Sandra Allison
2021-12-16
Bob Huerta
Beginner2021-12-17Added 41 answers
The answer is:
x=3+14t
y=11+14t
z=11+6t
The point
Now let's search the generic vector tangent to the curve:
x'=14
y'=14
So, for
So, remembering that given a point
y=y_p+bt
z=z_p+ct
so the tangent is
x=3+14t
y=11+14t
z=11+6t
N.B. The direction (14,14,6) is the same that
x=3+7t
y=11+7t
z=11+3t
That's simpler!
Mary Goodson
Beginner2021-12-18Added 37 answers
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