Answer
The following answers are given below:-
a. 2√2, π/3, 3π/4
b. 2,3π/4, π/4
c. 6,π/2, π/3
d. 10,π/2, π/2
Work Step by Step
Given:-a.(1, √3, -2)
here x=1, y=√3, z=-2
$\rho$=$\sqrt {1^2+√3^2+(-2)^2}$
=√8=2√2
tan$\theta$=$\frac{y}{x}$
=√3/1
$\theta$=π/3
cos$\phi$=$\frac{z}{\sqrt{x^2+y^2+z^2}}$
=$\frac{-2}{\sqrt{1^2+√3^2+(-2)^2}}$=-1/√2
$\phi$=3π/4
b. (1, -1, √2)
here x=1, y=-1, z=√2
$\rho$=$\sqrt {1^2+-1^2+(√2)^2}$
=√4=2
tan$\theta$=$\frac{y}{x}$
=-1/1=-1
$\theta$=3π/4
cos$\phi$=$\frac{z}{\sqrt{x^2+y^2+z^2}}$
=$\frac{√2}{\sqrt{1^2+(-1)^2+(√2)^2}}$=1/√2
$\phi$=π/4
c. (0, 3√3, 3)
here x=0, y=3√3, z=3
$\rho$=$\sqrt {0^2+3√3^2+3^2}$
=√36=6
tan$\theta$=$\frac{3√3}{0}$
=\infty
$\theta$=π/2
cos$\phi$=$\frac{z}{\sqrt{x^2+y^2+z^2}}$
=$\frac{3}{\sqrt{0^2+3√3^2+3^2}}$=1/2
$\phi$=π/3
d. (-5√3, 5,0)
here x=-5√3, y=5, z=0
$\rho$=$\sqrt {-5√3^2+5^2+0^2}$
=√100=10
tan$\theta$=$\frac{5}{-5√3}$
=-1/√3
$\theta$=π/2
cos$\phi$=$\frac{z}{\sqrt{x^2+y^2+z^2}}$
=$\frac{0}{\sqrt{-5√3^2+5^2+0^2}}$=0
$\phi$=π/2
Ans.
