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lecture_21

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lecture_21 [2016/04/14 09:52] – old revision restored (2016/04/14 10:50) rupertlecture_21 [2017/04/18 09:40] (current) – rupert
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 hence $\Pi$ has equation $-2x-13y+17z=d$, and subbing in the point $(0,3,2)$ gives $d=0-13(3)+17(2)=-39+34=-5$, so $\Pi$ has equation $-2x-13y+17z=-5$, or hence $\Pi$ has equation $-2x-13y+17z=d$, and subbing in the point $(0,3,2)$ gives $d=0-13(3)+17(2)=-39+34=-5$, so $\Pi$ has equation $-2x-13y+17z=-5$, or
 \[ 2x+13y-17z=5.\] \[ 2x+13y-17z=5.\]
 +
lecture_21.1460627546.txt.gz · Last modified: by rupert
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