6. Find the cartesian equation of the line which passes through the point (– 2, 4, – 5) and parallel to the line given by \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}.

Answers (1)

Given a line which passes through the point (– 2, 4, – 5) and is parallel to the line given by the \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6};

The direction ratios of the line, \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6} are 3,5 and 6.

So, the required line is parallel to the above line.

Therefore we can take direction ratios of the required line as 3k, 5k, and 6k, where k is a non-zero constant.

And we know that the equation of line passing through the point (x_{1},y_{1},z_{1}) and with direction ratios a, b, c is written by: \frac{x-x_{1}}{a} = \frac{y-y_{1}}{b} = \frac{z-z_{1}}{c}.

Therefore we have the equation of the required line:

\frac{x+2}{3k} = \frac{y-4}{5k} = \frac{z+5}{6k}   

or   \frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6} = k

The required line equation. 

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