Dot Product Question, Perpendicular

[brandon2743]

pretty much this is all i was given.
I have no idea how to even approach it. I do not need an answer (would be nice though), just an idea on how to go about starting it.

 

[vela]

If a vector x is perpendicular to (-3, 8, -1), you know x•(-3, 8, -1)=0. Write out the left-hand side, and solve the equation.

 

[HallsofIvy]

Since this is in R³ writing the vector x as <x, y, z>, that dot product will given you a single equation in the three unknowns, x, y, and z. You can solve for one of them in terms of other two. For example, you were to find that z= ax+ by, then you could write your vector as <x, y, z>= <x, y, ax+ by>= x<1, 0, a>+ y<0, 1, b>.

(On the other hand, if you solve for x, say, as x= py+ qz, then you could write the vector as <x, y, z>= <py+ qz, y, z>= y<p, 1, 0>+ z<q, 0, 1>. There are an infinite number of correct solutions to this problem.)

 

[brandon2743]

Since this is in R³ writing the vector x as <x, y, z>, that dot product will given you a single equation in the three unknowns, x, y, and z. You can solve for one of them in terms of other two. For example, you were to find that z= ax+ by, then you could write your vector as <x, y, z>= <x, y, ax+ by>= x<1, 0, a>+ y<0, 1, b>.

(On the other hand, if you solve for x, say, as x= py+ qz, then you could write the vector as <x, y, z>= <py+ qz, y, z>= y<p, 1, 0>+ z<q, 0, 1>. There are an infinite number of correct solutions to this problem.)

this makes sense thank you.

 

[brandon2743]

anyone else have an input just so i can be sure