Explanation: The Cylindrical coordinates is of the form ( ρ, φ, z) where \(ρ=\sqrt{x^2+y^2} \) and \( φ=tan^{-1}⁡(\frac{y}{x}) \) and z=z where (x, y, z) is the Cartesian coordinates. The Spherical coordinates is of the form (r, θ, φ) where \(r=\sqrt{(a^2+b^2 +c^2 )}, \, θ = tan^{-1}⁡(\frac{\sqrt{x^2+y^2}}{z}) \) and \( φ=tan^{-1}⁡(\frac{y}{x})\). Now, substituting the values for x as 2, y as 6 and z as 9, we get the answer as (6.32, 71.565., 9) and (11, 35.097., 71.565.).

z = 10 cos90 = 0.

Substituting the column matrix in right hand side by the given the vector, and solving the matrix we get a vector in cylindrical coordinates. Now change the point P which is in Cartesian coordinates to Cylindrical coordinates. Now, we should substitute the point P in X thus finding the value of X at P. Hence we get the value 100 ax – 398 ay + 108 az.

\(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{yz}{\sqrt{x^2+y^2 }})ax + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{xz}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} az] \).

Explanation: First convert each point which is in cylindrical coordinates to Cartesian coordinates. Now using the formula, distance \( = \sqrt{(x-x’)^2+(y-y’)^2 +(z-z’)^2} \), and substituting the values of x, y, and z in it, we get the required answer as 6.19 units. This sum can also be solved using a direct formula to find distance using two points in Cylindrical coordinates.

\end{bmatrix} \), we get cos⁡θ.

U = zρ2sin⁡φ cos⁡φ + ρsin⁡φ + zρcos⁡φ.

Explanation: We first have to convert the given points to Cartesian coordinates from Spherical coordinates. Then we have to apply the formula for distance between two points in Cartesian coordinates and find the distance.

\end{bmatrix} \), and doing dot product we get, sin⁡θ cos⁡φ.