Question

Find the equation of the plane passing through (-2, 1, 3) and having (3,-5,4) as d.r.'s of its normal
pls give me answer step by step​

Answer 1

Answer:

equation of plane having (3,−5,4) as d.r's is 3x−5y+4z=c

This plane pass through(−2,1,3) so the point will satisfy the equation

3(−2)−5(1)+4(3)=c

c=1

So, equation of plane is 3x−5y+4z=1

Answer 2

Answer:

equation of plane having (3,−5,4) as d.r's is 3x−5y+4z=c

This plane pass through(−2,1,3) so the point will satisfy the equation

3(−2)−5(1)+4(3)=c

c=1

So, equation of plane is 3x−5y+4z=1