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

Given, x - 1 2 = y + 1 3 = z - 1 4 . . . . 1 x - 2 1 = y + m 2 = z - 2 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 - 2 1 = y + m 2 = z - 2 1 = μ So, x = μ + 2 , y = 2 μ - m & z = μ + 2 Now if the line intersect each other then, 2 λ + 1 = μ + 2 , 3 λ - 1 = 2 μ - m , 4 λ + 1 = μ + 2 ⇒ 2 λ - μ = 1 , 3 λ - 2 μ = 1 - m , 4 λ - μ = 1 Now solving above equations we get, λ = 0 & μ = - 1 and putting the value in 3 λ - 2 μ = 1 - m we get, 3 × 0 - 2 - 1 = 1 - m ⇒ m = - 1

Find the direction cosines of the line r → = - 2 i ^ + 5 2 j ^ - k ^ + λ 2 i ^ + 3 j ^ .

Given, Equation of line r → = - 2 i ^ + 5 2 j ^ - k ^ + λ 2 i ^ + 3 j ^ Now on comparing with r → = a → + λ b → we get, b → = 2 i ^ + 3 j ^ , So direction ratio of line are 2 , 3 , 0 We know that if direction ratio are α , β , γ then direction cosines are given by α α 2 + β 2 + γ 2 , β α 2 + β 2 + γ 2 , γ α 2 + β 2 + γ 2 So direction cosines will be 2 2 2 + 3 2 + 0 2 , 3 2 2 + 3 2 + 0 2 , 0 2 2 + 3 2 + 0 2 = 2 13 , 3 13 , 0

Find the co-ordinates of points on the line x - 1 1 = y - 2 - 2 = z - 3 2 which are at the distance 3 unit from the base point A 1 , 2 , 3 .

Given, Equation of line x - 1 1 = y - 2 - 2 = z - 3 2 = t Now any point on the line will be denoted by x , y , z ≡ t + 1 , - 2 t + 2 , 2 t + 3 Now given that the point is at the distance of 3 unit from the base point A 1 , 2 , 3 , So by distance formula we get, t + 1 - 1 2 + - 2 t + 2 - 2 2 + 2 t + 3 - 3 2 = 3 ⇒ 9 t 2 = 3 ⇒ 3 t = 3 ⇒ t = ± 1 So the required point at t = 1 is 2 , 0 , 5 and at t = - 1 is 0 , 4 , 1

Find the Cartesian equation of a line passing through the points 3 , - 2 , - 5 and 3 , - 2 , 6 .

Let points A 3 , - 2 , - 5 and B 3 , - 2 , 6 . Equation of line passing through two points x 1 , y 1 , z 1 and x 2 , y 2 , z 2 is x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 = z - z 1 z 2 - z 1 . Equation of line passing through points A and B is x - 3 3 - 3 = y - - 2 - 2 - - 2 = z - - 5 6 - - 5 ⇒ x - 3 0 = y + 2 0 = z + 5 11

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.

E M B I B E

Mathematics & Statistics (Arts & Science) Part 1 Standard 12>Chapter 6 - Line and Plane>Miscellaneous Exercise 6A>Q 15

MEDIUM

12th Maharashtra Board

IMPORTANT

Earn 100

Find the co-ordinates of the foot of the perpendicular drawn from the point $(0, 2, 3)$ to the line $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ .

Important Questions on Line and Plane

MEDIUM

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 1 Standard 12>Chapter 6 - Line and Plane>Miscellaneous Exercise 6A>Q 16

By computing the shortest distance determine whether following lines intersect each other.

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