Answer
The required answer given below:-
a. 8√2, π/4, π/6
b. 2√2, 5π/6, 3π/4
c. 2,0,0
d. 4,π/6,π/6
Work Step by Step
Given:-a.(4, 4, 4√6)
here x=4, y=4, z=4√6
$\rho$=$\sqrt {4^2+4^2+4√6^2}$
=8√2
tan$\theta$=$\frac{4}{4}$
=1
$\theta$=π/4
cos$\phi$=$\frac{z}{\sqrt{x^2+y^2+z^2}}$
=$\frac{4√6}{\sqrt{4^2+4^2+4√6^2}}$=√3/2
$\phi$=π/6
b. (1, -√3, -2)
here x=1, y=-√3, z=-2
$\rho$=$\sqrt {1^2+(-√3)^2+(-2)^2}$
=√8=2√2
tan$\theta$=$\frac{-√3}{1}$
=-√3
$\theta$=5π/6
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
c. (2,0,0)
here x=2, y=0, z=0
$\rho$=$\sqrt {2^2+0^2+0^2}$
=√2^2=2
tan$\theta$=$\frac{0}{2}$
=0
$\theta$=0
cos$\phi$=$\frac{z}{\sqrt{x^2+y^2+z^2}}$
=$\frac{3}{\sqrt{2^2+0^2+0^2}}$=0
$\phi$=0
d. (√3, 1,2√3)
here x=√3, y=1, z=2√3
$\rho$=$\sqrt {-5√3^2+1^2+2√3^2}$
=√16=4
tan$\theta$=$\frac{1}{√3}$
=1/√3
$\theta$=π/6
cos$\phi$=$\frac{z}{\sqrt{x^2+y^2+z^2}}$
=$\frac{2√3}{\sqrt{5√3^2+1^2+2√3^2}}$=√3/2
$\phi$=π/6
Ans.
