Jensen Mclean

2022-10-06

Geometric probability (with regards to span)

Suppose we have two non-parallel vectors in ${\mathbb{R}}^{3}$.

Now, if we were to randomly select another vector in ${\mathbb{R}}^{3}$, what is the probability that that new vector lies in the span of the first two vectors?

Suppose we have two non-parallel vectors in ${\mathbb{R}}^{3}$.

Now, if we were to randomly select another vector in ${\mathbb{R}}^{3}$, what is the probability that that new vector lies in the span of the first two vectors?

smh3402en

Beginner2022-10-07Added 11 answers

Step 1

One way to model this, so everything becomes finite is to work on the unit sphere ${S}^{2}$. We may assume that the hyperplane is $H=\{z=0\}\subset {\mathbb{R}}^{3}$.

Step 2

The probability that a random vector of ${S}^{2}$ doesn't lie on the great circle $H\cap {S}^{2}$ would be exactly $\frac{|{S}^{2}\setminus (H\cap {S}^{2})|}{|{S}^{2}|}=1.$.

The only problem here is that we exclude the zero vector, but I assume that this is not a big deal.

One way to model this, so everything becomes finite is to work on the unit sphere ${S}^{2}$. We may assume that the hyperplane is $H=\{z=0\}\subset {\mathbb{R}}^{3}$.

Step 2

The probability that a random vector of ${S}^{2}$ doesn't lie on the great circle $H\cap {S}^{2}$ would be exactly $\frac{|{S}^{2}\setminus (H\cap {S}^{2})|}{|{S}^{2}|}=1.$.

The only problem here is that we exclude the zero vector, but I assume that this is not a big deal.

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