**Unformatted text preview: **Example 10. Find the Cartesian form of the plane x = 1 + A 1 + ,u 0 for A, p: E R. 2 1 2
371
SOLUTION. Let x = 3:2 . By comparing the components, we get the following parametric
333 equations
$1=1+A—n, $2=1+A and $3=2+A+2n. Hence from the second equation, we get A = 3:2 — 1. Substitute this into the ﬁrst equation, we have
$1 = 1 —|— (3:2 — 1) — ,u and that is ,u; = 332 — 391. Now substitute A and n which are in terms of 3:1
and 2:2 into the third parametric equation. The Cartesian form of the plane is, therefore, m3=2+(m2—1)+2($2—m1) or 2391—3m3+$3=1. 1.6 Vectors, Matrices / Arrays and MATLAB Tn +1‘1‘:C’1 nnn‘l‘lnn 11TH Trr'l-l-l nlnnfr'r larwr'r +f‘l I‘WT‘IT‘lfq-ln Trnn‘l‘n‘wc n‘nrq mn‘l‘v‘lnnn fnvvn‘trﬂ '1“ “1'1— fir-PT hp Vﬁ1'l ﬂ-l'lﬁ'l1llq ...

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- Fall '08
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- Linear Algebra, Algebra, Equations