Find the intercepts cut off by the plane 2x + y – z = 5.
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Step 1:
The given equation is 2x+y−z=52x+y−z=5
Divide throughout by 5,we get
x25x25+y5+y5+z−5+z−5=1=1
Step 2:
Equation of a plane in intercept form is xaxa+yb+yb+zc+zc=1=1
Where a,b,ca,b,c are the intercepts cut off by the plane at x,yx,y and zz axes respectively.
a=25a=25,b=5,b=5 and c=−5c=−5
Thus the intercepts cutoff by the plane are 5252,5,5 and −5
hope u like it!!!
The given equation is 2x+y−z=52x+y−z=5
Divide throughout by 5,we get
x25x25+y5+y5+z−5+z−5=1=1
Step 2:
Equation of a plane in intercept form is xaxa+yb+yb+zc+zc=1=1
Where a,b,ca,b,c are the intercepts cut off by the plane at x,yx,y and zz axes respectively.
a=25a=25,b=5,b=5 and c=−5c=−5
Thus the intercepts cutoff by the plane are 5252,5,5 and −5
hope u like it!!!
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