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Solution - A line passes through (2, −1, 3) and is perpendicular to the lines r=(i+j-k)+lambda(2i-2j+k) and vecr=(2i-j-3k)+mu(i+2j+2k) . Obtain its equation in vector and Cartesian from. - Equation of a Line in Space

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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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Solution for question: A line passes through (2, −1, 3) and is perpendicular to the lines r=(i+j-k)+lambda(2i-2j+k) and vecr=(2i-j-3k)+mu(i+2j+2k) . Obtain its equation in vector and Cartesian from. concept: Equation of a Line in Space. For the courses CBSE (Arts), CBSE (Commerce), CBSE (Science)
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