Question

Find the angle at which the normal vector to the plane $4x+8y+z=5$ is inclined to the coordinate axes.

Answer 1

Hint: The direction ratios of the normal to the plane $ax+by+cz=d$ are $a,b,c$. If $l,m,n$ are the direction cosines of the normal to the a plane then $l=\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$, $m=\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$ and $n=\dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$. Also if the normal to the plane makes an angle $\alpha $ with positive $x$- axis $\beta $ with positive $y$- axis and $\gamma $ with positive $z$- axis then the relation between direction cosines of normal to the plane and the angles made with the coordinate axes are $l=\cos \alpha $, $m=\cos \beta $ and $n=\cos \gamma $. In this question first find the direction ratios of the plane and use this result to solve it.

Complete step-by-step answer:
We have to find the angle at which the normal vector to the plane $4x+8y+z=5$ is inclined to the coordinate axes.
We know that the direction ratios of the normal to the plane $ax+by+cz=d$ are $a,b,c$.
So comparing the given plane $4x+8y+z=5$ with the general plane we get $a=4,b=8,c=1$.
Therefore the direction cosines of the normal to the plane are $4,8,1$.
Also we know that if $a,b,c$ are the direction ratios of the normal to the plane the direction cosines of the normal to plane are $l,m,n$ where $l=\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$, $m=\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$ and $n=\dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$.
Here \[\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}=\sqrt{{{4}^{2}}+{{8}^{2}}+{{1}^{2}}}\].
Calculating further we get \[\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}=\sqrt{16+64+1}=\sqrt{81}=9\].
Therefore $l=\dfrac{4}{9}$, $m=\dfrac{8}{9}$ and $n=\dfrac{1}{9}$.
Hence the direction cosines of the normal to the vector are $\dfrac{4}{9},\dfrac{8}{9},\dfrac{1}{9}$.
We know that if the normal to the plane makes an angle $\alpha $ with positive $x$- axis $\beta $ with positive $y$- axis and $\gamma $ with positive $z$- axis then $l=\cos \alpha $, $m=\cos \beta $ and $n=\cos \gamma $.
So using the formula $l=\cos \alpha $ we get $\dfrac{4}{9}=\cos \alpha $.
Using the formula $m=\cos \beta $ we get $\dfrac{8}{9}=\cos \beta $.
And using the formula $n=\cos \gamma $ we get $\dfrac{1}{9}=\cos \gamma $.
Now we know that if $x=\cos \theta $ then $\theta ={{\cos }^{-1}}x$.
Using this identity we get $\alpha ={{\cos }^{-1}}\left( \dfrac{4}{9} \right)$, $\beta ={{\cos }^{-1}}\left( \dfrac{8}{9} \right)$ and $\gamma ={{\cos }^{-1}}\left( \dfrac{1}{9} \right)$.
Hence the normal to the plane $4x+8y+z=5$ makes and angle ${{\cos }^{-1}}\left( \dfrac{4}{9} \right)$ with $x$ - axis ${{\cos }^{-1}}\left( \dfrac{8}{9} \right)$ with $y$- axis and ${{\cos }^{-1}}\left( \dfrac{1}{9} \right)$ with $z$ -axis.
This is the required solution.

Note: In this problem the main key is to convert the direction ratios to direction cosines. So students must be aware of the formulas and concepts. Also students must take care while using trigonometric identities. As there is a direct relation between direction cosines of the line normal to the plane and the angles inclined with coordinate axes so students must convert the direction ratios of the normal to the plane to the direction cosines of the normal to the plane.