# 5  Find the vector and cartesian equations of the planes (a) that passes through the point (1, 0, – 2) and the normal to the plane is $\widehat{i}+\widehat{j}-\widehat{k}.$

Given the point $A (1,0,-2)$ and the normal vector $\widehat{n}$ which is perpendicular to the plane is $\widehat{n} = \widehat{i}+\widehat{j}-\widehat{k}$

The position vector of point A is $\vec {a} = \widehat{i}-2\widehat{k}$

So, the vector equation of the plane would be given by,

$(\vec{r}-\vec{a}).\widehat{n} = 0$

Or $\left [ \vec{r}-(\widehat{i}-2\widehat{k}) \right ].(\widehat{i}+\widehat{j}-\widehat{k}) = 0$

where $\vec{r}$ is the position vector of any arbitrary point $A(x,y,z)$ in the plane.

$\therefore$ $\vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k}$

Therefore, the equation we get,

$\left [(x\widehat{i}+y\widehat{j}+z\widehat{k})-(\widehat{i}-2\widehat{k}) \right ].(\widehat{i}+\widehat{j}-\widehat{k}) = 0$

$\Rightarrow \left [(x-1)\widehat{i}+y\widehat{j}+(z+2)\widehat{k}\right ].(\widehat{i}+\widehat{j}-\widehat{k}) = 0$

$\Rightarrow(x-1)+y-(z+2) = 0$

$\Rightarrow x+y-z-3=0$    or   $x+y-z=3$

So, this is the required Cartesian equation of the plane.

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