27.16 - Angle Between Two Planes

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.

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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, $\small{\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}}$

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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:
$\small{\theta=\cos^{-1}\left(\frac{1}{\sqrt {2}} \right)=45^{\circ}}$

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In the next section, we will see distance of a point from a plane.

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27.15 - Coplanarity of Two Lines

In the previous section, we completed a discussion related to the intersection of two planes. In this section, we will see coplanarity of two lines.

The vector form can be obtained in 7 steps:
1. In fig.27.24 below, two magenta lines lie on the same plane.
• That is., the two lines are coplanar lines.

Fig.27.24

2. We are given the following information about the two lines:
(i) The first line passes through the point $\small{U_1}$.
    ♦ The position vector of $\small{U_1~\text{is}~\vec{u_1}}$
    ♦ This line is parallel to $\small{\vec{v_1}}$
(ii) The second line passes through the point $\small{U_2}$.
    ♦ The position vector of $\small{U_2~\text{is}~\vec{u_2}}$
    ♦ This line is parallel to $\small{\vec{v_2}}$

3. Now we write the vectors which lie flat on the plane:
(i) First line is parallel to $\small{\vec{v_1}}$. So this $\small{\vec{v_1}}$ can be shifted and made to lie flat on the plane.
(ii) The second line is parallel to $\small{\vec{v_2}}$. So this $\small{\vec{v_2}}$ can also be shifted and made to lie flat on the plane.

4. Since $\small{\vec{v_1}~\text{and}~\vec{v_2}}$ lie on the plane, the cross product $\small{\left(\vec{v_1}\times\vec{v_2} \right)}$ will be a vector perpendicular to the plane.

5. Based on the position vectors of the points $\small{U_1~\text{and}~U_2}$, we get:
$\small{\vec{U_1 U_2}=\vec{u_2}-\vec{u_1}}$

6. Now we see an interesting situation:
• $\small{\vec{U_1 U_2}}$ lies on the plane, while $\small{\left(\vec{v_1}\times\vec{v_2} \right)}$ is perpendicular to the plane.
• So their dot product must be zero.

7. Therefore we can write:
• The given lines are coplanar if and only if $\small{\vec{U_1 U_2}.\left(\vec{v_1}\times\vec{v_2} \right)=0}$
• In other words, the given lines are coplanar if and only if $\small{\left(\vec{u_2}-\vec{u_1} \right).\left(\vec{v_1}\times\vec{v_2} \right)=0}$
• This is the vector form of the condition

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The Cartesian form can be obtained in steps:
1. First we write the coordinates:
    ♦ The coordinates of $\small{U_1~\text{are}~\left(x_1,y_1,z_1 \right)}$
    ♦ The coordinates of $\small{U_2~\text{are}~\left(x_2,y_2,z_2 \right)}$
2. Based on the above coordinates, we can write $\small{\vec{U_1 U_2}}$ in component form:
$\small{\vec{U_1 U_2} = \left(x_2 - x_1 \right)\hat{i} + \left(y_2 - y_1 \right)\hat{j} + \left(z_2 - z_1 \right)\hat{k}}$
3. Next we write the direction ratios of $\small{\vec{v_1}~\text{and}~\vec{v_2}}$:
    ♦ Direction ratios of $\small{\vec{v_1}~\text{are}~a_1,~b_1,~c_1}$
    ♦ Direction ratios of $\small{\vec{v_2}~\text{are}~a_2,~b_2,~c_2}$
4. Based on the above direction ratios, we can write $\small{\vec{v_1}~\text{and}~\vec{v_2}}$ in component forms:
    ♦ $\small{\vec{v_1}=a_1\hat{i}+b_1\hat{j}+c_1\hat{k}}$
    ♦ $\small{\vec{v_2}=a_2\hat{i}+b_2\hat{j}+c_2\hat{k}}$

5. We obtained the condition in vector form as:
$\small{\left(\vec{u_2}-\vec{u_1} \right).\left(\vec{v_1}\times\vec{v_2} \right)=0}$

6. First we will write $\small{\left(\vec{u_2}-\vec{u_1} \right)}$
We have:
$\small{\left(\vec{u_2}-\vec{u_1} \right)=\vec{U_1 U_2} = \left(x_2 - x_1 \right)\hat{i} + \left(y_2 - y_1 \right)\hat{j} + \left(z_2 - z_1 \right)\hat{k}}$

7. Next we write $\small{\left(\vec{v_1}\times\vec{v_2} \right)}$
We have:
$\vec{v_1}\times\vec{v_2}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ a_{1}      &{    a_{2}     } &{   a_{3}      }     \\ b_{1}      &{    b_{2}     }    &{ b_{3}        }     \\ \end{array}\right|$
• Expanding this, we get:
$\small{\vec{v_1}\times\vec{v_2}=\left(a_2 b_3 - a_3 b_2 \right)\hat{i}-\left(a_1 b_3 - a_3 b_1 \right)\hat{j}+\left(a_1 b_2 - a_2 b_1 \right)\hat{k}}$

8. So the vector condition written in (5) is:
$\small{\left(x_2 - x_1 \right)\left(a_2 b_3 - a_3 b_2 \right)\hat{i}-\left(y_2 - y_1 \right)\left(a_1 b_3 - a_3 b_1 \right)\hat{j}+\left(z_2 - z_1 \right)\left(a_1 b_2 - a_2 b_1 \right)\hat{k}=0}$

9. By close examination, we see that, the L.H.S in the above result is a determinant. We can write:
$\left|\begin{array}{r}                             x_2 - x_1     &{    y_2 - y_1     }    &{    z_2 - z_1      }    \\ a_{1}      &{    a_{2}     }    &{   a_{3}      }     \\ b_{1}      &{    b_{2}     }    &{ b_{3}        }     \\ \end{array}\right|~=~0$
• This is the Cartesian form of the condition

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Now we will see a solved example

Solved example 27.53
Show that the lines
$\small{\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}~\text{and}~\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}}$ are coplanar.
Solution:
Part (a): Showing coplanarity in vector form
1. The given lines in vector form are:
• $\small{-3\hat{i}+\hat{j}+5\hat{k}+\lambda \left(-3\hat{i}+\hat{j}+5\hat{k} \right)}$
• $\small{-\hat{i}+2\hat{j}+5\hat{k}+\lambda \left(-\hat{i}+2\hat{j}+5\hat{k} \right)}$

2. Form the above vector forms, we get:
• $\small{\vec{u_1}= -3\hat{i}+\hat{j}+5\hat{k}}$
• $\small{\vec{u_2}= -\hat{i}+2\hat{j}+5\hat{k}}$
• $\small{\vec{v_1}= -3\hat{i}+\hat{j}+5\hat{k}}$
• $\small{\vec{v_2}= -\hat{i}+2\hat{j}+5\hat{k}}$

3. $\small{\vec{u_2}-\vec{u_1}=2\hat{i}+\hat{j}}$

4. $\small{\vec{v_1}\times\vec{v_2}=-5\hat{i}+10\hat{j}-5\hat{k}}$
(The reader should write all the steps related to the above cross product)

5. The given lines are coplanar if and only if the dot product of the vectors in (3) and (4) is zero.
• The dot product is:
(2)(−5)+(1)(10)+(0)(-5) = −10 + 10 = 0
• So the lines are coplanar.  

Part (b): Showing coplanarity in Cartesian form

1. From the given equations, we get:
• $\small{\left(x_1,y_1,z_1 \right)=\left(-3,1,5 \right)}$
• $\small{\left(x_2,y_2,z_2 \right)=\left(-1,2,5 \right)}$
• $\small{a_1,b_1,c_1~=~-3,1,5}$
• $\small{a_2,b_2,c_2~=~-1,2,5}$

2. The condition in Cartesian form is:
$\left|\begin{array}{r}                             x_2 - x_1     &{    y_2 - y_1     }    &{    z_2 - z_1      }    \\ a_{1}      &{    a_{2}     }    &{   a_{3}      }     \\ b_{1}      &{    b_{2}     }    &{ b_{3}        }     \\ \end{array}\right|~=~0$

3. Let us evaluate the determinant:
$\left|\begin{array}{r}                             (-1)-(-3)&{    2-1   }    &{    5-5    }    \\ -3     &{    1     }    &{   5      }     \\ -1     &{    2     }    &{ 5        }     \\ \end{array}\right|$

$\small{~=(2)(5-10)-(1)(-15+5)+(0)(-6+1)}$
$\small{~=(2)(-5)-(1)(-10)}$
$\small{~=-10+10=0}$

• So the given lines are coplanar

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In the next section, we will see angle between two planes.

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27.14 - Intersection of Planes - More cases

In the previous section, we saw case I related to the intersection of two planes. In this section, we will see the solved examples related to the remaining cases.

Solved example 27.50
Find the equation of the plane through the line of intersection of the planes
$\small{x+y+z=1~\text{and}~2x+3y+4z=5}$ which is perpendicular to the plane $\small{x-y+z=0}$
Solution:
This problem belongs to Case II
1. We are given two planes:
• $\small{\pi_1~\text{whose equation is}~x+y+z=1}$
• $\small{\pi_2~\text{whose equation is}~2x+3y+4z=5}$

2. We are asked to find the equation of the plane $\small{\pi_3}$ which satisfies two conditions:
(i) $\small{\pi_3}$ must pass through the intersection of $\small{\pi_1~\text{and}~\pi_2}$
(ii) $\small{\pi_3}$ must be perpendicular to a given plane.
• This given plane is:
$\small{\pi_4~\text{whose equation is}~x-y+z=0}$

• All the four planes are shown in fig.27.23 below:

Fig.27.23

• The blue plane is perpendicular to the brown plane.  

3. First we need two points on the line of intersection. They can be obtained in 3 steps:
(i) The Cartesian equation of the two intersecting planes are:
    ♦ $\small{x+y+z=1}$
    ♦ $\small{2x+3y+4z=5}$
• Any point (x,y,z) on the line of intersection, will satisfy the above two equations
(ii) Put x = 0. We get:
    ♦ $\small{y+z=1}$
    ♦ $\small{3y+4z=5}$
• Solving the two equations, we get: y = −1 and z = 2
• So (0,−1,2) is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

(iii) Put z = 0. We get:
    ♦ $\small{x+y=1}$
    ♦ $\small{2x+3y=5}$
• Solving the two equations, we get: x = −2 and y = 3
• So (−2,3,0) is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

4. If we know two points on a line, we can calculate the direction ratios of that line. So the direction ratios of the line of intersection are: −2, 4, −2
• Therefore, the vector $\small{-2\hat{i}+4\hat{j}-2\hat{k}}$ is parallel to the line of intersection.
• Dividing throughout by −2, we can write:
The vector $\small{\hat{i}-2\hat{j}+\hat{k}}$ is parallel to the line of intersection.
• This vector will lie flat entirely on the required plane $\small{\pi_3}$

5. Now we consider $\small{\pi_4}$. Based on it’s given equation, we can write:
The vector $\small{\hat{i}-\hat{j}+\hat{k}}$ is perpendicular to $\small{\pi_4}$
(Imagine a vector, perpendicular to the blue plane in fig.27.23 above)

6. $\small{\pi_4}$ is perpendicular to $\small{\pi_3}$. So the vector $\small{\hat{i}-\hat{j}+\hat{k}}$ is parallel to $\small{\pi_3}$. This vector can be shifted and can be made to lie flat on $\small{\pi_3}$

7. Now we have two vectors which lie flat on $\small{\pi_3}$:
    ♦ The vector obtained in step (4)
    ♦ The vector obtained in step (6)
• The cross product of those two vectors will be a vector perpendicular to $\small{\pi_3}$
• The cross product is: $\small{-\hat{i}+\hat{k}}$
(The reader should write all steps related to the cross product)

8. Now we can write the equation of $\small{\pi_3}$. We have two parameters:
(i) From step (7), the vector perpendicular to $\small{\pi_3}$
(ii) From step (3), we can pick any one of the two points. That point will lie on $\small{\pi_3}$
• So we can use the formula: $\small{A\left(x-x_1 \right)+A\left(y-y_1 \right)+A\left(z-z_1 \right)=0}$
See section 27.10
• In our present case,
    ♦ A = −1, B = 0 and C = 1
    ♦ $\small{\left(x_1,y_1,z_1 \right) = \left(0,-1,2 \right)}$

9. Therefore, the equation of $\small{\pi_3}$ is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{A\left(x-x_1 \right)+B\left(y-y_1 \right)+C\left(z-z_1 \right)}    & {~=~}    &{0} \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{(-1)\left(x-0 \right)+(0)\left(y-(-1) \right)+(1)\left(z-2 \right)}    & {~=~}    &{0} \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{-x+z-2}    & {~=~}    &{0} \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x-z+2}    & {~=~}    &{0} \\
\end{array}}$

Solved example 27.51
Find the equation of the plane through the line of intersection of the planes
$\small{\vec{r}.\left(\hat{i}+2\hat{j}+3\hat{k} \right)-4=0~\text{and}~\vec{r}.\left(2\hat{i}+\hat{j}-\hat{k} \right)+5=0}$ and which is perpendicular to the plane $\small{\vec{r}.\left(5\hat{i}+3\hat{j}-6\hat{k} \right)+8=0}$
Solution:
This problem belongs to Case II
1. We are given two planes:
• $\small{\pi_1~\text{whose equation is}~\vec{r}.\left(\hat{i}+2\hat{j}+3\hat{k} \right)-4=0}$
• $\small{\pi_2~\text{whose equation is}~\vec{r}.\left(2\hat{i}+\hat{j}-\hat{k} \right)+5=0}$

2. We are asked to find the equation of the plane $\small{\pi_3}$ which satisfies two conditions:
(i) $\small{\pi_3}$ must pass through the intersection of $\small{\pi_1~\text{and}~\pi_2}$
(ii) $\small{\pi_3}$ must be perpendicular to a given plane.
• This given plane is:
$\small{\pi_4~\text{whose equation is}~\vec{r}.\left(5\hat{i}+3\hat{j}-6\hat{k} \right)+8=0}$

3. First we need two points on the line of intersection. They can be obtained in 3 steps:
(i) The Cartesian equation of the two intersecting planes are:
    ♦ $\small{x+2y+3z=4}$
    ♦ $\small{2x+y-z=-5}$
• Any point (x,y,z) on the line of intersection, will satisfy the above two equations
(ii) Put x = 0. We get:
    ♦ $\small{2y+3z=4}$
    ♦ $\small{y-z=-5}$
• Solving the two equations, we get: y = -11/5 and z = 14/5
• So $\small{\left(0,\frac{-11}{5},\frac{14}{5} \right)}$ is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

(iii) Put z = 0. We get:
    ♦ $\small{x+2y=4}$
    ♦ $\small{2x+y=-5}$
• Solving the two equations, we get: x = −14/3 and y = 13/3
• So $\small{\left(\frac{-14}{3},\frac{13}{3},0 \right)}$ is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

4. If we know two points on a line, we can calculate the direction ratios of that line. So the direction ratios of the line of intersection are:
    ♦ $\small{\frac{-14}{3}-0=\frac{-14}{3}}$
    ♦ $\small{\frac{13}{3}-\frac{-11}{5}=\frac{98}{15}}$
    ♦ $\small{0-\frac{14}{5}=\frac{-14}{5}}$
• Therefore, the vector $\small{\left(\frac{-14}{3} \right)\hat{i}+\left(\frac{98}{15} \right)\hat{j}+\left(\frac{-14}{5} \right)\hat{k}}$ is parallel to the line of intersection.
• Multiplying throughout by $\small{\frac{-15}{2}}$, we can write:
The vector $\small{35\hat{i}-49\hat{j}+21\hat{k}}$ is parallel to the line of intersection.
• This vector will lie flat entirely on the required plane $\small{\pi_3}$

5. Now we consider $\small{\pi_4}$. Based on it’s given equation, we can write:
The vector $\small{5\hat{i}+3\hat{j}-6\hat{k}}$ is perpendicular to $\small{\pi_4}$

6. $\small{\pi_4}$ is perpendicular to $\small{\pi_3}$. So the vector $\small{5\hat{i}+3\hat{j}-6\hat{k}}$ is parallel to $\small{\pi_3}$. This vector can be shifted and can be made to lie flat on $\small{\pi_3}$

7. Now we have two vectors which lie flat on $\small{\pi_3}$:
    ♦ The vector obtained in step (4)
    ♦ The vector obtained in step (6)
• The cross product of those two vectors will be a vector perpendicular to $\small{\pi_3}$
• The cross product is: $\small{231\hat{i}+315\hat{j}+350\hat{k}}$
(The reader should write all steps related to the cross product)

8. Now we can write the equation of $\small{\pi_3}$. We have two parameters:
(i) From step (7), the vector perpendicular to $\small{\pi_3}$
(ii) From step (3), we can pick any one of the two points. That point will lie on $\small{\pi_3}$
• So we can use the formula: $\small{A\left(x-x_1 \right)+A\left(y-y_1 \right)+A\left(z-z_1 \right)=0}$
See section 27.10
• In our present case,
    ♦ A = 231, B = 315 and C = 350
    ♦ $\small{\left(x_1,y_1,z_1 \right) = \left(0,\frac{-11}{5},\frac{14}{5} \right)}$

9. Therefore, the equation of $\small{\pi_3}$ is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{A\left(x-x_1 \right)+B\left(y-y_1 \right)+C\left(z-z_1 \right)}    & {~=~}    &{0} \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{(231)\left(x-0 \right)+(315)\left(y-(\frac{-11}{5}) \right)+(350)\left(z-\frac{14}{5} \right)}    & {~=~}    &{0} \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{231x+315y+350z-287}    & {~=~}    &{0} \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{7(33x+45y+50z-41)}    & {~=~}    &{0} \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{33x+45y+50z-41}    & {~=~}    &{0} \\
\end{array}}$

Solved example 27.52
Find the equation of the plane through the line of intersection of the planes
$\small{\vec{r}.\left(\hat{i}+\hat{j}+\hat{k} \right)=1~\text{and}~\vec{r}.\left(2\hat{i}+3\hat{j}-\hat{k} \right)+4=0}$ and parallel to x-axis.
Solution:
This problem belongs to Case III
1. We are given two planes:
• $\small{\pi_1~\text{whose equation is}~\vec{r}.\left(\hat{i}+\hat{j}+\hat{k} \right)=1}$
• $\small{\pi_2~\text{whose equation is}~\vec{r}.\left(2\hat{i}+3\hat{j}-\hat{k} \right)+4=0}$

2. We are asked to find the equation of the plane $\small{\pi_3}$ which satisfies two conditions:
(i) $\small{\pi_3}$ must pass through the intersection of $\small{\pi_1~\text{and}~\pi_2}$
(ii) $\small{\pi_3}$ must be parallel to the x-axis.

3. First we need two points on the line of intersection. They can be obtained in 3 steps:
(i) The Cartesian equation of the two intersecting planes are:
    ♦ $\small{x+y+z=1}$
    ♦ $\small{2x+3y-z=-4}$
• Any point (x,y,z) on the line of intersection, will satisfy the above two equations
(ii) Put x = 0. We get:
    ♦ $\small{y+z=1}$
    ♦ $\small{3y-z=-4}$
• Solving the two equations, we get: y = -3/4 and z = 7/4
• So $\small{\left(0,\frac{-3}{4},\frac{7}{4} \right)}$ is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

(iii) Put z = 0. We get:
    ♦ $\small{x+y=1}$
    ♦ $\small{2x+3y=-4}$
• Solving the two equations, we get: x = 7 and y = -6
• So $\small{\left(7,-6,0 \right)}$ is a point on the line of intersection.
(The reader must write all steps related to the solution of the two equations and also the checking steps)

4. If we know two points on a line, we can calculate the direction ratios of that line. So the direction ratios of the line of intersection are:
    ♦ $\small{7-0=7}$
    ♦ $\small{-6-\frac{-3}{4}=\frac{-21}{4}}$
    ♦ $\small{0-\frac{7}{4}=\frac{-7}{4}}$
• Therefore, the vector $\small{\left(7 \right)\hat{i}-\left(\frac{21}{4} \right)\hat{j}-\left(\frac{7}{4} \right)\hat{k}}$ is parallel to the line of intersection.
• Multiplying throughout by $\small{4}$, we can write:
The vector $\small{28\hat{i}-21\hat{j}-7\hat{k}}$ is parallel to the line of intersection.
• This vector will lie flat entirely on the required plane $\small{\pi_3}$

5. Now we consider the x-axis.
The vector equation of x-axis is: $\small{\hat{i}}$

6. $\small{\pi_3}$ is to be parallel to the x-axis.
That is., $\small{\pi_3}$ will never intersect the x-axis. So the vector $\small{\hat{i}}$ is parallel to $\small{\pi_3}$. This vector can be shifted and can be made to lie flat on $\small{\pi_3}$

7. Now we have two vectors which lie flat on $\small{\pi_3}$:
    ♦ The vector obtained in step (4)
    ♦ The vector obtained in step (6)
• The cross product of those two vectors will be a vector perpendicular to $\small{\pi_3}$
• The cross product is: $\small{-7\hat{j}+21\hat{k}}$
(The reader should write all steps related to the cross product)

8. Now we can write the equation of $\small{\pi_3}$. We have two parameters:
(i) From step (7), the vector perpendicular to $\small{\pi_3}$
(ii) From step (3), we can pick any one of the two points. That point will lie on $\small{\pi_3}$
• So we can use the formula: $\small{A\left(x-x_1 \right)+A\left(y-y_1 \right)+A\left(z-z_1 \right)=0}$
See section 27.10
• In our present case,
    ♦ A = 0, B = −7 and C = 21
    ♦ $\small{\left(x_1,y_1,z_1 \right) = \left(7,-6,0 \right)}$

9. Therefore, the equation of $\small{\pi_3}$ is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{A\left(x-x_1 \right)+B\left(y-y_1 \right)+C\left(z-z_1 \right)}    & {~=~}    &{0} \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{(0)\left(x-7 \right)+(-7)\left(y-(-6) \right)+(21)\left(z-0 \right)}    & {~=~}    &{0} \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{-7y+21z-42}    & {~=~}    &{0} \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{-7(y-3z+6)}    & {~=~}    &{0} \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{y-3z+6}    & {~=~}    &{0} \\
\end{array}}$

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In the next section, we will see coplanarity of two lines.

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