how can we find a normal vector from the equ of aplane give reasons of ur ans
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rearrange the eqn of a plane in the form ax +by + cz +d=0
the normal vector is
give reasons why we do this
suppose any known vector k, and normal vector n
then the eqn of the plane is (r-k).n = 0
if you expand this, youll see the coefficients of x, y, z in cartesian form are the direction ratios of the normal vector
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PS : k is the position vector of a known pt on the plane
what are direction ratios??
consider a plane equation in its primary form (r-k).n=0 (where n->normal vector & k-> is the position vector of the plane) and when u expand this using simple dot product then finally u will get that coefficients of x,y,z gives u the dierection ratios
direction ratios are the ratio of the magnitude of vector in one direction to the magnitude in the resultant direction.