The Cartesian equation of a line is `(x-3)/2=(y+1)/(-2)=(z-3)/5` . Find the vector equation of the line.
The Cartesian equation of a line is `(x-3)/2=(y+1)/(-2)=(z-3)/5` . Find the vector equation of the line.
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The given line is `(x-3)/(2)=(y+1)/(-2)=(z-3)/(5)`
Note that it passes through (3, -13) and is parallel to the line whose direction ratios are 2, -2 and 5. Therefore, its vector equation is `vecr=3hati-hatj+3hatk+lamda(2hati-2hatj+5hatk)`, where `lamda` is a parameter.
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