Trigonometric EquationHard
Question
A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of the point of intersection are given by
Options
A.(3a, 3a, 3a), (a, a, a)
B.(3a, 2a, 3a), (a, a, a)
C.(3a, 2a, 3a), (a, a, 2a)
D.(2a, 3a, 3a), (2a, a, a)
Solution
Any point on the line 
= t1 (say) is (t1, t1 - a, t1) and any point on the line
= t2 (say) is (2t2 - a, t2, t2).
Now direction cosine of the lines intersecting the above lines is proportional to
(2t2 - a - t1, t2 - t1 + a, t2 - t1).
Hence 2t1 - a - t1 = 2k, t2 - t1 + a = k and t2 - t1 = 2k
On solving these, we get t1 = 3a , t2 = a.
Hence points are (3a, 2a, 3a) and (a, a, a).
= t1 (say) is (t1, t1 - a, t1) and any point on the line = t2 (say) is (2t2 - a, t2, t2).Now direction cosine of the lines intersecting the above lines is proportional to
(2t2 - a - t1, t2 - t1 + a, t2 - t1).
Hence 2t1 - a - t1 = 2k, t2 - t1 + a = k and t2 - t1 = 2k
On solving these, we get t1 = 3a , t2 = a.
Hence points are (3a, 2a, 3a) and (a, a, a).
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