Question

. (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.

Answer 1

a)

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

(P1) ⃗.$(n1) ⃗=0$

Here let $(P1) ⃗$ be the vector along the line $ax +by=0$

$ax=-by$

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

$p=-bi+aj$

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

$(n1) ⃗=ai+bj$

now $(P1) ⃗.(n1) ⃗=(-ba+aj).(ai+bj)$

$=-ba+ab$

$=0$

hence proved that $(P1) ⃗$ and $(n1) ⃗$ are orthogonal

b)

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

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

$(n2) ⃗.(x,y,z)=0$

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

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