In the previous section, we saw coplanarity of two lines. In this section, we will see the method to obtain the angle between two planes.
The method can be explained in 9 steps:
1. In fig.27.25 below, we have two planes $\small{\pi_1~\text{and}~\pi_2}$
• Equation of $\small{\pi_1~\text{is}~\vec{r}.\vec{n_1}=d_1}$
♦ This plane is shown in yellow color.
• Equation of $\small{\pi_2~\text{is}~\vec{r}.\vec{n_2}=d_2}$
♦ This plane is shown in brown color.
 |
| Fig.27.25 |
• The yellow vector is normal to the yellow plane
• The brown vector is normal to the brown plane
2. $\small{\angle{DAE}}$ is the acute angle $\small{\theta}$ between the two planes.
• Note that, $\small{\angle{DAF}}$ is equal to $\small{180-\theta}$. It can also be considered as the angle between the two planes. But we are interested in the acute angle $\small{\theta}$.
3. The brown vector is perpendicular to the brown plane. So we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\angle{CAE}} & {~=~} &{\angle{CAD}+\angle{EAD}} \\
{~\color{magenta} 2 } &{\Rightarrow} &{90} & {~=~} &{\angle{CAD}+\theta} \\
{~\color{magenta} 3 } &{\Rightarrow} &{\angle{CAD}} & {~=~} &{90-\theta} \\
\end{array}}$
4. The yellow vector is perpendicular to the yellow plane. So we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\angle{BAD}} & {~=~} &{\angle{BAC}+\angle{DAC}} \\
{~\color{magenta} 2 } &{\Rightarrow} &{90} & {~=~} &{\angle{BAC}+\left(90-\theta \right)} \\
{~\color{magenta} 3 } &{\Rightarrow} &{\angle{BAC}} & {~=~} &{\theta} \\
\end{array}}$
5. So we can write:
$\small{\angle{BAC} = \angle{EAD} = \theta}$
• Here,
♦ $\small{\angle{BAC}}$ is the angle between the two normals
♦ $\small{\angle{EAD}}$ is the angle between the two planes.
6. Therefore,
To find the angle between two planes, all we need to do, is to find the angle between their normals.
7. From the equation of the planes, we get the equation of the normals $\small{\vec{n_1}~\text{and}~\vec{n_2}}$
• So the angle between the normals is given by:
$\small{\cos \theta = \left|\frac{\vec{n_1}.\vec{n_2}}{\left(\left|\vec{n_1} \right| \right)\left(\left|\vec{n_2} \right| \right)} \right|}$
• This is the angle in vector form
8. Recall that, if two vectors are perpendicular to each other, their dot product will be zero.
• We know that, if the normals are perpendicular to each other, the planes will also be perpendicular to each other.
• So we can write:
If two planes are perpendicular to each other, $\small{\vec{n_1}.\vec{n_2}=0}$
9. Also, if two normals are parallel to each other, then the corresponding planes are parallel to each other.
The Cartesian form can be obtained in 5 steps:
1. Let the Cartesian equations of the two planes be:
$\small{A_1 x + B_1 y + C_1 z + D_1 = 0~\text{and}~A_2 x + B_2 y + C_2 z + D_2 = 0 }$
2. Then the component form of the normals are:
$\small{\vec{n_1}=A_1 \hat{i}+B_1 \hat{j}+C_1 \hat{k}}$
$\small{\vec{n_2}=A_2 \hat{i}+B_2 \hat{j}+C_2 \hat{k}}$
3. So we can write:
$\small{\cos \theta=\left|\frac{A_1 A_2 + B_1 B_2 + C_1 C_2}{\left(\sqrt{A_1^2 + B_1^2 + C_1^2} \right)\left(\sqrt{A_2^2 + B_2^2 + C_2^2} \right)} \right|}$
4. Recall that, if two vectors are perpendicular to each other, their dot product will be zero.
• We know that, if the normals are perpendicular to each other, the planes will also be perpendicular to each other.
• So we can write:
If two planes are perpendicular to each other, $\small{A_1 A_2 + B_1 B_2 + C_1 C_2=0}$
5. Recall that, if two vectors are parallel to each other, their direction ratios will be proportional.
• We know that, if the normals are parallel to each other, the planes will also be parallel to each other.
• So we can write:
If two planes are parallel to each other, $\small{\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}}$
Now we will see some solved example
Solved example 27.54
Find the angle between the two planes
$\small{2x + y - 2z = 5~\text{and}~3x - 6y - 2z = 7}$.
Solution:
From the given equations, we can write:
• Normals in component form:
♦ $\small{\vec{n_1}=2\hat{i}+\hat{j}-2\hat{k}}$
♦ $\small{\vec{n_2}=3\hat{i}-6\hat{j}-2\hat{k}}$
• Direction ratios of the normals:
♦ $\small{A_1,B_1,C_1~=~2,~1,~ -2}$
♦ $\small{A_2,B_2,C_2~=~3,~-6,~ -2}$
• Checking whether the planes are perpendicular:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{A_1 A_2 + B_1 B_2 + C_1 C_2} & {~=~} &{(2)(3)+(1)(-6)+(-2)(-2)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{6-6+4} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{4~\ne~0} \\
\end{array}}$
♦ So the planes are not perpendicular to each other.
• Checking whether the planes are parallel:
Proportionality check of the direction ratios gives:
$\small{\frac{2}{3}~\ne~\frac{1}{-6}~\ne~\frac{-2}{-2}}$
♦ So the planes are not parallel to each other.
Part (a): Finding angle by vector method
1. We have:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos \theta} & {~=~} &{\left|\frac{\vec{n_1}.\vec{n_2}}{\left(\left|\vec{n_1} \right| \right)\left(\left|\vec{n_2} \right| \right)} \right|} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\left|\frac{\left(2\hat{i}+\hat{j}-2\hat{k} \right).\left(3\hat{i}-6\hat{j}-2\hat{k} \right)}{\left(\left|\sqrt{2^2 + 1^2 +(-2)^2} \right| \right)\left(\sqrt{3^2 + (-6)^2 + (-2)^2} \right)} \right|} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\left|\frac{4}{\left(3 \right)\left(7 \right)} \right|} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{4}{21}} \\
\end{array}}$
2. So we get:
$\small{\theta=\cos^{-1}\left(\frac{4}{21} \right)}$
Part (a): Finding angle by Cartesian method
1. We have:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos \theta} & {~=~} &{\left|\frac{A_1 A_2 + B_1 B_2 + C_1 C_2}{\left(\sqrt{A_1^2 + B_1^2 + C_1^2} \right)\left(\sqrt{A_2^2 + B_2^2 + C_2^2} \right)} \right|} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\left|\frac{(2)(3)+(1)(-6)+(-2)(-2)}{\left(\left|\sqrt{2^2 + 1^2 +(-2)^2} \right| \right)\left(\sqrt{3^2 + (-6)^2 + (-2)^2} \right)} \right|} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\left|\frac{4}{\left(3 \right)\left(7 \right)} \right|} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{4}{21}} \\
\end{array}}$
2. So we get:
$\small{\theta=\cos^{-1}\left(\frac{4}{21} \right)}$
Solved example 27.55
Find the angle between the two planes
$\small{3x - 6y + 2z = 7~\text{and}~2x + 2y - 2z =5}$.
Solution:
From the given equations, we can write:
• Normals in component form:
♦ $\small{\vec{n_1}=3\hat{i}-6\hat{j}+2\hat{k}}$
♦ $\small{\vec{n_2}=2\hat{i}+2\hat{j}-2\hat{k}}$
• Direction ratios of the normals:
♦ $\small{A_1,B_1,C_1~=~3,~-6,~ 2}$
♦ $\small{A_2,B_2,C_2~=~2,~2,~ -2}$
• Checking whether the planes are perpendicular:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{A_1 A_2 + B_1 B_2 +
C_1 C_2} & {~=~} &{(3)(2)+(-6)(2)+(2)(-2)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{6-12-4} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{-10~\ne~0} \\
\end{array}}$
♦ So the planes are not perpendicular to each other.
• Checking whether the planes are parallel:
Proportionality check of the direction ratios gives:
$\small{\frac{3}{2}~\ne~\frac{-6}{2}~\ne~\frac{2}{-2}}$
♦ So the planes are not parallel to each other.
In the question, the method is not specified. So we will use the Cartesian method.
1. We have:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos \theta} & {~=~} &{\left|\frac{A_1 A_2 + B_1 B_2 + C_1 C_2}{\left(\sqrt{A_1^2 + B_1^2 + C_1^2} \right)\left(\sqrt{A_2^2 + B_2^2 + C_2^2} \right)} \right|} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\left|\frac{(3)(2)+(-6)(2)+(2)(-2)}{\left(\left|\sqrt{3^2 + (-6)^2 +(2)^2} \right| \right)\left(\sqrt{2^2 + 2^2 + (-2)^2} \right)} \right|} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\left|\frac{-10}{\left(7 \right)\left(\sqrt{12} \right)} \right|~=~\frac{10}{(7)\left(2 \sqrt{3} \right)}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{5\left(\sqrt{3} \right)}{(7)\left(\sqrt{3} \right)\left(\sqrt{3} \right)}~=~\frac{5\sqrt{3} }{21}} \\
\end{array}}$
2. So we get:
$\small{\theta=\cos^{-1}\left(\frac{5\sqrt{3} }{21} \right)}$
Solved example 27.56
In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.
$\small{\begin{array}{ll} {~\color{magenta} {\text{(a)}} } &{7x+5y+6z+30=0} & {\text{and}} &{3x-y-10z+4=0} \\
{~\color{magenta}{\text{(b)}} } &{2x+y+3z-2=0} & {\text{and}} &{x-2y+5=0} \\
{~\color{magenta} {\text{(c)}} } &{2x-2y+4z+5=0} & {\text{and}} &{3x-3y+6z-1=0} \\
{~\color{magenta} {\text{(d)}} } &{2x-y+3z-1=0} & {\text{and}} &{2x-y+3z+3=0} \\
{~\color{magenta} {\text{(e)}} } &{4x+8y+z-8=0} & {\text{and}} &{y+z-4=0} \\
\end{array}}$
Solution:
In the question, the method is not specified. So we will use the Cartesian method.
Case (a):
From the given equations, we can write:
• Direction ratios of the normals:
♦ $\small{A_1,B_1,C_1~=~7,~5,~ 6}$
♦ $\small{A_2,B_2,C_2~=~3,~-1,~ -10}$
• Checking whether the planes are perpendicular:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{A_1 A_2 + B_1 B_2 + C_1 C_2} & {~=~} &{(7)(3)+(5)(-1)+(6)(-10)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{21-5-60} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{-44~\ne~0} \\
\end{array}}$
♦ So the planes are not perpendicular to each other.
• Checking whether the planes are parallel:
Proportionality check of the direction ratios gives:
$\small{\frac{7}{3}~\ne~\frac{5}{-1}~\ne~\frac{6}{-10}}$
♦ So the planes are not parallel to each other.
1. We have:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos \theta} & {~=~} &{\left|\frac{A_1 A_2 + B_1 B_2 + C_1 C_2}{\left(\sqrt{A_1^2 + B_1^2 + C_1^2} \right)\left(\sqrt{A_2^2 + B_2^2 + C_2^2} \right)} \right|} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\left|\frac{-44}{\left(\left|\sqrt{(7)^2 + (5)^2 +(6)^2} \right| \right)\left(\sqrt{(3)^2 + (-1)^2 + (-10)^2} \right)} \right|} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\left|\frac{-44}{\left(\sqrt{110} \right)\left(\sqrt{110} \right)} \right|~=~\frac{44}{110}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{22}{55}~=~\frac{2 }{5}} \\
\end{array}}$
2. So we get:
$\small{\theta=\cos^{-1}\left(\frac{2 }{5} \right)}$
Case (b):
From the given equations, we can write:
• Direction ratios of the normals:
♦ $\small{A_1,B_1,C_1~=~2,~1,~ 3}$
♦ $\small{A_2,B_2,C_2~=~1,~-2,~ 5}$
• Checking whether the planes are perpendicular:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{A_1 A_2 + B_1 B_2 + C_1 C_2} & {~=~} &{(2)(1)+(1)(-2)+(3)(0)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{2-2+0} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{0} \\
\end{array}}$
♦ So the planes are perpendicular to each other.
Case (c):
From the given equations, we can write:
• Direction ratios of the normals:
♦ $\small{A_1,B_1,C_1~=~2,~-2,~4}$
♦ $\small{A_2,B_2,C_2~=~3,~-3,~ 6}$
• Checking whether the planes are perpendicular:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{A_1 A_2 + B_1 B_2 + C_1 C_2} & {~=~} &{(2)(3)+(-2)(-3)+(4)(6)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{6+6+24} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{36~\ne~0} \\
\end{array}}$
♦ So the planes are not perpendicular to each other.
• Checking whether the planes are parallel:
Proportionality check of the direction ratios gives:
$\small{\frac{2}{3}~=~\frac{-2}{-3}~=~\frac{4}{6}}$
♦ So the planes are parallel to each other.
(A close inspection of the given equations will reveal that, the direction ratios are proportional, which indicates that the planes are parallel. In such a situation, we can avoid the test for perpendicularity)
Case (d):
From the given equations, we can write:
• Direction ratios of the normals:
♦ $\small{A_1,B_1,C_1~=~2,~-1,~3}$
♦ $\small{A_2,B_2,C_2~=~2,~-1,~3}$
• Checking whether the planes are perpendicular:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{A_1 A_2 + B_1 B_2 +
C_1 C_2} & {~=~} &{(2)(2)+(-1)(-1)+(3)(3)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{4+1+9} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{14~\ne~0} \\
\end{array}}$
♦ So the planes are not perpendicular to each other.
• Checking whether the planes are parallel:
Proportionality check of the direction ratios gives:
$\small{\frac{2}{2}~=~\frac{-1}{-1}~=~\frac{3}{3}}$
♦ So the planes are parallel to each other.
(A close inspection of the given equations will reveal that, the
direction ratios are proportional, which indicates that the planes are
parallel. In such a situation, we can avoid the test for
perpendicularity)
Case (e):
From the given equations, we can write:
• Direction ratios of the normals:
♦ $\small{A_1,B_1,C_1~=~4,~8,~ 1}$
♦ $\small{A_2,B_2,C_2~=~0,~1,~ 1}$
• Checking whether the planes are perpendicular:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{A_1 A_2 + B_1 B_2 +
C_1 C_2} & {~=~} &{(4)(0)+(8)(1)+(1)(1)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{0+8+1} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{9~\ne~0} \\
\end{array}}$
♦ So the planes are not perpendicular to each other.
• Checking whether the planes are parallel:
Proportionality check of the direction ratios gives:
$\small{\frac{4}{0}~\ne~\frac{8}{1}~\ne~\frac{1}{1}}$
♦ So the planes are not parallel to each other.
1. We have:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos \theta} & {~=~} &{\left|\frac{A_1 A_2 + B_1 B_2 + C_1 C_2}{\left(\sqrt{A_1^2 + B_1^2 + C_1^2} \right)\left(\sqrt{A_2^2 + B_2^2 + C_2^2} \right)} \right|} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\left|\frac{9}{\left(\left|\sqrt{(4)^2 + (8)^2 +(1)^2} \right| \right)\left(\sqrt{(0)^2 + (1)^2 + (1)^2} \right)} \right|} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\left|\frac{9}{\left(\sqrt{81} \right)\left(\sqrt{2} \right)} \right|~=~\frac{9}{9 \sqrt{2}}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{1}{\sqrt {2}}} \\
\end{array}}$
2. So we get:
$\small{\theta=\cos^{-1}\left(\frac{1}{\sqrt {2}} \right)=45^{\circ}}$
In the next section, we will see distance of a point from a plane.
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