The plane which bisects the line segment join | vedclass

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Question

(A) Let the points be $A(4, -2, 3)$ and $B(2, 4, -1)$.The midpoint $M$ of the line segment $AB$ is given by:$M = \left( \frac{4+2}{2}, \frac{-2+4}{2},

Vedclass · Accepted Answer

$(4, 0, -1)$

Vedclass · Answer

(A) Let the points be $A(4, -2, 3)$ and $B(2, 4, -1)$.The midpoint $M$ of the line segment $AB$ is given by:$M = \left( \frac{4+2}{2}, \frac{-2+4}{2}, \frac{3-1}{2} ight) = (3, 1, 1)$.The normal vector $\vec{n}$ to the plane is the vector $\vec{AB}$:$\vec{n} = \vec{AB} = (2-4, 4-(-2), -1-3) = (-2, 6, -4)$.We can simplify the normal vector by dividing by $-2$,so we take $\vec{n} = (1, -3, 2)$.The equation of the plane passing through $M(3, 1, 1)$ with normal vector $\vec{n} = (1, -3, 2)$ is:$1(x - 3) - 3(y - 1) + 2(z - 1) = 0$$x - 3 - 3y + 3 + 2z - 2 = 0$$x - 3y + 2z - 2 = 0$.Now,we check which of the given points satisfies this equation:For $(4, 0, -1)$: $4 - 3(0) + 2(-1) - 2 = 4 - 0 - 2 - 2 = 0$. This point satisfies the equation.For $(4, 0, 1)$: $4 - 3(0) + 2(1) - 2 = 4 + 2 - 2 = 4 
eq 0$.For $(0, 1, -1)$: $0 - 3(1) + 2(-1) - 2 = -3 - 2 - 2 = -7 
eq 0$.For $(0, -1, 1)$: $0 - 3(-1) + 2(1) - 2 = 3 + 2 - 2 = 3 
eq 0$.Thus,the plane passes through the point $(4, 0, -1)$.

The plane which bisects the line segment joining the points $A(4, -2, 3)$ and $B(2, 4, -1)$ at right angles also passes through the point:

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