The equation of the plane passing through the intersection of planes 2 x − y + z = 3 , 4 x − 3 y + 5 z = − 9 and parallel to the line 2 x + 1 = 4 y + 3 = 5 z − 3 is...

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The equation of the plane passing through the intersection of planes 2 x − y + z = 3 , 4 x − 3 y + 5 z = − 9 and parallel to the line 2 x + 1 ​ = 4 y + 3 ​ = 5 z − 3 ​ is...

TestHub · Accepted Answer

The equation of the plane passing through the intersection of the given planes is

$(2 x - y + z - 3) + λ (4 x - 3 y + 5 z + 9) = 0$

$(2 + 4 λ) x - (1 + 3 λ) y + (1 + 5 λ) z - (3 - 9 λ) = 0$

Since this plane is parallel to the line $\frac{x + 1}{2} = \frac{y + 3}{4} = \frac{z - 3}{5}$ , its normal is perpendicular to the line. Therefore,

$2 (2 + 4 λ) - 4 (1 + 3 λ) + 5 (1 + 5 λ) = 0$

$4 + 8 λ - 4 - 12 λ + 5 + 25 λ = 0$

$21 λ + 5 = 0 \Rightarrow λ = - \frac{5}{21}$

Substituting $λ = - \frac{5}{21}$ in the equation of the plane, we get

$(2 x - y + z - 3) - \frac{5}{21} (4 x - 3 y + 5 z + 9) = 0$

$21 (2 x - y + z - 3) - 5 (4 x - 3 y + 5 z + 9) = 0$

$42 x - 21 y + 21 z - 63 - 20 x + 15 y - 25 z - 45 = 0$

$22 x - 6 y - 4 z - 108 = 0$

$11 x - 3 y - 2 z - 54 = 0$

Answer: $11 x - 3 y - 2 z - 54 = 0$

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