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Find the vector equation of a plane passing through a point having position vector `(2hati-hatj+hatk)` and perpendicular to the vector `(4hati+2hatj-3hatk)`. Also, reduce it to Cartesian form. |
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Answer» Here, `veca=(2hati-hatj+hatk)` and `vecn=(4hati+2hatj-3hatk)`. Clearly, the required vector equation is `(vecr-veca).vecn=0`. `rArr vecr.vecn=veca.vecn` `rArr vecr.(4hati+2hatj-3hatk)=(2hati-hatj+hatk).(4hati+2hatj-3hatk)` =`(8-2-3)=3`. Hence, the vector equation of the given plane is `vecr.(4hati+2hatj-3hatk)=3`..............(1) Reduction to Cartesian form: Putting `vecr=(xhati+yhatj+zhatk)`, we get `(xhati+yhatj+zhatk).(4hati+2hatj-3hatk)=3 rArr 4x+2y-3z=3`. Hence, the equation ofthe given plane in Cartesian form is `4x+2y-3z=3`. |
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