Question

Write the direction cosines of the normal to plane $3x+4y+12z=52$.

Answer 1

Hint: We can write the equation of the plane which Is normal to $ax+by+cz=d$ as $\bar{r}.\bar{n}=d$ where $\bar{n}=a\hat{i}+b\hat{j}+c\hat{k}$. Now the direction cosines of the normal plane can be calculated by calculating the unit vector of $\bar{n}$. The direction cosines are equal to the coefficient of the unit vector $\hat{n}$.

Complete step by step answer:
Given plane, $3x+4y+12z=52$
Comparing it with $ax+by+cz=d$ then $a=3$, $b=4$, $c=12$.
Let the equation of the plane which is normal to $3x+4y+12z=52$ can be calculated by
$\bar{r}.\bar{n}=d$ where $\bar{n}=a\hat{i}+b\hat{j}+c\hat{k}$. Substituting the values of $a$, $b$, $c$ in the above equation, then we will get the equation of the plane as $\Rightarrow \bar{r}.\left( 3\hat{i}+4\hat{j}+12\hat{k} \right)=52$ and $\bar{n}=3\hat{i}+4\hat{j}+12\hat{k}$
Now the magnitude the vector $\bar{n}$ is
$\begin{align}
  & \left| {\bar{n}} \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \\
 & \Rightarrow \left| {\bar{n}} \right|=\sqrt{{{3}^{2}}+{{4}^{2}}+{{12}^{2}}} \\
 & \Rightarrow \left| {\bar{n}} \right|=\sqrt{9+16+144} \\
 & \Rightarrow \left| {\bar{n}} \right|=\sqrt{169} \\
 & \Rightarrow \left| {\bar{n}} \right|=13 \\
\end{align}$
Now the unit vector of the vector $\bar{n}$ is given by
$\begin{align}
  & \hat{n}=\dfrac{{\bar{n}}}{\left| {\bar{n}} \right|} \\
 & \Rightarrow \hat{n}=\dfrac{\left( 3\hat{i}+4\hat{j}+12\hat{k} \right)}{13} \\
 & \Rightarrow \hat{n}=\left( \dfrac{3}{13} \right)\hat{i}+\left( \dfrac{4}{13} \right)\hat{j}+\left( \dfrac{12}{13} \right)\hat{k} \\
\end{align}$
Now the coefficients of the unit vector $\hat{n}$ are $\left( \dfrac{3}{13},\dfrac{4}{13},\dfrac{12}{13} \right)$

$\therefore $The direction cosines are $\left( \dfrac{3}{13},\dfrac{4}{13},\dfrac{12}{13} \right)$.

Note: We will also solve this problem in another method. In this method we will divide the equation of the plane $ax+by+cz=d$ with $\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$, then the equation of the plane converts into form $lx+my+nz=p$. Now the values of direction cosines are $\left( l,m,n \right)$.
Given plane, $3x+4y+12z=52$
Comparing it with $ax+by+cz=d$ then $a=3$, $b=4$, $c=12$.
Dividing the plane equation with $\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}=\sqrt{{{3}^{2}}+{{4}^{2}}+{{12}^{2}}}=13$, then we will get
$\begin{align}
  & \dfrac{3x+4y+12z}{13}=\dfrac{52}{13} \\
 & \Rightarrow \left( \dfrac{3}{13} \right)x+\left( \dfrac{4}{13} \right)y+\left( \dfrac{12}{13} \right)z=4 \\
\end{align}$
Comparing the above equation with $lx+my+nz=p$, then the values of $l$,$m$,$n$ are
$l=\dfrac{3}{13}$, $m=\dfrac{4}{13}$, $n=\dfrac{12}{13}$.
$\therefore $ The direction cosines are $\left( l,m,n \right)=\left( \dfrac{3}{13},\dfrac{4}{13},\dfrac{12}{13} \right)$
From both the methods we got the same result.