The equation of the plane which is bisecting | vedclass

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(B) The plane is perpendicular to the line segment $AB$ and passes through its midpoint. First,find the midpoint $M$ of the line segment $AB$: $M = \l

Vedclass · Accepted Answer

$3x+y+3z-2=0$

Vedclass · Answer

(B) The plane is perpendicular to the line segment $AB$ and passes through its midpoint. First,find the midpoint $M$ of the line segment $AB$: $M = \left( \frac{2 + (-4)}{2}, \frac{3 + 1}{2}, \frac{4 + (-2)}{2} \right) = (-1, 2, 1)$. Next,find the normal vector $\vec{n}$ to the plane,which is the vector $\vec{AB}$: $\vec{n} = \vec{AB} = (-4 - 2, 1 - 3, -2 - 4) = (-6, -2, -6)$. We can simplify the normal vector by dividing by $-2$: $\vec{n}' = (3, 1, 3)$. The equation of the plane passing through $(x_0, y_0, z_0)$ with normal vector $(a, b, c)$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$. Substituting the values: $3(x - (-1)) + 1(y - 2) + 3(z - 1) = 0$ $3(x + 1) + (y - 2) + 3(z - 1) = 0$ $3x + 3 + y - 2 + 3z - 3 = 0$ $3x + y + 3z - 2 = 0$. Thus,the correct option is $B$.

The equation of the plane which is bisecting the line segment joining the points $A(2,3,4)$ and $B(-4,1,-2)$ and is perpendicular to it,is

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