Since p is on the plane d 6 1 3 1 2 1 11 in conclusion

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Unformatted text preview: 1 2 0 � = 6ˆ + 3ˆ + 2k. ı j � � � −1 0 3 � → − � 1� 2 →� 1 �− 1√ 7 �− Then area(Δ) = �P Q × P R� = 6 + 32 + 22 = 49 = . 2 2 2 2 → − − →− → − b) A normal to the plane is given by N = P Q × P R = �6, 3, 2�. Hence the equation has the form 6x + 3y + 2z = d. Since P is on the plane d = 6 · 1 + 3 · 1 + 2 · 1 = 11. In conclusion the equation of the plane is 6x + 3y + 2z = 11. → − c) The line is parallel to �2 − 1, 2 − 2, 0 − 3� = �1, 0, −3�. Since N · �1, 0...
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This note was uploaded on 01/21/2014 for the course MATH 18.02C taught by Professor Denisauroux during the Fall '12 term at MIT.

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