Question

find the e question of the plane passing through (2,3,4) and perpendicular to x axis​

Answer 1

Our plane passes through the point (2, 3, 4). Let (x, y, z) be a point on our plane such that the vector $\left<x-2,\ y-3,\ z-4\right>$ is parallel to our plane.

Our plane is perpendicular to x axis, so that the vector $\left<1,\ 0,\ 0\right>$ is normal to our plane.

Now these two vectors are perpendicular to each other, so their dot product is zero, i.e.,

$\longrightarrow\left<x-2,\ y-3,\ z-4\right>\cdot\left<1,\ 0,\ 0\right>=0$

$\longrightarrow\underline{\underline{x-2=0}}$

This is the equation of our plane.

Equation of a Plane

Consider a plane passing through a known point with position vector $\vec{r_0}=\left<x_0,\ y_0,\ z_0\right>$ and perpendicular to the vector $\vec{n}=\left<a,\ b,\ c\right>.$

Let $\vec{r}=\left<x,\ y,\ z\right>$ be the position vector of an arbitrary point on the plane such that the vector $\vec{r}-\vec{r_0}=\left<x-x_0,\ y-y_0,\ z-z_0\right>$ is parallel to the plane.

Now $\vec{r}-\vec{r_0}$ is perpendicular to $\vec{n}$ so their dot product is zero, i.e.,

$\longrightarrow\left(\vec{r}-\vec{r_0}\right)\cdot\vec{n}=0$

$\longrightarrow\vec{r}\cdot\vec{n}-\vec{r_0}\cdot\vec{n}=0$

$\longrightarrow\vec{r}\cdot\vec{n}=\vec{r_0}\cdot\vec{n}$

Assume $\vec{r_0}\cdot\vec{n}=ax_0+by_0+cz_0=d.$ Then,

$\longrightarrow\boxed{\vec{r}\cdot\vec{n}=d}$

This is the equation of the plane in vector form.

Now,

$\longrightarrow\vec{r}\cdot\vec{n}=d$

$\longrightarrow\left<x,\ y,\ z\right>\cdot\left<a,\ b,\ c\right>=d$

$\longrightarrow\boxed{ax+by+cz=d}$

This is the equation of the plane in Cartesian form.