3 \hat{ı} + 4 \hat{k})$

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 2

Find the angle between the following pair of lines:

$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 4}{6}; x + 1 = \frac{y + 2}{2} = \frac{z - 4}{2}$

EASY

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 7 - Straight Lines>EXERCISE>Q 2

Find the angle between the following pair of lines

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

MEDIUM

JEE Advanced

IMPORTANT

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

Find the angle between the lines

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

MEDIUM

JEE Advanced

IMPORTANT

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

Find the angle between the lines

$\frac{x - 3}{1} = \frac{y - 2}{2} = \frac{z - 2}{- 4}$ and $\frac{x - 0}{3} = \frac{y - 5}{2} = \frac{z + 2}{- 6}$ .

HARD

JEE Advanced

IMPORTANT

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

Find the angle between the lines $\vec{r} = 4 \hat{i} - \hat{ȷ} + λ (\hat{i} + 2 \hat{j} - 2 \hat{k})$ and $\vec{r} = \hat{i} - \hat{ȷ} + 2 \hat{k} - μ (2 \hat{ı} + 4 \hat{ȷ} - 4 \hat{k})$ .