Perpendiculars are drawn from points on the line \frac{x + 2}{2} = \frac{y + 1}{- 1} = \frac{z}{3} to the plane x + y + z = 3. The feet of perpendiculars lie on the line

Engineering

Mathematics

Equation of Straight Line in 3D

Question

Perpendiculars are drawn from points on the line $\frac{x + 2}{2} = \frac{y + 1}{- 1} = \frac{z}{3}$ to the plane x + y + z = 3. The feet of perpendiculars lie on the line

$\frac{x}{4} = \frac{y - 1}{3} = \frac{z - 2}{- 7}$

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

$\frac{x}{5} = \frac{y - 1}{8} = \frac{z - 2}{- 13}$

$\frac{x}{2} = \frac{y - 1}{- 7} = \frac{z - 2}{5}$

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L : $\frac{x + 2}{2} = \frac{y + 1}{- 1} = \frac{z}{3}$ ….(1)

P : x + y + z = 6 ….(2)

Take 2 points A (– 2, – 1, 0) and B (0, – 2, 3) on line L.

$∴$ N₁ is (0, 1, 2)

[Foot of perpendicular of point (A) on plane P.]

and N₂ is $(\frac{2}{3}, \frac{- 4}{3}, \frac{11}{3})$

[Foot of perpendicular of point (B) on plane P.]

Hence, equation of line joining N₁ and N₂ is $\frac{x}{\frac{2}{3}} = \frac{y - 1}{\frac{- 7}{3}} = \frac{z - 2}{\frac{5}{3}} or \frac{x}{2} = \frac{y - 1}{- 7} = \frac{z - 2}{5}$