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Question
Show that the lines `(x - 3)/3 = (y - 3)/(-1), z - 1` = 0 and `(x - 6)/2 = (z - 1)/3, y - 2` = 0 intersect. Aslo find the point of intersection
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Solution
`(x - 3)/3 = (y - 3)/(-1), z - 1` = 0 ⇒ z = 1
`(x - 6)/2 = (z - 1)/3, y - 2` = 0 ⇒ y = 2
(x1, y1, z1) = (3, 3, 1) and (x2, y2, z2) = (6, 2, 1)
(b1, b2, b3) = (3, –1, 0) and (d1, d2, d3) = (2, 0, 3)
Condition for intersection of two lines
`|(x_2 - x_1, y_2 - y_1, z_2 - z_1),("b"_1, "b"_2, "b"_3),("d"_1, "d"_2, "d"_3)|` = 0
`|(3, -1, 0),(3, 1, 0),(2, 1, 3)|` = 0 Since (R1 = R2)
∴ Given two lines are intersecting lines.
Any point on the first time
`(x - 3)/3 = (y - 3)/(-1) = lambda` and z = 1
`(3lambda + 3, -lambda + 3, 1)`
Any point on the Second line
`(x - 6)/2 = (z - 1)/3 = mu` and y = 2
`(2mu + 6, 2, 3mu + 1)`
∴ `3mu + 1` = 1
`3mu` = 0
`mu` = 0
`-lambda + 3` = 2
`-lambda` = – 1
`lamda` = 1
∴ The required point of intersection is (6, 2, 1)
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