Q10 of 230 Page 29

Find the intercepts made on the coordinate axes by the plane 2x + y – 2z = 3 and also find the direction cosines of the normal to the plane.

The given equation of the plane is 2x + y – 2z = 3


Dividing by 3 on both the sides, we get




We know that, if a, b, c are the intercepts by the plane on the coordinate axes, new equation of the plane is



Comparing the equation (i) and (ii), we get



Again the given equation of the plane is


2x+y-2z=3


Writing this in the vector form, we get




So vector normal to the plane is given by






Direction vector of


Direction vector of


So,


Intercepts by the plane on the coordinate axes are


Direction cosines of normal to the plane are


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