Find the vector equation of the plane which is at a distance of 6 / √29 from the origin it's - Maths - Three Dimensional Geometry

Question

Find the vector equation of the plane which is at a distance of
6 / √29 from the origin it's normal vector from the origin
is
2 i^- 3 j^ + 4 k^ Also find it's cartesian form
Plzzz meritnation experts,answer to my question as soon as
possible!!!!!!!

Gursheen kaur. · Accepted Answer

The normal vector is,

$\vec{n} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k} t h e r e f o r e, \hat{n} = \frac{\vec{n}}{| \vec{n} |} = \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{4 + 9 + 16}} = \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{29}}$

It is known that the equation of the plane with position vector is given by,

$\Rightarrow \hat{r} . \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{29}} = \frac{6}{\sqrt{29}}$

This is the vector equation of the required plane.

is given by,

$\Rightarrow \hat{r} \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{29}} = \frac{6}{\sqrt{29}}$ This is the vector equation of the required plane

Find the vector equation of the plane which is at a distance of 6 / √29 from the origin it's normal vector from the origin is 2 i^- 3 j^ + 4 k^. Also find it's cartesian form. Plzzz meritnation experts,answer to my question as soon as possible!!!!!!!

Find the vector equation of the plane which is at a distance of 6 / √29 from the origin it's normal vector from the origin is

2i^- 3 j^ + 4 k^. Also find it's cartesian form.

Plzzz meritnation experts,answer to my question as soon as possible!!!!!!!