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Determine the point at which the function r(t) =
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normal vector of plane is (4,1,-2)
direction vector of tangent to r(t) is <t^2, 4, 2t> ?
\[\large \dfrac{t^2}{4} = \dfrac{4}{1} = \dfrac{2t}{-2}\] solve \(\large t\)
How do we solve for t? By putting in (4,1,-2)? @ganeshie8
take the last two ratios : \[\large \dfrac{4}{1} = \dfrac{2t}{-2}\] cross multiply and solve \(\large t\)
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t = -4?
yes! plug that in first two rations and see if it satisfies
The first two rations? You mean ratios?
yes
Okay I did the math and yes it does. :)
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Taking care of neighbors kid..sorry :/
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