. (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.
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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