1.

Find the vector equation of the plane which is at a distance of (6)/(sqrt(29)) the from the origin and its normal vector from the origin is 2hati-3hatj+4hatk. Also find its cartesian form.

Answer»


Answer :`vecr((2)/(sqrt(29))hati+((-3))/(sqrt(29))HATJ+(4)/(sqrt(29))hatk)=(6)/(sqrt(29))`


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