Question

Advance mathematics problem. Help needed. Question is attached.!? Thanks :)

Answer 1

See below

Explanation:

The pipeline parameterizes as:

`mathbf l(lambda) = ((0),(15),(40)) + lambda ((20),(5),(-10))`

The distance, `D(lambda)` between any point on that line and the origin is:

`D^2 = (0 + 20 lambda)^2 + (15 + 5 lambda)^2 + (40 - 10 lambda)^2 = 25 (21 lambda^2 - 26 lambda + 73) qquad square`

Differentiating `D^2(lambda)` wrt `lambda` in order to optimise:

`D_lambda^2 = 25 (42 lambda - 26) color(red)(= 0)`

`implies lambda = 13/21`

Plug that into `square`:

`D^2 = 34100/21 qquad implies D = 10 sqrt(341/21)`

The nearest point is:

• `mathbf l(13/21) = 10/21 ((26),(38),(71))`

We can incidentally verify that:

`((26),(38),(71)) cdot ((20),(5),(-10)) = 0`

This is the dot product of (1) the vector from O to the nearest point on the pipe and (2) the vector that runs along the pipe. These vectors should be, and are, orthogonal.

For the spherical coordinates, it is simply a matter of putting the results into these relationships:

• `r =\sqrt {x^{2}+y^{2}+z^{2}}`

Polar Angle:

• `theta = arccos (z/sqrt (x^2+y^2+z^2) ) = arccos (z/r)`

Azimuth:

• `varphi =arctan (y/x)`

[This is typically all done without the use of calculus but it is posted in this section.]