0
$\begingroup$

The question is in the following picture:

enter image description here

The answer is C.

I know the method of finding all vectors orthogonal to a vector,but it will take a long time if I applied for 2 vectors, as I must answer the question in only 2.5 minuites. this method if found here Finding all vectors orthogonal to a vector, could anyone give me a quicker method please?

$\endgroup$
1
  • $\begingroup$ This is a multiple choice question, so you don't need to compute the solution, as the solution is given. All you have do do is to eliminate the bogus solutions. $\endgroup$
    – Angina Seng
    Jul 23, 2017 at 4:47

2 Answers 2

Reset to default
3
$\begingroup$

The two given vectors are linearly independent, so they span a subspace of dimension $2$. Therefore their orthogonal complement has dimension $4-2=2$, and so each of its bases has two elements. That eliminates A, D and E. The first vector in B is $(1,0,0,0)$ which is not orthogonal to $(1,1,1,0)$. That eliminates B; only C remains.

$\endgroup$
1
  • $\begingroup$ I think the first vector in B is not orthogonal to the other vector not the one that you wrote. $\endgroup$
    – Emptymind
    Jul 23, 2017 at 9:14
1
$\begingroup$

If we define $V=\mathrm{span}\{(0,1,1,1)^T,(1,1,1,0)^T\}$, then we're looking for a basis for $V^{\perp}$.

Since $\dim V+\dim V^{\perp}=4$ and $\dim V=2$, it follows that $\dim V^{\perp}=2$. Hence there should be two vectors in a basis for $V^{\perp}$, which rules out (A),(D), and (E). In addition, the vectors in the basis for $V^{\perp}$ should actually be orthogonal to $V$, which rules out (B).

So that leaves (C) as the correct answer, and it is moreover easy to verify that the two vectors given in answer (C) are linearly independent and are orthogonal to $V$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .