Question

A plane bisects the line segment joining the points $\left( 1,2,3 \right)$ and $\left( -3,4,5 \right)$ at right angles. Then this plane also passes through the point:
A. $\left( -3,2,1 \right)$
B. $\left( 3,2,1 \right)$
C. $\left( 1,2,-3 \right)$
D. $\left( -1,2,3 \right)$

Answer 1

As we know that the plane bisect the line that is joining the points $\left( 1,2,3 \right)$ and $\left( -3,4,5 \right)$, then the plane must meet the line at the mid-point of the line. So first we will find out the mid-point then after getting the mid-point we will find the direction cosines i.e. also known as normal vector by using the formula$\left( \left( {{x}_{2}}-{{x}_{1}} \right),\left( {{y}_{2}}-{{y}_{1}} \right),\left( {{z}_{2}}-{{z}_{1}} \right) \right)$. Here $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$are the coordinates of the two given points. Then using the using of the plane i.e. $a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)$, we will substitute the values and here $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$are the coordinates of the mid-point and$\left( a,b,c \right)$is the coordinates of the direction cosines. Later substituting the values of the option taking one by one and in this we will get the required answer.

Formula used:
Mid-point of any given plane in three dimensional with two point is,
$mid-po\operatorname{int}=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2} \right)$, where $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$are the coordinates of the two given points.
The direction ratio of the plane with two given points are $\left( \left( {{x}_{2}}-{{x}_{1}} \right),\left( {{y}_{2}}-{{y}_{1}} \right),\left( {{z}_{2}}-{{z}_{1}} \right) \right)$.
The equation of the plane through a point ‘A’ that is equal to $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and have a normal vector that is equal to $\overrightarrow{n}=\left( a,b,c \right)$is given by;
$a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)$.

Complete step by step solution:
We have given that,
A plane bisects the line segment joining the points $\left( 1,2,3 \right)$ and $\left( -3,4,5 \right)$ at right angles.
As we know that,
The plane bisects the line that is joining the points $\left( 1,2,3 \right)$ and $\left( -3,4,5 \right)$, then the plane must meet the line at the mid-point of the line.
Now,
Mid-point of any given plane in three dimensional with two point is given by,
$mid-po\operatorname{int}=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2} \right)$, where $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$are the coordinates of the two given points.
Thus,
The plane must meet the line at the mid-point of the line is given by;
$mid-po\operatorname{int}=\left( \dfrac{1+\left( -3 \right)}{2},\dfrac{2+4}{2},\dfrac{3+5}{2} \right)=\left( \dfrac{-2}{2},\dfrac{6}{2},\dfrac{8}{2} \right)=\left( -1,3,4 \right)$,
Thus, the coordinate of the line that is perpendicular to the plane is $\left( -1,3,4 \right)$.
Now,
The direction ratio of the plane with two given points are $\left( \left( {{x}_{2}}-{{x}_{1}} \right),\left( {{y}_{2}}-{{y}_{1}} \right),\left( {{z}_{2}}-{{z}_{1}} \right) \right)$
Thus,
The direction ratio of the plane with the given points$\left( 1,2,3 \right)$and $\left( -3,4,5 \right)$is, $\Rightarrow \left( -3-1,4-2,5-3 \right)=\left( -4,2,2 \right)$
Therefore,
The equation of the plane through a point ‘A’ that is equal to $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and have a normal vector that is equal to $\overrightarrow{n}=\left( a,b,c \right)$ is given by;
$a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)$
Now,
The equation of plane must be,
As we know that the midpoint of the line is lying in the plane thus it must satisfy the plane.
Thus we have a point $\left( -1,3,4 \right)$ and the normal vector i.e. the direction cosines is equal to $\left( -4,2,2 \right)$.
So,
The equation of plane is given by,
$\Rightarrow -4\left( x+1 \right)+2\left( y-3 \right)+2\left( z-4 \right)=0$
Simplifying the above equation,
$\Rightarrow -4x-4+2y-6+2z-8=0$
Simplifying the numbers,
$\Rightarrow -4x+2y+2z-18=0$
Adding 18 to both the sides of the equation,
$\Rightarrow -4x+2y+2z=18$
Taking out 2 as a common term,
$\Rightarrow 2\left( -2x+y+z \right)=18$
Dividing both the sides of the equation by 2,
$\Rightarrow -2x+y+z=9$
$\therefore$ The equation of the plane is $-2x+y+z=9$
Thus,
Out of the four given options;
First option is $\left( -3,2,1 \right)$.
Substituting the values in the equation of the plane.
$\Rightarrow -2\left( -3 \right)+2+1=9\Rightarrow 6+2+1=9\Rightarrow 9=9$
So the point $\left( -3,2,1 \right)$ satisfies the equation of the plane.
No need to check for the others given options as we got the correct answer.

Hence, the option (A) is the correct answer.

Note: Students need to read the question carefully. Do not make mistakes, try to avoid making any calculation errors. Don’t get confused while solving the problem. Your concept regarding the planes should be clear. Do not entangle yourself while simplifying, taking care of each sign associated with the numbers.
While solving the questions, students should know the equation of the plane i.e. $a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)$ and the equation of the plane through a point ‘A’ that is equal to $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and have a normal vector that is equal to $\overrightarrow{n}=\left( a,b,c \right)$.
Students might get confused in putting or substituting the values so the chances of the silly mistakes are more here. Therefore they should be very careful about noting down or substituting the values.