Find the intercept cut off by the plane
2x+y-z =5 with the axes.
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equation of the plane in Intercept form is X/a+y/b+z/c=1
here a ,b ,c are intercept cut off of the plane x, y, z respectively
the given equation is 2x+y-z=5
divided by 5 ,
x/2/5+y/5+z/-5=1
intercept cut off plane a=5/2,b=5,c=-5
here a ,b ,c are intercept cut off of the plane x, y, z respectively
the given equation is 2x+y-z=5
divided by 5 ,
x/2/5+y/5+z/-5=1
intercept cut off plane a=5/2,b=5,c=-5
1RADHIKAA1:
If we divide the equation by two then it should be 2x/5 + y/5 + z/-5 = 1 naa?
Oh ok understood
But if it goes right side to the is equal to.. Then the a nd b should be in negative??
no sis . multiplication to division ,not add to subtract
already "Z" have negative , then that sign goes to constant value (-5)
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