Question

Find the equation of the plane which passes through the point (3,2,0) and contains the line $\dfrac{x-3}{1}=\dfrac{y-6}{5}=\dfrac{z-4}{4} ?$

Answer 1

An equation of the plane containing the point $(\mathrm{x}_0, \mathrm{y}_0, \mathrm{z}_0)$ with normal vector $\mathrm{N}=$ $\langle\mathrm{A}, \mathrm{B}, \mathrm{C}>$ is $\mathrm{A}(\mathrm{x}-\mathrm{x}_ 0)+\mathrm{B}(\mathrm{y}-\mathrm{y}_0)+\mathrm{C}(\mathrm{z}-\mathrm{z}_0)=0 .$ The equation of any plane can $\mathrm{be}$
expressed as $\mathrm{Ax}+\mathrm{By}+\mathrm{Cz}=\mathrm{D} .$ This is called the standard form of the equation of a plane. A plane is a two-dimensional doubly ruled surface spanned by two linearly independent vectors. The generalization of the plane to higher dimensions is called a hyperplane. The angle between two intersecting planes is known as the dihedral angle. The equation of a plane with nonzero normal vectors through the point is. Two planes are parallel if their normal vectors are parallel (constant multiples of one another). It is easy to recognize parallel planes written in the form $a x+b y+c z=d$ since a quick comparison of the normal vectors $\mathrm{n}=<\mathrm{a}, \mathrm{b}, \mathrm{c}>$ can be made.

Complete step-by-step answer:
If we know the normal vector of a plane and a point passing through the plane, the equation of the plane is established. a $(x-x_1)+b(y-y _1)+c(z-z_1)=0$
Let the equation of plane passing through (3,2,0) is
$\Rightarrow \mathrm{a}(\mathrm{x}-3)+\mathrm{b}(\mathrm{y}-2)+\mathrm{c}(\mathrm{z}-0)=0$…………………….. (1)
$\Rightarrow \dfrac{x-3}{1}=\dfrac{y-7}{5}=\dfrac{z-4}{4}$ pass through (3,2,0) and has direction ratios as 1,5,4
If the equation of plane contains this, it should pass from (3,7,4) and parallel to (1,5,4)
$\therefore \mathrm{a}(3-1)+\mathrm{b}(7-5)+\mathrm{c}(4-4)=0$
$\Rightarrow 2 \mathrm{a}+2 \mathrm{b}=0$
$\Rightarrow \mathrm{a}+\mathrm{b}=0$………………….. (2)
And
$\Rightarrow \mathrm{a}+5 \mathrm{b}+4 \mathrm{c}=0$………………… (3)
On solving equation (2) and (3) we get,
$\Rightarrow \mathrm{a}=\mathrm{c}$ and $4\text{b}=-\text{c}$
Hence, equation of plane will be,
$\Rightarrow \mathrm{c}(\mathrm{x}-3)-\mathrm{c}(\mathrm{y}-2)+\mathrm{cz}=0$, which is obtained from equation (1)
$\Rightarrow x-3-y+2+z=0$
$\Rightarrow \mathrm{x}-\mathrm{y}+\mathrm{z}-1=0$
$\Rightarrow \mathrm{x}-\mathrm{y}+\mathrm{z}=1$

Note: Similarly, the $\mathrm{y}$ -z-plane has standard equation $\mathrm{x}=0$ and the $\mathrm{x}$ -z-plane has standard equation $\mathrm{y}=0 .$ A plane parallel to the $\mathrm{x}$ -y-plane must have a standard equation $\mathrm{z}=$ d for some d, since it has normal vector k. A plane parallel to the $\mathrm{y}$ -z-plane has equation $x=d,$ and one parallel to the $x -z$-plane has equation $y=d$.In three dimensions, a surface normal, or simply normal, to a surface at point $\mathrm{P}$ is a vector perpendicular to the tangent plane of the surface at $\mathrm{P}$. The word "normal" is also used as an adjective: a line normal to a plane, the normal component of a force, the normal vector, etc.
Real-world examples of line segments are a pencil, a baseball bat, the cord to our cell phone charger, the edge of a table, etc. Think of a real-life quadrilateral, like a chessboard; it is made of four-line segments. Unlike line segments, examples of line segments in real life are endless.