A line with DCs proportional to 2, 1, 2 meets each of the lines x = y

1.	A line with DCs proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The coordinates of each of the points of intersection are(a) (3a, 2a, 3a), (a, a, 2a)(b) (3a, 2a, 3a), (a, a, a)(c) (3a, 3a, 3a), (a, a, a)(d) (2a, 3a, 3a), (2a, a, a)
Answer» Correct option is (b) (3a, 2a, 3a), (a, a, a) Explanation : The given line L₁ is x/1 = (y + a)/1 = z/1 P is a point on L₁. Therefore P = (t, -a + t, t) ...(1) and the given line L₂ is (x + a)/2 = y/1 = z/1 Q is a point on L₂. Therefore Q = (-a + 2s,s,s) ...(2) From Eqs. (1) and (2), the DRs of vector PQ are (2s - a - t,s - t + a,s - t) Therefore, by hypothesis, we have (2s - a - t)/2 = (s - t + a)/1 = (s - t)2 Hence (2s - a - t)/2 = (s - t + a)/1 t = 3a (s - t + a)/1 = (s - t)/2 s - t = - 2a s = a (∵ t = 3a) Therefore, P = (3a, 2a, 3a) and Q = (a, a, a)

A line with DCs proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The coordinates of each of the points of intersection are(a) (3a, 2a, 3a), (a, a, 2a)(b) (3a, 2a, 3a), (a, a, a)(c) (3a, 3a, 3a), (a, a, a)(d) (2a, 3a, 3a), (2a, a, a)

Answer»

Correct option is (b) (3a, 2a, 3a), (a, a, a)

Explanation :

The given line L₁ is

x/1 = (y + a)/1 = z/1

P is a point on L₁. Therefore

P = (t, -a + t, t) ...(1)

and the given line L₂ is

(x + a)/2 = y/1 = z/1

Q is a point on L₂. Therefore

Q = (-a + 2s,s,s) ...(2)

From Eqs. (1) and (2), the DRs of vector PQ are

(2s - a - t,s - t + a,s - t)

Therefore, by hypothesis, we have

(2s - a - t)/2 = (s - t + a)/1 = (s - t)2

Hence

(2s - a - t)/2 = (s - t + a)/1

t = 3a

(s - t + a)/1 = (s - t)/2

s - t = - 2a

s = a (∵ t = 3a)

Therefore, P = (3a, 2a, 3a) and Q = (a, a, a)

Discussion

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