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I was wondering if it is possible to determine that b does not span a1 a2 since these vectors are R3 and there only being two vectors. With two vectors in R3 doesn't that mean the span is basically a plane on R3? Is it possible to arrive at the conclusion b is not in the span by just looking at the question without solving it?

If I was given three vectors then b would always be in the span a1 a2 a3 since you can manipulate the vectors to cover the entire 3 dimensions, correct?

https://i.stack.imgur.com/FZZRU.jpg

EDIT: I'm more concerned about if my intuition is correct or not than the numerical process to solve the question.

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    You can compute the determinant of the three vectors: they're linearly dependent if and only if their determinant is $0$. – 2017-02-15
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    The two vectors given (if not scalar multiples of each other) span a plane. Their vector product of them gives us a normal vector to the plane. If a vector is in the plane then its scalar product with the said normal vector has to be $0$. – 2017-02-16
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    If you have three *linearly independent* vectors in $\mathbb R^3$, then yes, they would span the entire space. You can’t assume that an arbitrary set of three vectors is linearly independent, though. – 2017-02-16
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    Yea I was assuming three linearly independent vectors in R3, but can I assume that with two linearly independent vectors in R3 that when both are not a scalar of b or vice-versa, b cannot possibly be in the span of those two vectors? Since this means those two vectors only cover a plane in R3 with b not on the plane? I'm trying to better understand the intuition rather than the numerical process. – 2017-02-16

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If the two vectors given are

$$\vec x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} \text{ and } \vec y=\begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}$$ then their vector product, a normal vector to the plane they span is

$$\vec n=\vec x\times \vec y=\begin{bmatrix}x_2y_3-x_3y_2\\-(x_1y_3-x_3x_1)\\x_1y_2-x_2y_1\end{bmatrix}. $$

If

$$\vec b=\begin{bmatrix}b_1\\b_2\\x_3\end{bmatrix}$$

and the question is if $\vec b$ is in the plane then we have to calculate the scalar product

$$(\vec n\cdot \vec b)=b_1(x_2y_3-x_3y_2)-b_2(x_1y_3-x_3x_1)+b_3(x_1y_2-x_2y_1).$$

If $(\vec n\cdot \vec b)=0$ then $\vec b$ is in the said plane because it is perpendicular to $\vec n.$

The other possibility is to calculate the following determinant

$$\begin{vmatrix} b_1&b_2&b_3\\ x_1&x_2&x_3\\ y_í&y_2&y_3 \end{vmatrix}=b_1(x_2y_3-x_3y_2)-b_2(x_1y_3-x_3x_1)+b_3(x_1y_2-x_2y_1).$$

which results $$(\vec x\times \vec y\cdot \vec b).$$

So, the question is if the determinant above is $0$?

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    But intuitively, can I look at two vectors, assuming they both are not a scalar of another vector(b), and determine that vector(b) is not in the span of those two vectors? My thinking is that two linearly independent vectors in R3 span a plane and that if vector(b) is not a scalar of both those vectors than it is not in the span of both those vectors. Sorry my question at the top is worded very confusingly. – 2017-02-16