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Engineering Calculus Notes 391

# Engineering Calculus Notes 391 - 379 3.9 THE PRINCIPAL AXIS...

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Unformatted text preview: 379 3.9. THE PRINCIPAL AXIS THEOREM which amounts to v1 + v2 + v3 = 0. This deﬁnes a plane of solutions. If we set v3 = 0, we get 1 → − = √ (1, −1, 0). u2 2 → This is automatically perpendicular to − 1 (but check that it is!). We u → → need a third eigenvector perpendicular to both − 1 and − 2 . We can u u take their cross-product, which is → − =− ×− u 3 →1 →2 u u →→→ −−− ı k 1 =√ 1 1 1 6 1 −1 0 → − 1− − ı = √ (→ + → − 2 k ) 6 1 = √ (1, 1, −2). 6 → You should check that − 3 is an eigenvector with λ3 = λ2 = −1. u The weighted-sums expression for Q, then, is Q(x, y, z ) = 2 = x+y+z √ 3 2 − x−y √ 2 2 − x + y − 2z √ 6 2 1 1 2 (x + y + z )2 − (x − y )2 − (x + y − 2z )2 . 3 2 6 2. As another example, the form Q(x, y, z ) = 4x2 − y 2 − z 2 − 4xy + 4xz − 6yz has matrix representative [Q] = 4 −2 2 −2 −1 −3 2 −3 −1 ...
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