Let the plane 2x + 3y + z + 20 = 0 be rotated | vedclass

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(A) The equation of the family of planes passing through the intersection of $2x + 3y + z + 20 = 0$ and $x - 3y + 5z - 8 = 0$ is given by $(2x + 3y +

Vedclass · Accepted Answer

$\frac{a}{8} = \frac{b}{5} = \frac{c}{-4}$

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(A) The equation of the family of planes passing through the intersection of $2x + 3y + z + 20 = 0$ and $x - 3y + 5z - 8 = 0$ is given by $(2x + 3y + z + 20) + \lambda(x - 3y + 5z - 8) = 0$,which simplifies to $(2 + \lambda)x + (3 - 3\lambda)y + (1 + 5\lambda)z + (20 - 8\lambda) = 0$.Since the plane is rotated by a right angle,the new plane is perpendicular to the original plane $2x + 3y + z + 20 = 0$. Thus,the dot product of their normals is zero:$2(2 + \lambda) + 3(3 - 3\lambda) + 1(1 + 5\lambda) = 0$$4 + 2\lambda + 9 - 9\lambda + 1 + 5\lambda = 0$$14 - 2\lambda = 0 \Rightarrow \lambda = 7$.Substituting $\lambda = 7$ into the family equation,we get the rotated plane: $9x - 18y + 36z - 36 = 0$,or $x - 2y + 4z - 4 = 0$.The mirror image $B(a, b, c)$ of point $A(2, -1/2, 2)$ in the plane $x - 2y + 4z - 4 = 0$ is found using the formula $\frac{a - 2}{1} = \frac{b + 1/2}{-2} = \frac{c - 2}{4} = -2 \frac{2 - 2(-1/2) + 4(2) - 4}{1^2 + (-2)^2 + 4^2} = -2 \frac{2 + 1 + 8 - 4}{1 + 4 + 16} = -2 \frac{7}{21} = -2/3$.Solving for $a, b, c$:$a = 2 - 2/3 = 4/3$$b = -1/2 + 4/3 = 5/6$$c = 2 - 8/3 = -2/3$.Thus,$B = (4/3, 5/6, -2/3) = (8/6, 5/6, -4/6)$.Comparing with the options,$\frac{a}{8} = \frac{b}{5} = \frac{c}{-4} = 1/6$.

Let the plane $2x + 3y + z + 20 = 0$ be rotated through a right angle about its line of intersection with the plane $x - 3y + 5z = 8$. If the mirror image of the point $(2, -1/2, 2)$ in the rotated plane is $B(a, b, c)$,then:

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