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}.

Answers (1)

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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