Find the vector equation of the plane which is at a distance of `6/(sqrt(29))`from the origin and its normal vector from the origin is `2 hat i-3 hat j+4 hat k`. Also find its cartesian form.

Explanation

,Here , It is given that

`d=frac{6}{sqrt{29}}`


Let `vec{N}=2{ }_{1}^{a}-3 vec{j}-4 vec{k}`

So, `{f}=frac{vec{N}}{|vec{N}|}=frac{2 vec{{a}}-3 vec{{j}}-4 vec{{k}}}{sqrt{29}}`

So, the vector equation of the plane is `{vec{r}} cdot {n}={d}`

`{vec{r}} cdot frac{2 vec{i}-3 vec{j}-4 vec{k}}{sqrt{29}}`
`=frac{6}{sqrt{29}}`
`{vec{r}} cdot(2_{1}^{vec{A}}-3 vec{j}-4 vec{k})`
`=6`