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Convert the vector P to Cartesian coordinates where P = r ar + cos⁡θ aφ.

Posted: Thu Jul 14, 2022 9:41 am
by answerhappygod
a) \(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{xyz}{\sqrt{x^2+y^2 }})az + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{xyz}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} ax] \)
b) \(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{yz}{\sqrt{x^2+y^2 }})ax + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{xz}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} az] \)
c) \(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{y}{\sqrt{x^2+y^2 }})ax + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{z}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} az] \)
d) \(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{y}{\sqrt{x^2+y^2 }})ax + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{x}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} az] \)