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Equation of a Plane Passing through a Point and Parallel to the Two Given Vectors

The equation ...

Question

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

14 x + 2 y - 15 z = 31

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14 x - 2 y + 15 z = 31

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14 x + 2 y + 15 z = 31

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14 x - 2 y - 15 z = 31

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Solution

The correct option is C $14 x + 2 y + 15 z = 31$
Given:
$P_{1} : 2 x + y - 2 z = 5$ and $P_{2} : 3 x - 6 y - 2 z = 7$
$D . r^{'} s$ of normal to $P_{1} : (a_{1}, b_{1}, c_{1}) = (2, 1, - 2)$
$D . r^{'} s$ of normal to $P_{2} : (a_{2}, b_{2}, c_{2}) = (3, - 6, - 2)$
The required equation of plane is given by:
$∣ ∣ ∣ ∣ \begin{matrix} x - x_{1} & y - y_{1} & z - z_{1} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} \end{matrix} ∣ ∣ ∣ ∣$
$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} x - 1 & y - 1 & z - 1 2 & 1 & - 2 3 & - 6 & - 2 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow (x - 1) (- 14) - (y - 1) (2) + (z - 1) (- 15) = 0 \Rightarrow 14 x - 14 + 2 y - 2 + 15 z - 15 = 0 \Rightarrow 14 x + 2 y + 15 z = 31$

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Similar questions

(1, 1, 1)

2 x + y - 2 z = 5

3 x - 6 y - 2 z = 7

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

(1, 1, 1)

2 x + y - 2 z = 5 and 3 x - 6 y - 2 z = 7

(- 1, 3, 2)

x + 2 y + 2 z = 5

3 x + 3 y + 2 z = 8

2 x + y - 2 z = 5

3 x - 6 y - 2 z = 7

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