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I really need help with this calculus related question?
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- ?Lv 75 days ago
Consider all vectors (a,b,c) perpendicular to L2. Thus,
(1,4,-1) • (a,b,c) = 0.
The point of L1 minus (1,13,-3) is such a sought vector.
That is, (-2+s, 1+4s, 0-s) - (1,13,-3) is your (a,b,c).
Plug, solve for s. You have your point of L1.
Done!
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