Find the eqaaution of the line which intersect the lines

Question 1:- Find the eqaaution of the line which intersect the lines $\dfrac{x + 2}{1} = \dfrac{y - 3}{2} = \dfrac{z + 1}{4}$ and $\dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z - 3}{4}$ passes through the point (1, 1, 1)

Solution:- Let dr’s of the line (a, b, c) and Equation of line passes through the point P(1, 1, 1)

$\dfrac{x - 1}{a} = \dfrac{y - 1}{b} = \dfrac{z - 1}{c}$ . . . (i)

Let line $l_1: \dfrac{x + 2}{1} = \dfrac{y - 3}{2} = \dfrac{z + 1}{4}$ . . .(ii)

and lline $l_2: \dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z - 3}{4}$ . . .(iii)

Let point Q and R lies on the line l1 and l2

From line $l_1 : \dfrac{x + 2}{1} = \dfrac{y - 3}{2} = \dfrac{z + 1}{4} = m$

x = m – 2, y = 2m + 3, z = 4m – 1

point Q(m – 2, 2m + 3, 4m -1)

Dr’s of PQ (m -2 -1, 2m + 3 -1, 4m – 1 – 1)

⇒ (m -3, 2m + 2, 4m -2)

From line $l_2 : \dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z - 3}{4} = n$

x = 2n + 1, y = 3n + 2, z = 4n + 3

R (2n +1 , 3n + 2, 4n + 3)

Dr’s of PR (2n +1 -1, 3n + 2 -1, 4n + 3 – 1)

⇒ (2n, 3n + 1, 4n + 2)

Since PQ and PR are parallel

$\dfrac{m- 3}{2n} = \dfrac{2m + 2}{3n + 1} = \dfrac{4m -2}{4n + 2}$

$\dfrac{m- 3}{2n} = \dfrac{2m + 2}{3n + 1}$

⇒ (m-3)(3n + 1) = 2n(2m + 2)

⇒ 3mn – 9n + m -3 = 4mn + 4n

⇒ m -13n = mn + 3 . . . (iv)

Again $\dfrac{2m + 2}{3n + 1} = \dfrac{4m -2}{4n + 2}$

⇒ (2m+2)(4n + 2) = (4m -2)(3n + 1)

⇒ 8mn + 8n + 4m + 4 = 12mn +4m -6n -2

⇒ 8n + 6n + 6 = 4mn

⇒ 14n – 4 mn +6 =0

⇒ n(14 – 4m) = -6

⇒ n = -6/ (14-4m) . . .(v)

Solving eq (iv) and (v)

m -13[-6/(14-4m)] = m[-6/(14-4m)] + 3

⇒ $\dfrac{m(14-4m) + 78}{14- 4m} =\dfrac{-6m + 3(14-4m)}{14-4m}$

⇒ 14m -4m² + 78 = -6m + 42 -12m

⇒ -4m² + 32m + 36 = 0

Divide by -4

⇒ m² -8m – 9= 0

⇒ m² -9m + m -9 = 0

⇒ m(m – 9 ) + 1 (m – 9) = 0

⇒ (m – 9)(m +1) = 0

⇒ m = 9, -1

Taaking m= 9

Dr’s of the line

(m -3, 2m + 2, 4m -2)

(9 -3, 2(9) + 2, 4(9) -2)

⇒ (6, 20, 34)

Equation of line Passing through (1, 1, 1)

$\dfrac{x - 1}{6} = \dfrac{y - 1}{20} = \dfrac{z - 1}{34}$

⇒ $\dfrac{x - 1}{3} = \dfrac{y - 1}{10} = \dfrac{z - 1}{17}$

Find the eqaaution of the line which intersect the lines

Other Three Dimensional Geometry Question

Leave a Comment Cancel reply