Question

Find a vector in V^3 that is perpendicular to u (5,-2,3)

Answer 1

(1, 1, -1)

Answer 2

How did you get that? Please can you show me your work?

Answer 3

Do you know about dot products? Its sorta like a guessing game where you want the dot (or inner product) to come out to 0.

Answer 4

If ventor u(a, b, c) and v(a', b', c') are perpendicular to each other, a x a' + b x b' + c x c' = 0. In a mathematical expression, " u·v = 0 "

Answer 5

Also there are many vectors that is perpendicular to one vector in V^3

Answer 6

So should I say (1,-1,1) is one of the solution?

Answer 7

Yes. It is one of the infinite number of solutions.

There is an entire 2-dim plane of vectors perpendicular to any (non-zero) vector in R^3.

Answer 8

Consider the generic solution with components x, y, and z. We know their dot product must be zero since they are perpendicular (this is our constraint).\[\underbrace{\langle 5,-2,3\rangle}_{\text{given}} \cdot \underbrace{\langle x,y,z \rangle}_{\text{solution}} = 0\]\[5x-2y+3z=0\]Solve for one of the variables, let's say x:\[x = \frac{2y-3z}{5}\]Pick any values of y and z you want, and just plug back into above to find the value of x that will work.