The image of the line \frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{- 5} in the plane 2x

Engineering

Mathematics

Equation of Straight Line in 3D

Question

The image of the line $\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{- 5}$ in the plane 2x – y + z + 3 = 0 is the line

$\frac{x - 3}{- 3} = \frac{y + 5}{- 1} = \frac{z - 2}{5}$

$\frac{x + 3}{- 3} = \frac{y - 5}{- 1} = \frac{z + 2}{5}$

$\frac{x - 3}{3} = \frac{y + 5}{1} = \frac{z - 2}{- 5}$

$\frac{x + 3}{3} = \frac{y - 5}{1} = \frac{z - 2}{- 5}$

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Line is parallel to the given plane

$ℓ_{1}' \frac{x + 3}{3} = \frac{y - 5}{1} = \frac{z - 2}{- 5}$