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# I really need help with this calculus related question?

### 1 Answer

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- rotchmLv 72 weeks 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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