# What is the equation of the plane that cuts the coordinate axes (a,0,0) (0,b,0) and (0,0,c)

Vikas TU
14149 Points
5 years ago
Dear Student,
ATQ:3 points in plane are ( a, 0, 0 ), ( 0, b, 0 ) and ( 0, 0, c ).
Let a be the vector from ( a, 0, 0 ) to ( 0, b, 0 ) :
a = = so x/a +y/b +z/c =1 is the reqd equation of the plane. [Ans]
=>bc(x-a)+acy+abz=0
=0. =0=>n.
General eqn of plane:
Now finding the normal between a and b=n=a x b =bc i+ ac j+ ab k
= b =
similarly let b be vector from ( a, 0, 0 ) to ( 0, 0, c ) :
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
Vikas TU
14149 Points
5 years ago
Dear Student,
ATQ:3 points in plane are ( a, 0, 0 ), ( 0, b, 0 ) and ( 0, 0, c ).
Let a be the vector from ( a, 0, 0 ) to ( 0, b, 0 ) :
similarly let b be vector from ( a, 0, 0 ) to ( 0, 0, c ) :
Now finding the normal between a and b=n=a x b =bc i+ ac j+ ab k
General eqn of plane:
=>bc(x-a)+acy+abz=0
so x/a +y/b +z/c =1 is the reqd equation of the plane. [Ans]
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)