The equation of the plane bisecting the line | vedclass

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(C) Let the points be $A(2, 0, 6)$ and $B(-6, 2, 4)$.The plane bisects the line segment $AB$ and is perpendicular to it,meaning it passes through the

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$4x-y+z+4=0$

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(C) Let the points be $A(2, 0, 6)$ and $B(-6, 2, 4)$.The plane bisects the line segment $AB$ and is perpendicular to it,meaning it passes through the midpoint of $AB$ and the vector $\vec{AB}$ is the normal to the plane.The midpoint $M$ of $AB$ is $\left(\frac{2-6}{2}, \frac{0+2}{2}, \frac{6+4}{2}\right) = (-2, 1, 5)$.The normal vector $\vec{n}$ to the plane is $\vec{AB} = (-6-2, 2-0, 4-6) = (-8, 2, -2)$.We can simplify the normal vector by dividing by $-2$,giving $\vec{n}' = (4, -1, 1)$.The equation of the plane passing through $(x_0, y_0, z_0)$ with normal $(a, b, c)$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.Substituting the values: $4(x - (-2)) - 1(y - 1) + 1(z - 5) = 0$.$4(x+2) - y + 1 + z - 5 = 0$.$4x + 8 - y + z - 4 = 0$.$4x - y + z + 4 = 0$.

The equation of the plane bisecting the line segment joining the points $(2,0,6)$ and $(-6,2,4)$ and perpendicular to it,is

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