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(A) The equation of a plane passing through the intersection of planes $P_1: 2x - y + z - 3 = 0$ and $P_2: 4x - 3y + 5z + 9 = 0$ is given by $(2x - y

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equation of a plane passing through the intersection of planes $P_1: 2x - y + z - 3 = 0$ and $P_2: 4x - 3y + 5z + 9 = 0$ is given by $(2x - y + z - 3) + \lambda(4x - 3y + 5z + 9) = 0$.Rearranging the terms: $(2 + 4\lambda)x + (-1 - 3\lambda)y + (1 + 5\lambda)z + (-3 + 9\lambda) = 0$.The normal vector to this plane is $\vec{n} = (2 + 4\lambda)\hat{i} + (-1 - 3\lambda)\hat{j} + (1 + 5\lambda)\hat{k}$.Since the plane is parallel to the line with direction vector $\vec{v} = 2\hat{i} + 4\hat{j} + 5\hat{k}$, the normal vector $\vec{n}$ must be perpendicular to $\vec{v}$, so $\vec{n} \cdot \vec{v} = 0$.$2(2 + 4\lambda) + 4(-1 - 3\lambda) + 5(1 + 5\lambda) = 0$.$4 + 8\lambda - 4 - 12\lambda + 5 + 25\lambda = 0$.$21\lambda + 5 = 0 \implies \lambda = -5/21$.Substituting $\lambda = -5/21$ into the plane equation: $(2 - 20/21)x + (-1 + 15/21)y + (1 - 25/21)z + (-3 - 45/21) = 0$.$(42 - 20)/21 x + (-21 + 15)/21 y + (21 - 25)/21 z + (-63 - 45)/21 = 0$.$22x - 6y - 4z - 108 = 0$.Dividing by $2$: $11x - 3y - 2z - 54 = 0$.

The equation of the plane passing through the intersection of planes $2x - y + z = 3$ and $4x - 3y + 5z = -9$ and parallel to the line $\frac{x + 1}{2} = \frac{y + 3}{4} = \frac{z - 3}{5}$ is

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