Engineering Calculus Notes 113

# Engineering Calculus Notes 113 - 1.7 APPLICATIONS OF CROSS...

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Unformatted text preview: 1.7. APPLICATIONS OF CROSS PRODUCTS 101 and hence parallel to their cross-product →→ − =− ×− . v n 1 →2 n Thus, given any one point P0 (x0 , y0 , z0 ) on ℓ (i.e., one solution of the pair of equations) the line ℓ can be parametrized using P0 as a basepoint and →− − = → × − as a direction vector. v n 1 →2 n For example, consider the two planes 3x − 2y + z = 1 2x + y − z = 0. The ﬁrst has normal vector → − = 3− − 2− + − → →→ n1 ı k while the second has → − = 2− + − − − . →→→ n2 ı k Thus, a direction vector for the intersection line is →→ − =− ×− v n 1 →2 n →→→ −−− ı k = 3 −2 1 2 1 −1 −2 1 →3 1 −− 2 −1 1 −1 → − → → = − + 5− + 7 k . ı → =− ı → − 3 −2 +k 21 One point of intersection can be found by adding the equations to eliminate z 5x − y = 1 and, for example, picking x=1 which forces y = 4. ...
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