The equation of the plane passing through the line of intersection of the planes x + y + z = 1 and 3 x + 4 y + 5 z = 2 and perpendicular to the XY- plane is

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

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

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

Since this plane is perpendicular to the XY-plane, its normal is perpendicular to the normal of the XY-plane. The normal to the XY-plane is along the Z-axis, which has direction ratios $(0, 0, 1)$ .

Therefore, the dot product of their normals is zero:

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

$1 + 5 λ = 0 \Rightarrow λ = - \frac{1}{5}$

Substituting $λ = - \frac{1}{5}$ in the equation of the plane, we get

$(1 - \frac{3}{5}) x + (1 - \frac{4}{5}) y + (1 - \frac{5}{5}) z - (1 - \frac{2}{5}) = 0$

$\frac{2}{5} x + \frac{1}{5} y - \frac{3}{5} = 0$

$2 x + y - 3 = 0$

Answer: $2 x + y - 3 = 0$

Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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