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

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Question

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

Vedclass · Accepted Answer

$3x + 9y + z = 17$

Vedclass · Answer

(B) The equation of any plane passing through the line of intersection of the planes $P_1: 2x - y + 3z + 5 = 0$ and $P_2: x + y + z - 1 = 0$ is given by $P_1 + \lambda P_2 = 0$. $(2x - y + 3z + 5) + \lambda(x + y + z - 1) = 0$ $(2 + \lambda)x + (-1 + \lambda)y + (3 + \lambda)z + (5 - \lambda) = 0$. Since the plane is rotated by $90^{\circ}$ about the line of intersection,the new plane must be perpendicular to the original plane $2x - y + 3z + 5 = 0$. The normal vectors are $\vec{n_1} = (2 + \lambda, -1 + \lambda, 3 + \lambda)$ and $\vec{n_2} = (2, -1, 3)$. For perpendicular planes,the dot product of their normals is zero: $(2 + \lambda)(2) + (-1 + \lambda)(-1) + (3 + \lambda)(3) = 0$ $4 + 2\lambda + 1 - \lambda + 9 + 3\lambda = 0$ $4\lambda + 14 = 0 \Rightarrow \lambda = -\frac{14}{4} = -\frac{7}{2}$. Substituting $\lambda = -\frac{7}{2}$ into the equation: $(2 - \frac{7}{2})x + (-1 - \frac{7}{2})y + (3 - \frac{7}{2})z + (5 + \frac{7}{2}) = 0$ $-\frac{3}{2}x - \frac{9}{2}y - \frac{1}{2}z + \frac{17}{2} = 0$ Multiplying by $-2$,we get $3x + 9y + z - 17 = 0$,or $3x + 9y + z = 17$.

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

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