The plane 2x - y + 3z + 5 = 0 is rotated thro | vedclass

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(B) The equation of any plane passing through the line of intersection of the planes $P_1: 2x - y + 3z + 5 = 0$ and $P_2: 5x - 4y - 2z + 1 = 0$ is giv

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$27x - 24y - 26z - 13 = 0$

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(B) The equation of any plane passing through the line of intersection of the planes $P_1: 2x - y + 3z + 5 = 0$ and $P_2: 5x - 4y - 2z + 1 = 0$ is given by $P_1 + \lambda P_2 = 0$.$(2x - y + 3z + 5) + \lambda(5x - 4y - 2z + 1) = 0$$(2 + 5\lambda)x - (1 + 4\lambda)y + (3 - 2\lambda)z + (5 + \lambda) = 0$ ... $(1)$Since the plane is rotated by $90^o$ about the line of intersection,the new plane must be perpendicular to the original plane $2x - y + 3z + 5 = 0$.The normal vectors are $\vec{n_1} = (2 + 5\lambda, -(1 + 4\lambda), 3 - 2\lambda)$ and $\vec{n_2} = (2, -1, 3)$.For perpendicular planes,the dot product of their normals is zero:$2(2 + 5\lambda) - 1(-(1 + 4\lambda)) + 3(3 - 2\lambda) = 0$$4 + 10\lambda + 1 + 4\lambda + 9 - 6\lambda = 0$$8\lambda + 14 = 0 \Rightarrow \lambda = -\frac{14}{8} = -\frac{7}{4}$.Substituting $\lambda = -\frac{7}{4}$ into equation $(1)$:$(2 + 5(-\frac{7}{4}))x - (1 + 4(-\frac{7}{4}))y + (3 - 2(-\frac{7}{4}))z + (5 - \frac{7}{4}) = 0$$(\frac{8 - 35}{4})x - (1 - 7)y + (3 + \frac{7}{2})z + (\frac{20 - 7}{4}) = 0$$-\frac{27}{4}x + 6y + \frac{13}{2}z + \frac{13}{4} = 0$Multiplying by $-4$:$27x - 24y - 26z - 13 = 0$.

The plane $2x - y + 3z + 5 = 0$ is rotated through $90^o$ about its line of intersection with the plane $5x - 4y - 2z + 1 = 0$. The equation of the plane in its new position is:

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