Perpendiculars are drawn from points on the line 2x+2=−1y+1=3z to the plane x+y+z=3 The feet of perpendiculars lie on the line (a) 5x=8y−1=−13z−2 (b) 2x=3y−1=−5z−2 (c) 4x=3y−1=−7z−2 (d) 2x=−7y−1=5z−2

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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 (a) $\frac{x}{5} = \frac{y - 1}{8} = \frac{z - 2}{- 13}$ (b) $\frac{x}{2} = \frac{y - 1}{3} = \frac{z - 2}{- 5}$ (c) $\frac{x}{4} = \frac{y - 1}{3} = \frac{z - 2}{- 7}$ (d) $\frac{x}{2} = \frac{y - 1}{- 7} = \frac{z - 2}{5}$

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Owen Discussed

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 (a)

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

(b)

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

(c)

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

(d)

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

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Question Text	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 (a) $\frac{x}{5} = \frac{y - 1}{8} = \frac{z - 2}{- 13}$ (b) $\frac{x}{2} = \frac{y - 1}{3} = \frac{z - 2}{- 5}$ (c) $\frac{x}{4} = \frac{y - 1}{3} = \frac{z - 2}{- 7}$ (d) $\frac{x}{2} = \frac{y - 1}{- 7} = \frac{z - 2}{5}$
Answer Type	Video solution: 1
Upvotes	151