Q. 105.0( 1 Vote )

Find the vector equation of the plane which is at a distance of from the origin and its normal vector from the origin is . Also, find its Cartesian form.

Answer :

Given, normal vector

Now,


The equation of the plane in normal form is


…… (i)


(where d is the distance of the plane from the origin)


Substituting and d = in (i)


……(ii)


Cartesian Form


For Cartesian Form, substituting in (ii), we get




So, 2x – 3y + 4z = 6, is the Cartesian form


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