I understand there are questions on the Math Exchange already, but upon analyzing them I have still not figured out how to solve my own problem.

My problem is to find all vectors that are perpendicular to the vector $(1, -2, 5)$, have the y-components be equal to 3 times the x-components, and have a length of 5.

I know that the first step is to set the dot product of my vector and another vector equal to zero. And the result is $i-2j+5k=0$, correct?

After that my method falls apart as I am trying to find ALL vectors perpendicular to my vector. Help would be appreciated!

  • $\begingroup$ Your first equation is correct. Any vector satisfying that will be perpendicular to your given vector. Now create equations for the other conditions and solve. $\endgroup$ Oct 23, 2019 at 21:54
  • $\begingroup$ Despite the title you’ve given this question, the problem doesn’t want you to find a unit vector. $\endgroup$
    – amd
    Oct 23, 2019 at 22:31
  • $\begingroup$ You do realize that there’s an infinite number of vectors perpendicular to $(1,-2,5)$, yes? It looks like you’ve picked out just a single solution to the first equation that you formulated. The probability is zero that it happened to be the right one. $\endgroup$
    – amd
    Oct 23, 2019 at 22:32
  • $\begingroup$ @amd I believe my answer below is sufficient, thanks! $\endgroup$ Oct 23, 2019 at 22:36

1 Answer 1


A hint. The thing is, you need all prependicular vectors.

Start with all vectors possible, let's designate them $(x;y;z)$. Now choose all vectors that are perpendicular to your vector $0 = (1; -2; 5) \cdot (x;y;z) = x -2y + 5z$. This gives you one equation. The other two are $y = 3x$ and $5^2 = x^2 + y^2 + z^2$. Solve the equations and you'll get the result.

  • $\begingroup$ Would solving this require a system of equations? $\endgroup$ Oct 23, 2019 at 22:06
  • $\begingroup$ @ThatOneNerdyBoy , yes. We can call 3 equations a system. But in this case the system is quite simple. I can add a full solution for you if you feel need. $\endgroup$
    – guest
    Oct 23, 2019 at 22:15
  • $\begingroup$ I solved it and got x = sqrt(25/11), y = 3sqrt(25/11), and z = sqrt(25/11). Does that seem right? $\endgroup$ Oct 23, 2019 at 22:24
  • 2
    $\begingroup$ @ThatOneNerdyBoy You should have at least two solutions. $\endgroup$
    – amd
    Oct 23, 2019 at 22:26
  • $\begingroup$ @ThatOneNerdyBoy , looks good for me. But there is one more solution to this system of equations. Try vector directed just opposite to the one you found. $\endgroup$
    – guest
    Oct 23, 2019 at 22:32

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