/sci/ - Science & Math

>>14881527
We choose [math]t = 0, 1[/math] to obtain the points in the line [math](1, 5, 0)[/math] and [math](3, 4, 1)[/math]. So any point in the plane can be expressed as [math](0, 0, 1) + u ((1, 5, 0) - (0, 0, 1)) + v((3, 4, 1) - (0, 0, 1)) = (0, 0, 1) + u (1, 5, -1) + v (3, 4, 0)[/math]. The normal vector to the plane is orthogonal to both [math](1, 5, -1)[/math] and [math](3, 4, 0)[/math]. Then you can take the difference between [math](1, 2, 3)[/math] and any point in the plane (i.e. (0, 0, 1)), project the difference onto the normal vector and compute the norm of the result to get the distance.

I don't remember if there was a simpler way to do this.

>/sci/ meta thread
>turns into /pol/ this /pol/ that
Rent free.

>>11519587
>Are they frauds
They're Bogdanot.

>>11484795
Let [math]V[/math] be a vector space over a field [math]K[/math]. We say that a set [math]L[/math] is linearly independent if, for any [math]\{ l_1, \cdots , l_n \} \subseteq L[/math] and [math]\{ k_1 , \cdots , k_n \} \subseteq K[/math], [math]\Sigma_{i=1}^n k_il_i = 0[/math] imples that [math]k_i=0[/math] for every [math]i[/math].
We call a maximal linearly independent subset of [math]V[/math] a basis.
Theorem One: For any basis [math]B[/math], any element [math]v \in V[/math] can be written as a finite linear combination of elements of the basis.
Proof: Assume [math]v[/math] cannot be written as a finite linear combination of elements of [math]B[/math]. We will show that [math]B \cup \{ v \}[/math] is a linearly independent set.
Assume that, for some [math]\{ b_1 , \cdots , b_n \} \subseteq B \cup \{ v \}[/math] and some [math]\{ k_1 , \cdots , k_n \}[/math], not all [math]k_i =0[/math], we have [math]\Sigma_{i=1}^n k_ib_i = 0[/math]. If [math]v = b_j[/math] with [math]k_j \neq 0[/math] for some [math]j[/math], we then have [math]\Sigma_{i=1, ~ i \neq j}^n k_i b_i = -k_j v[/math], and [math]v =[ \Sigma_{i=1, ~ i \neq j}^n k_i b_i ]/k_j[/math], and thus [math]v[/math] can actually be written as a finite linear combination of elements of [math]B[/math], which violates the hypothesis.
Assume none of the [math]b_j=0[/math] or that [math]b_j = v[/math] and [math]k_j = 0[/math]. Then, by removing [math]v[/math] from the sum if necessary (since it doesn't actually contribute to it), we have [math]\Sigma_{i=1}^n k_ib_i = 0[/math] with all the [math]b_i \in B[/math], and thus, by the linear independence of [math]B[/math], all [math]k_i[/math] equal zero.
Theorem two: For a vector space [math]V[/math] and a basis [math]B[/math], any element [math]v \in V[/math] can be written as a unique finite linear combination of elements of [math]B[/math].