Question

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

Answer 1

Compare the equation of the given planes with the general form: ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ respectively. Find all the corresponding coefficients of x, y , z and the constant term. Now, use the formula for the angle between the two planes given as: $\cos \theta =\left| \dfrac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}+{{c}_{1}}^{2}}.\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}+{{c}_{2}}^{2}}} \right|$, where $\theta$ is the angle between the two planes. Substitute all the values in the formula to get the answer.

Complete step by step solution:
Here, we have been provided with two planes: $x+2y-z+1=0$ and $x-y+3z+4=0$ and we are asked to find the angle between the two planes.
Now, the angle between the two planes in their general form ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ is defined as the angle between their normal vectors ${{\vec{n}}_{1}}$ and ${{\vec{n}}_{2}}$ respectively. For the given general equation of the two planes, these normal vectors are given as: ${{\vec{n}}_{1}}={{a}_{1}}\hat{i}+{{b}_{1}}\hat{j}+{{c}_{1}}\hat{k}$ and ${{\vec{n}}_{1}}={{a}_{2}}\hat{i}+{{b}_{2}}\hat{j}+{{c}_{2}}\hat{k}$. Now, considering the angle between the two planes as $\theta$ we have the relation: $\cos \theta =\left| \dfrac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}+{{c}_{1}}^{2}}.\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}+{{c}_{2}}^{2}}} \right|$ to find the angle.
Now, comparing the equation of the given planes with their general form, we have,$$$$
$\Rightarrow {{a}_{1}}=1,{{b}_{1}}=2,{{c}_{1}}=-1$ and ${{a}_{2}}=1,{{b}_{2}}=-1,{{c}_{2}}=3$
Substituting all the values in the above mentioned formula, we get,
$\begin{aligned} & \Rightarrow \cos \theta =\left| \dfrac{\left( 1\times 1 \right)+\left( 2\times \left( -1 \right) \right)+\left( \left( -1 \right)\times 3 \right)}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{\left( -1 \right)}^{2}}}.\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}+{{3}^{2}}}} \right| \\\ & \Rightarrow \cos \theta =\left| \dfrac{1-2-3}{\sqrt{6}.\sqrt{11}} \right| \\\ & \Rightarrow \cos \theta =\left| \dfrac{-4}{\sqrt{6}.\sqrt{11}} \right| \\\ & \Rightarrow \cos \theta =\dfrac{4}{\sqrt{6}.\sqrt{11}} \\\ & \Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{4}{\sqrt{6}.\sqrt{11}} \right) \\\ \end{aligned}$
Hence, the above relation is required answer

Note: One may note that we cannot find the value of the angle $\theta$ obtained above without using the cosine table or calculator. Note that here the constant terms ${{d}_{1}}$ and ${{d}_{2}}$ will have no effect on the value of the angle $\theta$. You must remember the formula $\cos \theta =\left| \dfrac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}+{{c}_{1}}^{2}}.\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}+{{c}_{2}}^{2}}} \right|$ to solve the question in less time. You can easily predict that this formula is derived using the dot product formula.