# If x = r sinA cosC, y= r sinA sinC and z = r cosA , then prove that ${{r}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}$.

Last updated date: 25th Mar 2023

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Answer

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**Hint:**Evaluate the sum ${{x}^{2}}+{{y}^{2}}$ first. Then add it to ${{z}^{2}}$ to get the result. Use the trigonometric identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$.

**Complete step-by-step answer:**

We have x = r sinA cosC

Squaring both sides, we get

${{x}^{2}}={{(r\sin A\cos C)}^{2}}$

We know that ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$

Using the above formula, we get

\[{{x}^{2}}={{r}^{2}}{{\sin }^{2}}A{{\cos }^{2}}C\text{ (i)}\]

y= r sinA sinC

Squaring both sides, we get

${{y}^{2}}={{\left( r\sin A\sin C \right)}^{2}}$

We know that ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$

Using the above formula, we get

\[{{y}^{2}}={{r}^{2}}{{\sin }^{2}}A{{\sin }^{2}}C\text{ (ii)}\]

Adding equation (i) and equation (ii), we get

${{x}^{2}}+{{y}^{2}}={{r}^{2}}{{\sin }^{2}}A{{\cos }^{2}}C+{{r}^{2}}{{\sin }^{2}}A{{\sin }^{2}}C$

Taking \[{{r}^{2}}{{\sin }^{2}}A\] common, we get

${{x}^{2}}+{{y}^{2}}={{r}^{2}}{{\sin }^{2}}A\left( {{\cos }^{2}}C+{{\sin }^{2}}C \right)$

We know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$.

Using the above formula, we get

$ {{x}^{2}}+{{y}^{2}}={{r}^{2}}{{\sin }^{2}}A\left( 1 \right) $

$\Rightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}}{{\sin }^{2}}A\text{ (iii)} \\ $

Also, we know have z = r cosA

Squaring both sides, we get

${{z}^{2}}={{\left( r\cos A \right)}^{2}}$

We know that ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$

Using the above formula, we get

${{z}^{2}}={{r}^{2}}{{\cos }^{2}}A\text{ (iv)}$

Adding equation (iii) and equation (iv), we get

${{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}}{{\sin }^{2}}A+{{r}^{2}}{{\cos }^{2}}A$

Taking ${{r}^{2}}$ common in RHS, we get

${{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}}\left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)$

We know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$.

Using the above formula, we get

$ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}}\left( 1 \right) $

$ \Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}} $

Hence proved.

**Note:**[1] If $\overset{\to }{\mathop{v}}\,$ is any vector in 3D-space of magnitude r making an angle C with z-axis, and projection of the vector v in x-y plane making an angle A with x-axis, then we have $\overrightarrow{v}=r\sin A\cos C\widehat{i}+r\sin A\sin C\widehat{j}+r\cos A\widehat{k}=x\widehat{i}+y\widehat{j}+z\widehat{k}$

Taking modulus on both sides we get

\[\left| \overrightarrow{v} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}\]

But magnitude of |v| = r.

Substituting the value of |v| we get

$r=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$

Squaring both sides, we get

$ {{r}^{2}}={{\left( \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}} \right)}^{2}} $

$ \Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}} $

Hence proved.

[2] Here x, y and z can be considered as the x-coordinate, y-coordinate and the z-coordinate of a point P and r as the distance of P from origin in 3-D plane.

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