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Family of Planes Passing through the Intersection of Two Planes

Parametric fo...

Question

Parametric form of the equation of the line $3 x - 6 y - 2 z - 15 = 0, 2 x + y - 2 z - 5 = 0$ is

\frac{x - 5}{14} = \frac{y}{2} = \frac{z}{15}

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\frac{x - 1}{14} = \frac{y - 5}{2} = \frac{z - 1}{15}

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\frac{x - 3}{14} = \frac{y + 1}{2} = \frac{z}{15}

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\frac{x + 5}{14} = \frac{y}{2} = \frac{z}{15}

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Solution

The correct option is A $\frac{x - 5}{14} = \frac{y}{2} = \frac{z}{15}$
$3 x - 6 y - 2 z - 15 = 0$

$\Rightarrow \frac{3 x - 6 y - 15}{2} = z$ ..........(1)

$2 x + y - 2 z - 5 = 0$

$\Rightarrow \frac{2 x + y - 5}{2} = 2$ ........(ii)

from (i) and (ii)

$\frac{3 x - 6 y - 15}{2} = \frac{2 x + y - 5}{2}$

$3 x - 2 x - 15 = 6 y + y - 5$

$\Rightarrow x - 15 = 7 y - 5 \Rightarrow x - 10 = 7 y$

$\Rightarrow y = \frac{x - 9}{7}$

$\frac{y}{2} = \frac{x - 5}{14}$

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Similar questions

The equation of the plane passing through the point (1,1,1) and perpendicular to the planes 2x+y-2z=5 and 3x-6y-2z=7 is

3 x - 6 y - 2 z = 15

2 x + y - 2 z = 5

1

x = 3 + 14 t

y = 1 + 2 t,

z = 15 t

2

14^i + 2^j + 15^k

3 x - y + 2 z - 1 = 0 = x + 2 y - z - 2

3 x + 2 y + z = 0

3 x - 6 y - 2 z = 15

2 x + y - 2 z = 5

x = 3 + 14 t, y = 1 + 2 t, z = 15 t

14 i + 2 j + 15 k

3 x - 6 y - 2 z = 15

2 x + y - 2 z = 5.

x = 3 + 14 t, y = 1 + 2 t, z = 15 t .

14^i + 2^j + 15^k

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