Answer to Question #185570 in Analytic Geometry for AGIE

Question #185570

. (a) The equation ax + by = 0 represents a line through the origin in R2. Show that the vector n1 = (a, b) formed from the coefficients of the equation is orthogonal to the line, that is, orthogonal to every vector along the line. (b) The equation ax + by + cz = 0 represents a plane through the origin in R3. Show that the vector n2 = (a, b, c) formed from the coefficients of the equation is orthogonal to the plane, that is, orthogonal to every vector that lies in the plane.


1
Expert's answer
2021-04-27T12:29:14-0400

a)

If "(P1) \u20d7" is orthogonal to "(n1) \u20d7" then, dot product of "(P1) \u20d7" and "(n1) \u20d7" will be zero.


(P1) ⃗."(n1) \u20d7=0"


Here let "(P1) \u20d7" be the vector along the line "ax +by=0"


"ax=-by"


"\\frac{x}{y}=\\frac{-b}{a}"


"p=-bi+aj"


let "(n1) \u20d7" be the vector formed from the coefficient (a,b) of the line equation .


"(n1) \u20d7=ai+bj"


now "(P1) \u20d7.(n1) \u20d7=(-ba+aj).(ai+bj)"

"=-ba+ab"

"=0"

hence proved that "(P1) \u20d7" and "(n1) \u20d7" are orthogonal


b)

To prove "(n2) \u20d7=(a,b,c)" formed from the coefficients of the equation is orthogonal to the plane,

"(a,b,c).(x,y,z)=0"


"(n2) \u20d7.(x,y,z)=0"


"(n2) \u20d7" is orthogonal to entry vector "(x,y,z)" in the plane .


"ax+by+cz=0" is a homogeneous equation.




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