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+1 vote
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in Three Dimensional Geometry by (28.9k points)
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Consider a straight line through a fixed point with position vector 2i – 2j + 3k and parallel to i – j + 4k.

  1. Write down the vector equation of the straight line. 
  2. Show that the straight line is parallel to the plane \(\bar r\).(i + 5y + k) = 5
  3. Find the distance between the line and plane.

1 Answer

+1 vote
by (28.2k points)
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Best answer

1. Vector equation of a straight line is \(\bar r\) = \(\bar a\) + λ\(\bar b\) where a is \(\bar a\) fixed point and \(\bar b\) is a vector parallel to the line. Here \(\bar a\) = 2i – 2y + 3 it and \(\bar b\) = i – j + 4k. Therefore vector equation of the line \(\bar r\) = 2i – 2j + 3k + λ(i – j + 4k).

2. The vector parallel to the line is i – j + 4k and vector normal to the plane is i + 5j + k.
Then, (i – j + 4k). (i + 5j + k) = 1 – 5 + 4 = 0
implies that straight line and plane are parallel.

3. A point on the line is 2i – 2j + 3k. Then the distance of 2i – 2j + 3k to the given plane

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