Question

prove tha vector equation of the tangent line at p is R-r=λr'

Answer 1

may we know what are the coordinates of the point P?

Answer 2

Hint
if the equation of your linr, is:
\[r = r\left( t \right) = \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\]
then the tangent unit vector T, is:
\[T = \frac{{r'\left( t \right)}}{{\left\| {r'\left( t \right)} \right\|}}\]
where :
\[{\left\| {r'\left( t \right)} \right\|}\]
is the length of \[r\left( t \right)\]

Answer 3

oops... is the length of:
\[{r'\left( t \right)}\]

Answer 4

You don't need the coords.

My problem with it is this - is a proof an appropriate idea. To me it's just obvious that the tangent line is this.

By definition the tangent line touches the curve at R and runs along the tangent vector, r'. What more is there to it?

Answer 5

ok! In general a line which passes at poin R and is parallel to a vector A, has the subsequent equation:
\[X = R + \lambda A\]

Answer 6

where \[\lambda \] is a real parameter

Answer 7

and X is the column vector of the variable x, y and z:
\[X = \left( {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right)\]

Answer 8

so substituting vector T above, we have:
\[X = R + \lambda T = R + \lambda \frac{{r'\left( t \right)}}{{\left\| {r'\left( t \right)} \right\|}}\]

Answer 9

thank you