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# If x = r sinA cosC, y= r sinA sinC and z = r cosA , then prove that ${{r}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}$.  Answer Verified
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
\begin{align} & {{x}^{2}}+{{y}^{2}}={{r}^{2}}{{\sin }^{2}}A\left( 1 \right) \\ & \Rightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}}{{\sin }^{2}}A\text{ (iii)} \\ \end{align}
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
\begin{align} & {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}}\left( 1 \right) \\ & \Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}} \\ \end{align}
Hence proved.

Note:  If $\overset{\to }{\mathop{v}}\,$ is any vector in 3D-space of magnitude r making angle C and A with x and z-axis respectively 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
\begin{align} & {{r}^{2}}={{\left( \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}} \right)}^{2}} \\ & \Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}} \\ \end{align}
Hence proved.
 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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