Question

The equation of the line passing through a point (x₀, y₀, z₀) and having DRs (l, m, n) is (x - x₀)/l = (y - y₀)/m = (z - z₀)/n. Also the DRs of the line represented by the planes a₁x + b₁y + c₁z + d₁ = 0 = a₂x + b₂y + c₂z + d₂  is the cross product of the normals (a₁, b₁, c₁) and (a₂, b₂, c₂) of the given planes. Answer the following question

(i) The equation of the line drawn from (1, −1, 0) to intersect the lines (x - 2)/2 = (y - 1)/3 = (z - 3)/4 and (x - 4)/4 = y/5 = (z + 1)/2 orthogonally is

(a) (x - 1)/14 = (y + 1)/-12 = (z - 1)/2

(b) (x - 1)/14 = (y + 1)/-12 = z/2

(c) (x + 1)/14 = (y - 1)/-12 = z/2

(d) (x + 1)/14 = (y - 1)-12 = (z - 1)/2

(ii) The symmetric form of the equation of the line 4x − 4y − z + 11= 0 = x + 2y − z − 1 is

(a) (x - 2)/1 = y/1 = (z - 3)/4

(b) (x - 2)/2 = (y - 2)/1 = z/4

(c) x/2 = (y - 2)/1 = (z - 3)/4

(d) ((x + 3)/2)/2 = ((y - 5)/4)/1 = z/4

Answer 1

(i) → (b), (ii) → (d)

Explanation :

(i) The DRs of the required line are given by

|(vector i, vector j , vector k),(2, 3, 4),(4, 5, 2)| = - 14 vector i + 12 vector j - 2 vector k

Therefore, the DRs of the line are (14, −12, 2). The line passes through (1, −1, 0). Hence, its equation is

(x - 1)/14 = (y + 1)/-12 = z/2

(ii) The DRs of the line of intersection are

|(vector i,vector j,vector k),(4,-4,-1),(1,2,-1)| = 6 vector i + 3 vector j + 12 vector k

= 3(2 vector i + vector j + 4 vector k)

The DRs are (2, 1, 4). In the given equations, substituting z = 0, we get

4x - 4y = -11

and x + 2y = 1

Solving the equations, we get

x = -3/2, y = 5/4

Hence, (-3/2,5/4,0) is a point on the line. Therefore, its equation is

(x +(3/2)/2 = (y - (5/4)/1 = z/4