Finding Perpendicular Vectors in R^4 Using Gaussian Elimination

[Punkyc7]

Find all vectors in R^4 that are perpendicular to the three coloumn vectors

[1 1 1 1], [1 2 3 4], [1 9 9 7]

sorry i don't know how to make them verticle

guassian elimination or Ax=b


I know that the dot product must equal 0 but I am not sure how to solve it

 

[Char. Limit]

You have three vectors. Invent a fourth vector...

[x₁ x₂ x₃ x₄]

Take the dot product of your three vectors with this fourth vector. You'll end up with three equations and four unknowns.

 

[Punkyc7]

ok so you get x1 +18x2+27x3+28x4
plug in 1 for x 4 -1 for x3 0 for x2 and -1 for x

so some constant multipled by[-1 0 -1 1]

 

[SammyS]

ok so you get x1 +18x2+27x3+28x4
plug in 1 for x 4 -1 for x3 0 for x2 and -1 for x1

so some constant multipled by[-1 0 -1 1]

[-1 0 -1 1] is perpendicular to [1 2 3 4], but not perpendicular to the other two.

If [x₁ x₂ x₃ x₄] is perpendicular to [1 1 1 1],
then [x₁ x₂ x₃ x₄]·[1 1 1 1] = 0. As Char. Limit points out, that gives you one equation (in 4 unknowns), in this case: x₁+x₂+x₃+x₄=0 .

Do the same for the for the other two, ([1 2 3 4] and [1 9 9 7])Then use elimination, or some other method to get one equation with two unknowns. You are free pick one component, but choosing it wisely will make the other components work out nicely.

By the way, the 3 given vectors are not mutually perpendicular.

 

[SrEstroncio]

As other people pointed out, do the dot product for all three vectors and you'll get a system of 4 unknowns and 3 equations whose answer is most likely a 4-3=1-dimensional subspace of R^4. The answer should be a line.