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Let the coordinates of the foot of perpendicular P from the origin to the plane be  Given a plane equation , Or,  The direction ratios of the normal of the plane are . Therefore  So, now dividing both sides of the equation by  we will obtain, This equation is similar to  where,  are the directions cosines of normal to the plane and d is the distance of normal from the origin. Then finding the...
Given the equation of plane  So we have to find the Cartesian equation, Any point  on this plane will satisfy the equation and its position vector given by,   Hence we have,   Or,   Therefore this is the required Cartesian equation of the plane.
Given the equation of plane  So we have to find the Cartesian equation, Any point  on this plane will satisfy the equation and its position vector given by, Hence we have,   Or,   Therefore this is the required Cartesian equation of the plane.
Given the equation of the plane  So we have to find the Cartesian equation, Any point  on this plane will satisfy the equation and its position vector given by, Hence we have,   Or,   Therefore this is the required Cartesian equation of the plane.  
We have given the distance between the plane and origin equal to 7 units and normal to the vector . So, it is known that the equation of the plane with position vector  is given by, the relation,  , where d is the distance of the plane from the origin. Calculating ;   is the vector equation of the required plane.
Given the equation of plane is   or we can write So, the direction ratios of normal from the above equation are, . Therefore  Then dividing both sides of the plane equation by , we get So, this is the form of   the plane, where  are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.  The direction cosines of the given line are  and...
Given the equation of plane is    So, the direction ratios of normal from the above equation are, . Therefore  Then dividing both sides of the plane equation by , we get So, this is the form of   the plane, where  are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.  The direction cosines of the given line are  and the distance of...
Given the equation of the plane is  or we can write  So, the direction ratios of normal from the above equation are, . Therefore  Then dividing both sides of the plane equation by , we get So, this is the form of   the plane, where  are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.  The direction cosines of the given line...
Equation of plane Z=2, i.e.   The direction ratio of normal is 0,0,1 Divide equation  by 1 from both side  We get,       Hence, direction cosins are 0,0,1. The distance of the plane from the origin is 2.
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