Find the cartesian equation of the line which | vedclass

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(A) The given line is $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.The direction ratios of this line are $a = 3, b = 5, c = 6$.Since the required l

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

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(A) The given line is $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.The direction ratios of this line are $a = 3, b = 5, c = 6$.Since the required line is parallel to the given line,its direction ratios will also be proportional to $3, 5, 6$.The line passes through the point $(x_1, y_1, z_1) = (-2, 4, -5)$.The cartesian equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $a, b, c$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Substituting the values,we get $\frac{x - (-2)}{3} = \frac{y - 4}{5} = \frac{z - (-5)}{6}$.Thus,the required equation is $\frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6}$.

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

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