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Show that the lines x-33=y-3-1,z-1 = 0 and x-62=z-13,y-2 = 0 intersect. Aslo find the point of intersection - Mathematics

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प्रश्न

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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उत्तर

`(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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Application of Vectors to 3-dimensional Geometry
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पाठ 6: Applications of Vector Algebra - Exercise 6.5 [पृष्ठ २५५]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 6 Applications of Vector Algebra
Exercise 6.5 | Q 4 | पृष्ठ २५५

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