The Cartesian equation of the line which is p | vedclass

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Question

(B) The equation of a line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b}

Vedclass · Accepted Answer

$\frac{x-1}{3}=\frac{y+3}{5}=\frac{z-5}{6}$

Vedclass · Answer

(B) The equation of a line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Since the required line is parallel to the line $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$,it will have the same direction ratios,which are $(3, 5, 6)$.Given that the line passes through the point $(1, -3, 5)$,we substitute these values into the formula:$\frac{x-1}{3} = \frac{y-(-3)}{5} = \frac{z-5}{6}$This simplifies to $\frac{x-1}{3} = \frac{y+3}{5} = \frac{z-5}{6}$.Thus,the correct option is $B$.

The Cartesian equation of the line which is parallel to the line $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ and passing through the point $(1, -3, 5)$ is:

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