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Find a vector perpendicular to the plane containing the given points. (1, 2, 3), (-4,...
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You need to remember that the cross product of the vectors formed from the given points represents the orthogonal vector to the plane.
You should form two vector such that:
`bar u = (-4-1)bar i + (2-2) bar j + (-1-3)bar k`
`bar u = -5 bar i - 4 bar k`
`bar v = (5-1)bar i + (-3-2) bar j + (0 - 3)bar k`
`bar v = 4bar i - 5 bar j - 3 bar k`
Using the cross product as normal vector yields:
`bar n = bar u X bar v = [[bar i , bar j , bar k],[-5 , 0 , -4],[4 , -5 , -3]]`
`bar n = 25bar k - 16 bar j - 20bar i - 15 bar j`
`bar n = - 20bar i - 31 bar k + 25 bar k`
Hence, evaluating the perpendicular vector to the plane containing the given points yields `bar n = - 20bar i - 31 bar k + 25 bar k.`
Posted by sciencesolve on July 8, 2012 at 10:50 AM (Answer #1)
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