# Algebre lineaire by Florent.

By Florent.

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Of course, planes can be described in different ways. 14 Chapter 1 Vectors and Matrices EXAMPLE 9 We wish to ﬁnd a parametric equation of the plane that contains the points P = (1, 2, 1) −→ and Q = (2, 4, 0) and is parallel to the vector (1, 1, 3). We take x0 = (1, 2, 1), u = P Q = (1, 2, −1), and v = (1, 1, 3), so the plane consists of all points of the form x = (1, 2, 1) + s(1, 2, −1) + t (1, 1, 3), s, t ∈ R. Finally, note that three noncollinear points P , Q, R ∈ R3 determine a plane. To get a −→ −→ −→ parametric equation of this plane, we simply take x0 = OP , u = P Q, and v = P R.

If c1 , . . , ck ∈ R, the vector v = c1 v1 + c2 v2 + · · · + ck vk is called a linear combination of v1 , . . , vk . 15 12 Chapter 1 Vectors and Matrices Definition. Let v1 , . . , vk ∈ Rn . The set of all linear combinations of v1 , . . , vk is called their span, denoted Span (v1 , . . , vk ). That is, Span (v1 , . . , vk ) = {v ∈ Rn : v = c1 v1 + c2 v2 + · · · + ck vk for some scalars c1 , . . , ck }. In terms of our new language, then, the span of two nonparallel vectors u, v ∈ Rn is a plane through the origin.

16 Chapter 1 Vectors and Matrices 6. Find a parametric equation of each of the following lines: a. 3x1 + 4x2 = 6 ∗ b. the line with slope 1/3 that passes through A = (−1, 2) c. the line with slope 2/5 that passes through A = (3, 1) d. the line through A = (−2, 1) parallel to x = (1, 4) + t (3, 5) e. the line through A = (−2, 1) perpendicular to x = (1, 4) + t (3, 5) ∗ f. the line through A = (1, 2, 1) and B = (2, 1, 0) g. the line through A = (1, −2, 1) and B = (2, 1, −1) ∗ h. the line through (1, 1, 0, −1) parallel to x = (2 + t, 1 − 2t, 3t, 4 − t) 7.