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Equation of Perpendicular from a Point on a Line

If from the p...

Question

If from the point P(a, b, c) perpendicular PL, PM be drawn to YOZ and ZOX planes, then the equation of the plane OLM is

A

\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 0

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B

\frac{x}{a} - \frac{y}{b} + \frac{z}{c} = 0

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C

\frac{x}{a} + \frac{y}{b} - \frac{z}{c} = 0

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D

\frac{x}{a} - \frac{y}{b} - \frac{z}{c} = 0

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Solution

The correct option is C $\frac{x}{a} + \frac{y}{b} - \frac{z}{c} = 0$
Coordinates of L and M are (0, b, c) and (a, 0, c) respectively.
$∴$ Equation of plane through (0, 0, 0), (0, b, c) and (a, 0, c) is
$∣ ∣ ∣ ∣ \begin{matrix} x - 0 & y - 0 & z - 0 0 & b & c a & 0 & c \end{matrix} ∣ ∣ ∣ ∣ = 0 i e, \frac{x}{a} + \frac{y}{b} - \frac{z}{c} = 0$

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Similar questions

Q. If from the point P(a, b, c) perpendiculars PL, PM be drawn to YOZ and ZOX planes, then the equation of the plane OLM is

\frac{x - a + d}{α - δ} = \frac{y - a}{α} = \frac{z - a - d}{α + δ}

\frac{x - b + c}{β - γ} = \frac{y - b}{β} = \frac{z - b - c}{β + γ}

are coplanar and then equation to the plane in which they lie, is

Q. From the point

P (a, b, c)

, let perpendiculars

P L

P M

Y O Z

Z O X

planes, respectively. Then the equation of the plane

O L M

Q. If the equation of the plane containing the lines

x - y - z - 4 = 0, x + y + 2 z - 4 = 0

and parallel to the line of intersection of the planes

2 x + 3 y + z = 1

x + 3 y + 2 z = 2

x + A y + B z + C = 0

, then the value of

| A + B + C |

(i) If a(y+z) = b(z+x) = c(x+y) and out of a, b, c no two of them are equal then show that,

\frac{y - z}{a (b - c)} = \frac{z - x}{b (c - a)} = \frac{x - y}{c (a - b)}

\frac{x}{3 x - y - z} = \frac{y}{3 y - z - x} = \frac{z}{3 z - x - y}

x + y + z \neq 0

then show that the value of each ratio is equal to 1.

(iii) If

\frac{a x + b y}{x + y} = \frac{b x + a z}{x + z} = \frac{a y + b z}{y + z}

x + y + z \neq 0

\frac{a + b}{2}

\frac{y + z}{a} = \frac{z + x}{b} = \frac{x + y}{c}

\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c}

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

then show that every ratio =

\frac{x}{y}

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