# 2  Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3\widehat{i}+5\widehat{j}-6\widehat{k}$.

We have given the distance between the plane and origin equal to 7 units and normal to the vector $3\widehat{i}+5\widehat{j}-6\widehat{k}$.

So, it is known that the equation of the plane with position vector $\vec{r}$ is given by, the relation,

$\vec{r}.\widehat{n} =d$ , where d is the distance of the plane from the origin.

Calculating $\widehat{n}$;

$\widehat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{3\widehat{i}+5\widehat{j}-6\widehat{k}}{\sqrt{(3)^2+(5)^2+(6)^2}} = \frac{3\widehat{i}+5\widehat{j}-6\widehat{k}}{\sqrt{70}}$

$\vec{r}.\left ( \frac{3\widehat{i}+5\widehat{j}-6\widehat{k}}{\sqrt{70}} \right ) = 7$  is the vector equation of the required plane.

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