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Hint: To find the equation of plane, we know that plane bisect the line so first find the midpoint. After that, find the direction cosines of the plane and you will get the equation and constant as $\lambda $ . To find $\lambda $ substitute midpoint in the equation of plane. So, you will get the answer.

__Complete step-by-step answer:__

The equation of a plane in the three-dimensional space is defined with the normal vector and the known point on the plane. A vector is a physical quantity that with its magnitude also has a direction attached to it. A normal vector means the line which is perpendicular to the plane. With reference to an origin, the position vector basically denotes the location or position (in a 3D Cartesian system) of a point. The Cartesian equation of a plane in 3 Dimensional space and vectors are explained in this article.

Space is the set of all points in the three dimensions – length, width, and height. It is made up of an infinite number of planes. Figures in space are called solids.

A plane, in geometry, prolongs infinitely in two dimensions. It has no width. We can see an example of a plane in coordinate geometry. The coordinates define the position of points in a plane.

In Maths, a plane is a flat, two-dimensional surface that prolongs infinitely far. A plane is a two-dimensional geometry which could consist of a point, a line and three-dimensional space. Planes can appear as subspaces of few higher-dimensional spaces, like the walls of room extended extremely faraway, or these walls experience an independent existence on their own, as in the framework of Euclidean geometry.

The method to get the equation of the line of intersection connecting two planes is to determine the set of points that satisfies the equations of both the planes. Since the equation of a plane comprises three variables and there are two equations (because of two planes), solving the concurrent equations will give a relationship between the three variables, which is equal to the equation of the intersection line.

Now we are given that the plane bisects the line joining the points, then the plane must meet at the midpoint of the line.

So, midpoint if $(1,2,3)$ and $(3,4,5)$ ,

$=(\dfrac{1+3}{2},\dfrac{2+4}{2},\dfrac{5+3}{2})=(2,3,4)$ ……………………. (Line is perpendicular to the plane.)

Now, the direction cosines of the plane are,

$(3-1,4-2,5-3)=(2,2,2)$

So the equation of plane is,

$2x+2y+2z=\lambda $

We know that, midpoint lies between the plane so it will satisfy the equation of the plane.

$\begin{align}

& 2(2)+2(3)+2(4)=\lambda \\

& \lambda =18 \\

\end{align}$

So we get the equation of the plane as $2x+2y+2z=18$ i.e. $x+y+z=9$ .

Note: Read the question carefully. Do not make silly mistakes. Don’t get confused while solving the problem. Your concept regarding planes should be clear. Do not jumble yourself while simplifying. While solving, do not miss any sign.

The equation of a plane in the three-dimensional space is defined with the normal vector and the known point on the plane. A vector is a physical quantity that with its magnitude also has a direction attached to it. A normal vector means the line which is perpendicular to the plane. With reference to an origin, the position vector basically denotes the location or position (in a 3D Cartesian system) of a point. The Cartesian equation of a plane in 3 Dimensional space and vectors are explained in this article.

Space is the set of all points in the three dimensions – length, width, and height. It is made up of an infinite number of planes. Figures in space are called solids.

A plane, in geometry, prolongs infinitely in two dimensions. It has no width. We can see an example of a plane in coordinate geometry. The coordinates define the position of points in a plane.

In Maths, a plane is a flat, two-dimensional surface that prolongs infinitely far. A plane is a two-dimensional geometry which could consist of a point, a line and three-dimensional space. Planes can appear as subspaces of few higher-dimensional spaces, like the walls of room extended extremely faraway, or these walls experience an independent existence on their own, as in the framework of Euclidean geometry.

The method to get the equation of the line of intersection connecting two planes is to determine the set of points that satisfies the equations of both the planes. Since the equation of a plane comprises three variables and there are two equations (because of two planes), solving the concurrent equations will give a relationship between the three variables, which is equal to the equation of the intersection line.

Now we are given that the plane bisects the line joining the points, then the plane must meet at the midpoint of the line.

So, midpoint if $(1,2,3)$ and $(3,4,5)$ ,

$=(\dfrac{1+3}{2},\dfrac{2+4}{2},\dfrac{5+3}{2})=(2,3,4)$ ……………………. (Line is perpendicular to the plane.)

Now, the direction cosines of the plane are,

$(3-1,4-2,5-3)=(2,2,2)$

So the equation of plane is,

$2x+2y+2z=\lambda $

We know that, midpoint lies between the plane so it will satisfy the equation of the plane.

$\begin{align}

& 2(2)+2(3)+2(4)=\lambda \\

& \lambda =18 \\

\end{align}$

So we get the equation of the plane as $2x+2y+2z=18$ i.e. $x+y+z=9$ .

Note: Read the question carefully. Do not make silly mistakes. Don’t get confused while solving the problem. Your concept regarding planes should be clear. Do not jumble yourself while simplifying. While solving, do not miss any sign.