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. $\endgroup$ – Don Thousand Oct 23 '19 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 '19 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 '19 at 22:32
  • $\begingroup$ @amd I believe my answer below is sufficient, thanks! $\endgroup$ – ThatOneNerdyBoy Oct 23 '19 at 22:36

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.

| cite | improve this answer | |
  • $\begingroup$ Would solving this require a system of equations? $\endgroup$ – ThatOneNerdyBoy Oct 23 '19 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 '19 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? $\endgroup$ – ThatOneNerdyBoy Oct 23 '19 at 22:24
  • 2
    $\begingroup$ @ThatOneNerdyBoy You should have at least two solutions. $\endgroup$ – amd Oct 23 '19 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 '19 at 22:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.