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$\textbf{Question}$: Find all vectors in $\mathbb{R}^3$ that are orthogonal to both $(1,2,3)$ and $(4,5,6)$.

$\textbf{My Attempt}$: We can write two equations: $$x+2y+3z=0\\4x+5y+6z=0$$ And now we can combine and simplify:$$x+2y+3z=4x+5y+6z\\3x+3y+3z=0\\x+y+z=0$$ Which means that all vectors that lie on the plane $x+y+z=0$ are orthogonal to both of the original vectors, or any vector of the form $(-y-z,y,z)$.

Is this solution correct? I feel like I might be going wrong when I say that every vector in the $x+y+z=0$ plane is orthogonal. Can anyone think of a different method of approaching this problem?

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    $\begingroup$ As a sanity check on your work, observe that the two vectors are linearly independent, so their orthogonal complement must be one-dimensional. The plane $x+y+z=0$ is two-dimensional, however, so that can’t be the correct answer. $\endgroup$ – amd Nov 21 '16 at 1:11
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You started out on the right track by setting up a system of two linear equations, but then you took a bit of a left turn and didn’t actually solve the system. Each equation defines a plane orthogonal to the corresponding vector, and as the two vectors aren’t colinear, the intersection of these planes should be a line, not a plane as in your solution. This is consistent with a simple dimension check: the span of the vectors is two-dimensional, so its orthogonal complement in $\mathbb R^3$ will be one-dimensional. Specifically, your mistake was concluding that all of the solutions of $x+y+z=0$ are also solutions of the original two equations.

Since we’re working in $\mathbb R^3$, a simple way to solve this problem is to take the cross product of the two vectors. Every scalar multiple of this product is orthogonal to both vectors, and these are the only vectors orthogonal to them both.

Working instead with the equations that you set up, we have $$\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix},$$ i.e., we seek the kernel (null space) of the coefficient matrix. Row-reducing this matrix produces $$\begin{bmatrix}1&0&-1\\0&1&2\end{bmatrix},$$ from which we can read directly that the kernel is spanned by $\begin{bmatrix}1&-2&1\end{bmatrix}^T$.

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  • $\begingroup$ I understand how you got the first "1" and "-2" in the vector that spans the kernel, but how did you get the final "1" at the end? $\endgroup$ – riley lyman Nov 21 '16 at 2:51
  • $\begingroup$ @rileylyman The rref tells us that $x-z=0$ and $y+2z=0$. Setting $z=1$ gives us the other two components. Equivalently, you can simply read the basis from the columns of the rref that don’t have pivots. See math.stackexchange.com/a/1521354/265466 for a detailed explanation, but basically you negate the entries of each pivotless column and set the element that corresponds to the column index to $1$. $\endgroup$ – amd Nov 21 '16 at 7:42
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Incorrect. The vector $(1, -1,0)$ is in the plane $x+y+z=0$, but it is not orthogonal to either of the vectors you are given.

The space of vectors orthogonal to two given vectors is one dimensional, not two dimensional.

The vector cross product of the two given vectors is orthogonal to each, and any vector which is orthogonal to each of the two given vectors is a scalar multiple of this cross product.

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  • $\begingroup$ Oh, okay. Thanks for your answer! How did setting those two equations equal to each other lead me down the wrong path? $\endgroup$ – riley lyman Nov 21 '16 at 0:39
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    $\begingroup$ The cross product of $(1,2,3)$ and $(4,5,6)$ is $(-3,+6, -3)$. This is is the plane $x+y+z=0$, but is the only direction in this plane which is orthogonal to the given vectors. $\endgroup$ – DCarter Nov 21 '16 at 0:43

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