Question

Find the angle between the planes.
(i) 2x − y + z = 4 and x + y + 2z = 3
(ii) x + y − 2z = 3 and 2x − 2y + z = 5
(iii) x − y + z = 5 and x + 2y + z = 9
(iv) 2x − 3y + 4z = 1 and − x + y = 4
(v) 2x + y − 2z = 5 and 3x − 6y − 2z = 7

Answer 1

$$\begin{gathered} \left(i\right)\text{We know that the angle between the planes}{a}_{1}x+b_{1}y+c_{1}z+d_1=0\text{and}{a}_{2}x+b_{2}y+c_{2}z+d_2=0\text{is given by} \\ \cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{{a_1}^2+{b_1}^2+{c_1}^2}\sqrt{{a_2}^2+{b_2}^2+{c_2}^2}} \\ \text{So, the angle between}2x-y+z=4\text{and}x+y+2z=3\text{is given by} \\ \cos \theta =\frac{\left(2\right)\left(1\right)+\left(-1\right)\left(1\right)+\left(1\right)\left(2\right)}{\sqrt{2^2+{\left(-1\right)}^2+1^2}\sqrt{1^2+1^2+2^2}}=\frac{2-1+2}{\sqrt{4+1+1}\sqrt{1+1+4}}=\frac{3}{\sqrt{6}\sqrt{6}}=\frac{3}{6}=\frac{1}{2} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\mathrm{\pi }}{3} \end{gathered}$$

$$\begin{gathered} \left(ii\right)\text{We know that the angle between the planes}{a}_{1}x+b_{1}y+c_{1}z+d_1=0\text{and}{a}_{2}x+b_{2}y+c_{2}z+d_2=0\text{is given by} \\ \cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{{a_1}^2+{b_1}^2+{c_1}^2}\sqrt{{a_2}^2+{b_2}^2+{c_2}^2}} \\ \text{So, the angle between}x+y-2z=3\text{and 2}x-2y+z=5\text{is given by} \\ \cos \theta =\frac{\left(1\right)\left(2\right)+\left(1\right)\left(-2\right)+\left(-2\right)\left(1\right)}{\sqrt{1^2+1^2+{\left(-2\right)}^2}\sqrt{2^2+{\left(-2\right)}^2+1^2}}=\frac{2-2-2}{\sqrt{1+1+4}\sqrt{4+4+1}}=\frac{-2}{\sqrt{6}\sqrt{9}}=\frac{-2}{3\sqrt{6}} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{-2}{3\sqrt{6}}\right) \end{gathered}$$
$$\begin{gathered} \left(iii\right)\text{We know that the angle between the planes}{a}_{1}x+b_{1}y+c_{1}z+d_1=0\text{and}{a}_{2}x+b_{2}y+c_{2}z+d_2=0\text{is given by} \\ \cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{{a_1}^2+{b_1}^2+{c_1}^2}\sqrt{{a_2}^2+{b_2}^2+{c_2}^2}} \\ \text{So, the angle between}x-y+z=5\text{and}x+2y+z=9\text{is given by} \\ \cos \theta =\frac{\left(1\right)\left(1\right)+\left(-1\right)\left(2\right)+\left(1\right)\left(1\right)}{\sqrt{1^2+{\left(-1\right)}^2+1^2}\sqrt{1^2+2^2+1^2}}=\frac{1-2+1}{\sqrt{1+1+1}\sqrt{1+4+1}}=\frac{0}{\sqrt{3}\sqrt{6}}=0 \\ \Rightarrow \theta ={\cos }^{-1}\left(0\right)=\frac{\mathrm{\pi }}{2} \end{gathered}$$

$$\begin{gathered} \left(iv\right)\text{We know that the angle between the planes}{a}_{1}x+b_{1}y+c_{1}z+d_1=0\text{and}{a}_{2}x+b_{2}y+c_{2}z+d_2=0\text{is given by} \\ \cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{{a_1}^2+{b_1}^2+{c_1}^2}\sqrt{{a_2}^2+{b_2}^2+{c_2}^2}} \\ \text{So, the angle between 2}x-3y+4z=1\text{and -}x+y+0z=4\text{is given by} \\ \cos \theta =\frac{\left(2\right)\left(-1\right)+\left(-3\right)\left(1\right)+\left(4\right)\left(0\right)}{\sqrt{2^2+{\left(-3\right)}^2+4^2}\sqrt{{\left(-1\right)}^2+1^2+0^2}}=\frac{-2-3+0}{\sqrt{4+9+16}\sqrt{1+1+0}}=\frac{-5}{\sqrt{29}\sqrt{2}}=\frac{-5}{\sqrt{58}} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{-5}{\sqrt{58}}\right) \end{gathered}$$

$$\begin{gathered} \left(v\right)\text{We know that the angle between the planes}{a}_{1}x+b_{1}y+c_{1}z+d_1=0\text{and}{a}_{2}x+b_{2}y+c_{2}z+d_2=0\text{is given by} \\ \cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{{a_1}^2+{b_1}^2+{c_1}^2}\sqrt{{a_2}^2+{b_2}^2+{c_2}^2}} \\ \text{So, the angle between 2}x+y-2z=5\text{and 3}x-6y-2z=7\text{is given by} \\ \cos \theta =\frac{\left(2\right)\left(3\right)+\left(1\right)\left(-6\right)+\left(-2\right)\left(-2\right)}{\sqrt{2^2+1^2+{\left(-2\right)}^2}\sqrt{3^2+{\left(-6\right)}^2+{\left(-2\right)}^2}}=\frac{6-6+4}{\sqrt{4+1+4}\sqrt{9+36+4}}=\frac{4}{\left(3\right)\left(7\right)}=\frac{4}{21} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{4}{21}\right) \end{gathered}$$