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3 Find the Cartesian equation of the following planes:

(b)     \overrightarrow{r}.(2\widehat{i}+3\widehat{i}-4\widehat{k})=1

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Given the equation of plane \overrightarrow{r}.(2\widehat{i}+3\widehat{i}-4\widehat{k})=1

So we have to find the Cartesian equation,

Any point A (x,y,z) on this plane will satisfy the equation and its position vector given by,

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

Hence we have,

 (x\widehat{i}+y\widehat{j}+z\widehat{k}).(2\widehat{i}+3\widehat{j}-4\widehat{k}) =1

Or,  2x+3y-4z=1

Therefore this is the required Cartesian equation of the plane.

Posted by

Divya Prakash Singh

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