Engineering Calculus Notes 111

Engineering Calculus Notes 111 - 1.7 APPLICATIONS OF CROSS...

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Unformatted text preview: 1.7. APPLICATIONS OF CROSS PRODUCTS 99 For example, an equation for the plane P containing △P QR with vertices P (1, −2, 3), Q(−2, 4, −1) and R(5, 3, 1) can be found using → → − = − − 2− + 3− → p0 → ı k → − − − → → → P Q = −3− + 6− − 4 k ı − → − → → → P R = 4− + 5− − 2 k ı − → P→ P→ − = −Q × −R n →→→ −−− ı k = −3 6 −4 4 5 −2 → − −3 6 +k 45 → → − (−12 + 20) − − (6 + 4) + − (−15 − 10) → =ı k → − → → = 8− − 10− − 25 k ı → =− ı 6 −4 5 −2 → −− −3 −4 1 −2 so the equation for P is 8(x − 1) − 10(y + 2) − 25(z − 3) = 0 or 8x − 10y − 25z = −47. As another example, consider the plane P ′ parametrized by x= 3 −2s +t y = −1 +2s −2t z= 3s −t. We can read oﬀ that → − = 3− − − →→ p0 ı → is the position vector of − (0, 0) (corresponding to s = 0, t = 0), and two p vectors parallel to the plane are → → − = −2− + − + 3− →→ vs ı k → − → − = − − 2− − k . → vt → ı ...
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