Given points A(3, 2, -1) and B(1, 4, 3),find | vedclass

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(D) The plane bisects the segment $AB$ perpendicularly,meaning it passes through the midpoint of $AB$ and the vector $\vec{AB}$ is normal to the plane

Vedclass · Accepted Answer

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

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(D) The plane bisects the segment $AB$ perpendicularly,meaning it passes through the midpoint of $AB$ and the vector $\vec{AB}$ is normal to the plane. Midpoint $M = \left(\frac{3+1}{2}, \frac{2+4}{2}, \frac{-1+3}{2}\right) = (2, 3, 1)$. The normal vector $\vec{n}$ is the direction ratio of line $AB$: $\vec{n} = (1-3, 4-2, 3-(-1)) = (-2, 2, 4)$. We can simplify the normal vector by dividing by $-2$,giving $\vec{n}' = (1, -1, -2)$. 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: $1(x-2) - 1(y-3) - 2(z-1) = 0$. $x - 2 - y + 3 - 2z + 2 = 0$. $x - y - 2z + 3 = 0$.

Given points $A(3, 2, -1)$ and $B(1, 4, 3)$,find the equation of the plane that bisects the segment $AB$ perpendicularly.

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