5 Find the vector and cartesian equations of the planes 

    (b)     that passes through the point (1,4, 6) and the normal vector to the plane is \widehat{i}-2\widehat{j}+\widehat{k}.

Answers (1)

Given the point A (1,4,6) and the normal vector \widehat{n} which is perpendicular to the plane is \widehat{n} = \widehat{i}-2\widehat{j}+\widehat{k}

The position vector of point A is \vec {a} = \widehat{i}+4\widehat{j}+6\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}+4\widehat{j}+6\widehat{k}) \right ].(\widehat{i}-2\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}+4\widehat{j}+6\widehat{k}) \right ].(\widehat{i}-2\widehat{j}+\widehat{k}) = 0

\Rightarrow \left [(x-1)\widehat{i}+(y-4)\widehat{j}+(z-6)\widehat{k}\right ].(\widehat{i}-2\widehat{j}+\widehat{k}) = 0

(x-1)-2(y-4)+(z-6)=0

\Rightarrow x-2y+z+1=0    

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

 

 

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