Question

How do I find the angle between the planes $x+2y-z+1=0$ and $x-y+3z+4=0$?

Answer 1

Hint: We start solving the problem by assuming the variable for the angle between the given two planes. We then make use of the fact that the angle between the planes ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ is ${{\cos }^{-1}}\left( \dfrac{{{a}_{1}}.{{a}_{2}}+{{b}_{1}}.{{b}_{2}}+{{c}_{1}}.{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \right)$ to proceed through the problem. We then make the necessary calculations and make use of the result that $\sqrt{a}\sqrt{b}=\sqrt{ab}$ to get the required answer for the given problem.

Complete step by step answer:
According to the problem, we are asked to find the angle between the planes $x+2y-z+1=0$ and $x-y+3z+4=0$.
Let us assume the angle between the planes is $\theta $.
We know that the angle between the planes ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ is ${{\cos }^{-1}}\left( \dfrac{{{a}_{1}}.{{a}_{2}}+{{b}_{1}}.{{b}_{2}}+{{c}_{1}}.{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \right)$. Let us use this result to find the angle between the two given planes.
So, we have $\theta ={{\cos }^{-1}}\left( \dfrac{\left( 1\times 1 \right)+\left( 2\times -1 \right)+\left( -1\times 3 \right)}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{\left( -1 \right)}^{2}}}\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}+{{3}^{2}}}} \right)$.
$\Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{1-2-3}{\sqrt{1+4+1}\sqrt{1+1+9}} \right)$.
$\Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{-4}{\sqrt{6}\sqrt{11}} \right)$ ---(1).
We know that $\sqrt{a}\sqrt{b}=\sqrt{ab}$. Let us use this result in equation (1).
$\Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{-4}{\sqrt{66}} \right)$.
So, we have found the angle between the given two planes $x+2y-z+1=0$ and $x-y+3z+4=0$ as ${{\cos }^{-1}}\left( \dfrac{-4}{\sqrt{66}} \right)$.

$\therefore $ The angle between the given two planes $x+2y-z+1=0$ and $x-y+3z+4=0$ is ${{\cos }^{-1}}\left( \dfrac{-4}{\sqrt{66}} \right)$.

Note: We can also solve the given problem by first finding the normal vectors of the given planes and then finding the angle between the normal vectors by making use of the fact that the angle between the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ as $\dfrac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|}$. We should keep in mind that the angle between any two planes lies between ${{0}^{\circ }}$ and ${{180}^{\circ }}$. Similarly, we can expect problems to find the line of intersection of the planes $x+2y-z+1=0$ and $x-y+3z+4=0$.