unit vector orthogol to (4,2,3)

Question

let A=(4,2,3) , B= (a,b,,c) be orthogonal to A.

Then (A,B)=0

4a+2b+3c=o

By inspection, a solution is a=2, b=_1 c=_2

how we are getting a,b,c values.

we take B=(2,_1,_2) then norm B=sqr(4+1+4)=3

unit vector orthogonal to A =B /normB = (2/3 , -1/3 ,-2/3)

Answer 1

A=<4,2,3> (positional vector of point A), B=<a,b,c>.

A.B=0 if A and B are orthogonal, so 4a+2b+3c=0, which has many solutions. Imagine a line to represent A, then there are many lines which could be at right angles to it in 3-space.

|B|=√(a²+b²+c²) is the magnitude of B. Its unit vector is B/|B|=<a,b,c>/√(a²+b²+c²).

b=-(4a+3c)/2, so the unit vector is <a,-(4a+3c)/2,c>/√(a²+(4a+3c)²/4+c²), which depends on values of a and c. If a=0 and c=4, for example, the unit vector would be <0,-6,4>/(2√13)=<0,-3/√13,2√13).

If a=1, c=2, then b=-5, and the unit vector would be <1,-5,2>/√30.

There is no constant unit vector. Three equations are needed to determine a, b, c, and we can derive only one from the given information.