Solution - Equation of a Line in Space



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Find the Cartesian equation of the line which passes through the point (−2, 4, −5) and is parallel to the line ``


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Find the vector and cartesian equations of the line passing through the point (2, 1, 3) and perpendicular to the lines

`(x-1)/1=(y-2)/2=(z-3)/3 and x/(-3)=y/2=z/5`

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A line passes through (2, −1, 3) and is perpendicular to the lines `vecr=(hati+hatj-hatk)+lambda(2hati-2hatj+hatk) and vecr=(2hati-hatj-3hatk)+mu(hati+2hatj+2hatk)` . Obtain its equation in vector and Cartesian from. 

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The Cartesian equations of line are 3x -1 = 6y + 2 = 1 - z. Find the vector equation of line.

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Find the vector and Cartesian equations of the line through the point (1, 2, −4) and perpendicular to the two lines. 

`vecr=(8hati-19hatj+10hatk)+lambda(3hati-16hatj+7hatk) " and "vecr=(15hati+29hatj+5hatk)+mu(3hati+8hatj-5hatk)`



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The joint equation of the pair of lines passing through (2,3) and parallel to the coordinate axes is

  1.  xy -3x - 2y + 6 = 0
  2. xy +3x + 2y + 6 = 0
  3. xy = 0
  4. xy - 3x - 2y - 6 = 0
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Reference Material

Solution for concept: Equation of a Line in Space. For the courses 12th CBSE (Arts), 12th CBSE (Commerce), 12th CBSE (Science), PUC Karnataka Science