27.11 - Plane Passing Through Three Non Collinear Points

In the previous section, we saw the plane perpendicular to a given vector and passing through a given point. In this section, we will see plane passing through three non collinear points.

A plane can be uniquely defined if we have three non collinear points on that plane. The vector equation of such a plane, can be obtained in 6 steps:

1. First we write the points on the plane
    ♦ U, V and W are three points on the plane. They are the given points.
    ♦ P is an arbitrary point on the plane.
• Those four points are shown in fig.27.18 below:

Fig.27.18

• In the above fig., some vectors are drawn in bold lines, while the remaining vectors are drawn in dashed lines. The dashed lines indicate that, those vectors are hidden from view. The plane is obstructing us from viewing those vectors.
• Note that, the x and y-axes are also drawn in dashed lines. They are below the plane and hence hidden from view.

2. Now we write the position vectors of the above points
    ♦ $\small{\vec{u}}$ is the position vector of point U
    ♦ $\small{\vec{v}}$ is the position vector of point V
    ♦ $\small{\vec{w}}$ is the position vector of point W
    ♦ $\small{\vec{r}}$ is the position vector of point P

3. Consider the vectors $\small{\vec{UV}~\text{and}~\vec{UW}}$
Both of them lie on the plane. So their cross product will be perpendicular to the plane. That is:
$\small{\left(\vec{UV}\times\vec{UW}\right)}$ is perpendicular to the plane.

4. Now consider the two vectors:
• $\small{\vec{UP}~\text{and}~\left(\vec{UV}\times\vec{UW}\right)}$
    ♦ $\small{\vec{UP}}$ lies on the plane
    ♦ $\small{\left(\vec{UV}\times\vec{UW}\right)}$ is perpendicular to the plane.
• So the dot product of the two vectors will be zero. That is:
$\small{\vec{UP}.\left(\vec{UV}\times\vec{UW}\right)=0}$

5. Our next task is to find the vectors in the above equation.
• From triangle OUP, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u}+\vec{UP}}    & {~=~}    &{\vec{r}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\vec{UP}}    & {~=~}    &{\vec{r}-\vec{u}}
\\ \end{array}}$
• From triangle OUV, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u}+\vec{UV}}    & {~=~}    &{\vec{v}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\vec{UV}}    & {~=~}    &{\vec{v}-\vec{u}}
\\ \end{array}}$
• From triangle OUW, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u}+\vec{UW}}    & {~=~}    &{\vec{w}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\vec{UW}}    & {~=~}    &{\vec{w}-\vec{u}}
\\ \end{array}}$

6. Now we can substitute in (4). We get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{UP}.\left(\vec{UV}\times\vec{UW}\right)}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r}-\vec{u} \right).\left[\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)\right]}    & {~=~}    &{0}
\\ \end{array}}$
• This is the vector equation of the plane.

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We are discussing about plane passing through three given points. For such a discussion, it is very important to specify that, the three points are non collinear. The reason is simple: There are infinite number of planes which can pass through three collinear points. This is shown in fig.27.19 below:

Fig.27.19

    ♦ Plane $\small{\pi_1}$ is drawn in blue color
    ♦ Plane $\small{\pi_2}$ is drawn in green color
    ♦ Plane $\small{\pi_3}$ is drawn in orange color
• All three planes contain the three yellow points. The three yellow points are collinear, as indicated by the magenta line.
• It is like making multiple copies of the original plane and rotating the copies by using the magenta line as axis. All the copies will contain the three points.
• So it is clear that, to define a unique plane using three points, those three points should be non collinear.

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The Cartesian equation of the plane can be derived in 6 steps:
1. First we write the coordinates of the four points U, V, W and P:
• $\small{U\left(x_1,y_1,z_1 \right)}$
• $\small{V\left(x_2,y_2,z_2 \right)}$
• $\small{W\left(x_3,y_3,z_3 \right)}$
• $\small{P\left(x,y,z\right)}$

2. Based on the coordinates, we can write the position vectors:
• $\small{\vec{u}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}}$
• $\small{\vec{v}=x_2\hat{i}+y_2\hat{j}+z_2\hat{k}}$
• $\small{\vec{w}=x_3\hat{i}+y_3\hat{j}+z_3\hat{k}}$
• $\small{\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}}$

3. Based on the position vectors, we get:
$\small{\vec{r}-\vec{u} = \left(x-x_1 \right)\hat{i}+\left(y-y_1 \right)\hat{j}+\left(z-z_1 \right)\hat{k}}$
$\small{\vec{v}-\vec{u} = \left(x_2-x_1 \right)\hat{i}+\left(y_2-y_1 \right)\hat{j}+\left(z_2-z_1 \right)\hat{k}}$
$\small{\vec{w}-\vec{u} = \left(x_3-x_1 \right)\hat{i}+\left(y_3-y_1 \right)\hat{j}+\left(z_3-z_1 \right)\hat{k}}$

4. Now we can calculate the cross product:
$\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ x_2-x_1      &{    y_2-y_1     }    &{   z_2-z_1      }     \\ x_3-x_1      &{    y_3-y_1     }    &{ z_3-z_1        }     \\ \end{array}\right|$
• Expanding the determinant, we get:
$\small{\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)}$

$\small{~=\left[\left(y_2-y_1 \right)\left(z_3-z_1 \right)-\left(z_2-z_1 \right)\left(x_3-x_1 \right) \right]\hat{i}}$

$\small{~-\left[\left(x_2-x_1 \right)\left(z_3-z_1 \right)-\left(z_2-z_1 \right)\left(x_3-x_1 \right) \right]\hat{j}}$

$\small{~+\left[\left(x_2-x_1 \right)\left(y_3-y_1 \right)-\left(y_2-y_1 \right)\left(x_3-x_1 \right) \right]\hat{k}}$

5. Now we can substitute the various vectors in the vector equation:
$\small{\left(\vec{r}-\vec{u} \right).\left[\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)\right]}$

• The L.H.S involves the dot product of two vectors:
(i) $\small{\left(\vec{r}-\vec{u} \right)}$. We calculated this in (3) above
(ii) $\small{\left[\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)\right]}$. We calculated this in (4) above.

• So the dot product is:
$\small{\left[\left(y_2-y_1 \right)\left(z_3-z_1 \right)-\left(z_2-z_1 \right)\left(x_3-x_1 \right) \right]\left(x-x_1 \right)}$

$\small{~-\left[\left(x_2-x_1 \right)\left(z_3-z_1 \right)-\left(z_2-z_1 \right)\left(x_3-x_1 \right) \right]\left(y-y_1 \right)}$

$\small{~+\left[\left(x_2-x_1 \right)\left(y_3-y_1 \right)-\left(y_2-y_1 \right)\left(x_3-x_1 \right) \right]\left(z-z_1 \right)}$

6. Now we can compile the results:
• The dot product obtained in (5) above, is the L.H.S of the vector equation. It can be written in determinant form.
• The R.H.S of the vector equation is zero.
• So substituting in the vector equation, we get:
$\left|\begin{array}{r}                             x-x_1     &{    y-y_1     }    &{    z-z_1      }    \\ x_2-x_1      &{    y_2-y_1     }    &{   z_2-z_1      }     \\ x_3-x_1      &{    y_3-y_1     }    &{ z_3-z_1        }     \\ \end{array}\right|~=~0$
• This is the Cartesian form

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Now we will see some solved examples

Solved example 27.41
Find the vector and Cartesian equations of the planes that passes through three points (2,5,−3), (−2,−3,5), (5,3,−3)
Solution:
1. Let the points be:
U(2,5,−3), V(−2,−3,5), W(5,3,−3)
• First we check whether the three points are collinear:
(i) For the line UV, the direction ratios are: −4,−8,8
(ii) For the line VW, the direction ratios are: 7,6,−8
(iii) Taking ratios, we get:
$\small{\frac{-4}{7}\ne\frac{-8}{6}\ne\frac{8}{-8}}$
(iv) The ratio is not a constant. That means, the two lines are not parallel.
(v) But point V is common to both the lines. That means, we cannot travel from U to W through V, along a straight line. That means, the three points U, V and W are non collinear.

2. Since the given three points are non collinear, there will be a unique plane.

3. Based on the coordinates, we get the position vectors:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u}}    & {~=~}    &{2\hat{i}+5\hat{j}-3\hat{k}}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{v}}    & {~=~}    &{-2\hat{i}-3\hat{j}+5\hat{k}}
\\ {~\color{magenta}    3    }    &{{}}    &{\vec{w}}    & {~=~}    &{5\hat{i}+3\hat{j}-3\hat{k}}
\\ \end{array}}$

4. Based on the position vectors, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{v}-\vec{u}}    & {~=~}    &{-4\hat{i}-8\hat{j}+8\hat{k}}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{w}-\vec{u}}    & {~=~}    &{3\hat{i}-2\hat{j}}
\\ \end{array}}$

5. So the vector form is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r}-\vec{u} \right).\left[\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left[\vec{r}-\left(2\hat{i}+5\hat{j}-3\hat{k} \right) \right].\left[\left(-4\hat{i}-8\hat{j}+8\hat{k} \right)\times\left(3\hat{i}-2\hat{j} \right)\right]}    & {~=~}    &{0}
\\ \end{array}}$

6. The above vector form can be expanded to get the Cartesian form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r}-\vec{u} \right).\left[\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left[\vec{r}-\left(2\hat{i}+5\hat{j}-3\hat{k} \right) \right].\left[\left(-4\hat{i}-8\hat{j}+8\hat{k} \right)\times\left(3\hat{i}-2\hat{j} \right)\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left[(x-2)\hat{i}+(y-5)\hat{j}+(z+3)\hat{k} \right].\left[16\hat{i}+24\hat{j}+32\hat{k}\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{(16)(x-2)+(24)(y-5)+(32)(z+3)}    & {~=~}    &{0}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{(2)(x-2)+(3)(y-5)+(4)(z+3)}    & {~=~}    &{0}
\\ {~\color{magenta}    6    }    &{{\Rightarrow}}    &{2x - 4+3y-15+4z+12}    & {~=~}    &{0}
\\ {~\color{magenta}    7    }    &{{\Rightarrow}}    &{2x+3y+4z-7}    & {~=~}    &{0}
\\ {~\color{magenta}    8    }    &{{\Rightarrow}}    &{2x+3y+4z}    & {~=~}    &{7}
\\ \end{array}}$

7. If we are asked to find only the Cartesian form, we can avoid all the above six steps and apply the determinant form directly:
$\left|\begin{array}{r}                             x-x_1     &{    y-y_1     }    &{    z-z_1      }    \\ x_2-x_1      &{    y_2-y_1     }    &{   z_2-z_1      }     \\ x_3-x_1      &{    y_3-y_1     }    &{ z_3-z_1        }     \\ \end{array}\right|~=~0$

$\Rightarrow\left|\begin{array}{r}                             x-2     &{    y-5     }    &{    z+3      }    
\\ -2-2      &{    -3-5     }    &{   5+3      }     
\\ 5-2      &{    3-5     }    &{ -3+3        }     \\ \end{array}\right|~=~0$

$\Rightarrow\left|\begin{array}{r}                             x-2     &{    y-5     }    &{    z+3      }    
\\ -4      &{    -8     }    &{   8      }     
\\ 3      &{    -2     }    &{ 0        }     \\ \end{array}\right|~=~0$

$\small{\Rightarrow(x-2)(0+16)-(y-5)(0-24)+(z+3)(8+24)=0}$

$\small{\Rightarrow(x-2)(16)-(y-5)(-24)+(z+3)(32)=0}$

$\small{\Rightarrow(x-2)(2)-(y-5)(-3)+(z+3)(4)=0}$

$\small{\Rightarrow 2x-4+3y-15+4z+12=0}$

$\small{\Rightarrow 2x+3y+4z=7}$

Solved example 27.42
Find the vector and Cartesian equations of the planes that passes through three points
(a) (1,1,−1), (6,4,−5), (−4,−2,3)
(b) (1,1,0), (1,2,1), (−2,2,−1)
Solution:
Part (a):
1. Let the points be:
U(1,1,−1), V(6,4,−5), W(−4,−2,3)
• First we check whether the three points are collinear:
(i) For the line UV, the direction ratios are: 5,3,−4
(ii) For the line VW, the direction ratios are: −10,−6,8
(iii) Taking ratios, we get:
$\small{\frac{5}{-10}=\frac{3}{-6}=\frac{-4}{8}=\frac{-1}{2}}$
(iv) The ratio is a constant. That means, the two lines are parallel.
(v) But point V is common to both the lines. That means, the three points U, V and W are collinear.

2. Since the given three points are collinear, there will be infinite planes passing through them. We cannot write a unique plane.

Part (b):
1. Let the points be:
U(1,1,0), V(1,2,1), W(−2,2,−1)
• First we check whether the three points are collinear:
(i) For the line UV, the direction ratios are: 0,1,1
(ii) For the line VW, the direction ratios are: −3,0,−2
(iii) Taking ratios, we get:
$\small{\frac{0}{-3}\ne\frac{1}{0}\ne\frac{1}{-2}}$
(iv) The ratio is not a constant. That means, the two lines are not parallel.
(v) But point V is common to both the lines. That means, we cannot travel from U to W through V, along a straight line. That means, the three points U, V and W are non collinear.

2. Since the given three points are non collinear, there will be a unique plane.

3. Based on the coordinates, we get the position vectors:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u}}    & {~=~}    &{\hat{i}+\hat{j}}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{v}}    & {~=~}    &{\hat{i}+2\hat{j}+\hat{k}}
\\ {~\color{magenta}    3    }    &{{}}    &{\vec{w}}    & {~=~}    &{-2\hat{i}+2\hat{j}-\hat{k}}
\\ \end{array}}$

4. Based on the position vectors, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{v}-\vec{u}}    & {~=~}    &{\hat{j}+\hat{k}}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{w}-\vec{u}}    & {~=~}    &{-3\hat{i}+\hat{j}-\hat{k}}
\\ \end{array}}$

5. So the vector form is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r}-\vec{u} \right).\left[\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left[\vec{r}-\left(\hat{i}+\hat{j} \right) \right].\left[\left(\hat{j}+\hat{k} \right)\times\left(-3\hat{i}+\hat{j}-\hat{k} \right)\right]}    & {~=~}    &{0}
\\ \end{array}}$

6. The above vector form can be expanded to get the Cartesian form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r}-\vec{u} \right).\left[\left(\vec{v}-\vec{u} \right)\times\left(\vec{w}-\vec{u} \right)\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left[\vec{r}-\left(\hat{i}+\hat{j} \right) \right].\left[\left(\hat{j}+\hat{k} \right)\times\left(-3\hat{i}+\hat{j}-\hat{k} \right)\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left[(x-1)\hat{i}+(y-1)\hat{j}+z\hat{k} \right].\left[-2\hat{i}-3\hat{j}+3\hat{k}\right]}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{(-2)(x-1)+(-3)(y-1)+3z}    & {~=~}    &{0}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{-2x + 2-3y+3+3z}    & {~=~}    &{0}
\\ {~\color{magenta}    6    }    &{{\Rightarrow}}    &{-2x-3y+3z+5}    & {~=~}    &{0}
\\ {~\color{magenta}    7    }    &{{\Rightarrow}}    &{2x+3y-3z}    & {~=~}    &{5}
\\ \end{array}}$

7. If we are asked to find only the Cartesian form, we can avoid all the above six steps and apply the determinant form directly:
$\left|\begin{array}{r}                             x-x_1     &{    y-y_1     }    &{    z-z_1      }    \\ x_2-x_1      &{    y_2-y_1     }    &{   z_2-z_1      }     \\ x_3-x_1      &{    y_3-y_1     }    &{ z_3-z_1        }     \\ \end{array}\right|~=~0$

$\Rightarrow\left|\begin{array}{r}                             x-1     &{    y-1     }    &{    z-0      }    
\\ 1-1      &{    2-1     }    &{   1-0      }     
\\ -2-1      &{    2-1     }    &{ -1-0        }     \\ \end{array}\right|~=~0$

$\Rightarrow\left|\begin{array}{r}                             x-1     &{    y-1     }    &{    z-0      }    
\\ 0      &{    1     }    &{   1      }     
\\ -3      &{    1     }    &{ -1        }     \\ \end{array}\right|~=~0$

$\small{\Rightarrow(x-1)(-1-1)-(y-1)(0+3)+z(0+3)=0}$

$\small{\Rightarrow(x-1)(-2)-(y-1)(3)+z(3)=0}$

$\small{\Rightarrow -2x+2-3y+3+3z=0}$

$\small{\Rightarrow 2x+3y-3z=5}$

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In the next section, we will see intercept form.

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27.10 - Plane Perpendicular To a Given Vector and Passing Through Given Point

In the previous section, we completed a discussion on normal form. In this section, we will see the plane perpendicular to a given vector and passing through a given point.

Some basic details can be written in 3 steps:
1. In fig.27.14 below, $\small{\vec{N}}$ is a vector in 3D space.

Fig.27.14

• Infinite number of planes are possible perpendicular to $\small{\vec{N}}$.

2. Now suppose that, in addition to $\small{\vec{N}}$, we are given a point U also. We want a plane which satisfies two conditions:
(i) The plane must be perpendicular to $\small{\vec{N}}$
(ii) The plane must pass through U.
• Only one plane will satisfy both the above conditions.

3. We are trying to find the vector and Cartesian equations of a plane which satisfies both the conditions.

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The vector equation can be obtained in 5 steps:

1. In fig.27.15 below, the plane is perpendicular to $\small{\vec{N}}$.
• Also, the plane passes through a given point $\small{U\left(x_1,y_1,z_1 \right)}$

Fig.27.15

2. Mark any convenient point $\small{P(x,y,z)}$ on the plane.
• Any vector lying on the plane will be perpendicular to $\small{\vec{N}}$.
• So $\small{\vec{UP}}$ will be perpendicular to $\small{\vec{N}}$.
• So we get: $\small{\vec{UP}.\vec{N}=0}$

3. Now we write the position vectors:
    ♦ $\small{\vec{r}}$ is the position vector of $\small{P}$
    ♦ $\small{\vec{u}}$ is the position vector of $\small{U}$

4. Applying the triangle law of vector addition, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{u}+\vec{UP}}    & {~=~}    &{\vec{r}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\vec{UP}}    & {~=~}    &{\vec{r}-\vec{u}}
\\ \end{array}}$

5. Substituting in (2), we get: $\small{\left(\vec{r}-\vec{u} \right).\vec{N}=0}$
• This is the vector equation of the plane.

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The Cartesian equation can be derived in 4 steps:
1. In the fig.27.15 above, $\small{P}$ is an arbitrary point. So we can write the component form of $\small{\vec{r}}$:
$\small{\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}}$

2. The coordinates of $\small{U~\text{are}~\left(x_1,y_1,z_1 \right)}$. So we can write the component form of $\small{\vec{u}}$:
$\small{\vec{u}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}}$

3. Let the direction ratios of $\small{\vec{N}}$ be: A, B and C
• Then we can write the component form of $\small{\vec{N}}$:
$\small{\vec{N}=A\hat{i}+B\hat{j}+C\hat{k}}$

4. Substituting the above values in the vector equation, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r}-\vec{u} \right).\vec{N}}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left[\left(x\hat{i}+y\hat{j}+z\hat{k} \right)-\left(x_1\hat{i}+y_1\hat{j}+z_1\hat{k} \right) \right].\left[A\hat{i}+B\hat{j}+C\hat{k} \right]}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left[\left(x-x_1 \right)\hat{i}+\left(y-y_1 \right)\hat{j}+\left(z-z_1 \right)\hat{k} \right].\left[A\hat{i}+B\hat{j}+C\hat{k} \right]}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{A\left(x-x_1 \right)+B\left(y-y_1 \right)+C\left(z-z_1 \right)}    & {~=~}    &{0}
\\ \end{array}}$
• This is the Cartesian form.

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Now we will see some solved examples

Solved example 27.39
Find the vector and Cartesian equations of the planes
(a) that passes through the point (1,0,−2) and the normal to the plane is $\small{\hat{i}+\hat{j}-\hat{k}}$
(b) that passes through the point (1,4,6) and the normal to the plane is $\small{\hat{i}-2\hat{j}+\hat{k}}$
Solution:
Part (a):
1. From the given data, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{N}}    & {~=~}    &{\hat{i}+\hat{j}-\hat{k}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{A,~B,~C}    & {~=~}    &{1,~1,~-1}
\\ {~\color{magenta}    3    }    &{}    &{x_1,~y_1,z_1}    & {~=~}    &{1,~0,~-2}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\vec{u}}    & {~=~}    &{\hat{i}+0\hat{j}-2\hat{k}}
\\ \end{array}}$

2. So the vector form is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r}-\vec{u} \right).\vec{N}}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left[\vec{r} - \left(\hat{i}-2\hat{k} \right) \right].\left(\hat{i}+\hat{j}-\hat{k} \right)}    & {~=~}    &{0}
\\ \end{array}}$

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

Part (b):
1. From the given data, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{N}}    & {~=~}    &{\hat{i}-2\hat{j}+\hat{k}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{A,~B,~C}    & {~=~}    &{1,~-2,~1}
\\ {~\color{magenta}    3    }    &{}    &{x_1,~y_1,z_1}    & {~=~}    &{1,~4,~6}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\vec{u}}    & {~=~}    &{\hat{i}+4\hat{j}+6\hat{k}}
\\ \end{array}}$

2. So the vector form is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r}-\vec{u} \right).\vec{N}}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left[\vec{r} - \left(\hat{i}+4\hat{j}+6\hat{k} \right) \right].\left(\hat{i}-2\hat{j}+\hat{k} \right)}    & {~=~}    &{0}
\\ \end{array}}$

3. Also the Cartesian form is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{A\left(x-x_1 \right)+B\left(y-y_1 \right)+C\left(z-z_1 \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{(1)\left(x-1 \right)+(-2)\left(y-4 \right)+(1)\left(z-6 \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{x-1+2y+8+z-6}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{x+2y+z+1}    & {~=~}    &{0}
\\ \end{array}}$

Solved example 27.40
Find the vector and Cartesian equations of the planes that passes through the point (5,2,−4) and perpendicular to the line with direction ratios 2, 3, −1
Solution:
1. From the given data, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{N}}    & {~=~}    &{2\hat{i}+3\hat{j}-\hat{k}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{A,~B,~C}    & {~=~}    &{2,~3,~-1}
\\ {~\color{magenta}    3    }    &{}    &{x_1,~y_1,z_1}    & {~=~}    &{5,~2,~-4}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\vec{u}}    & {~=~}    &{5\hat{i}+2\hat{j}-4\hat{k}}
\\ \end{array}}$

◼ Remarks:
1 (magenta color):
line with direction ratios 2, 3, −1 will be parallel to the vector $\small{2\hat{i}+3\hat{j}-\hat{k}}$

2. So the vector form is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r}-\vec{u} \right).\vec{N}}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left[\vec{r} - \left(5\hat{i}+2\hat{j}-4\hat{k} \right) \right].\left(2\hat{i}+3\hat{j}-\hat{k} \right)}    & {~=~}    &{0}
\\ \end{array}}$

3. Also the Cartesian form is:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{A\left(x-x_1 \right)+B\left(y-y_1 \right)+C\left(z-z_1 \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{(2)\left(x-5 \right)+(3)\left(y-2 \right)+(-1)\left(z-(-4) \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{2x-10+3y-6-z-4}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{2x+3y-z}    & {~=~}    &{20}
\\ \end{array}}$

---

Now we will see an interesting point.  It can be written in 6 steps:
1. Let us first compare the discussions:
• In the discussions in this section, we used vector normal to the plane.
• In the discussions in the previous section, we used the distance of the plane from the origin.

2. So are there two types of plane?
The answer is: No, both planes are related.

3. This can be easily shown in the case of lines in 2D.
• In fig.27.16(a) below, while discussing about the green line, we can use the magenta vector. This magenta vector is perpendicular to the green line.

Fig.27.16

• In fig.b, the same green line is extended to a convenient length. To the extended line, we can easily drop a perpendicular from the origin.

4. Fig.27.17 below shows another example:

Fig.27.17

5. In the same way, in the case of a plane, at first glance, we may get the impression that, it is impossible to drop a perpendicular from the origin. But it can be achieved by extending the plane suitably.

• For any plane, infinite number of perpendicular lines/vectors can be drawn. One of those lines/vectors, will surely pass through the origin.

6. So the planes in the two discussions are not two different types of planes.

---

In the next section, we will see plane passing through three non collinear points.

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27.9 - Solved Examples on Normal Form

In the previous section, we saw equation of a plane in normal form. We saw some solved examples also. In this section, we will see a few more solved examples. Later in this section, we will see the method to find the coordinates of the foot of the perpendicular from the origin.

Solved example 27.35
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $\small{3\hat{i}+5\hat{j}-6\hat{k}}$
Solution:
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. Substituting the known values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{d}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{3\hat{i}+5\hat{j}-6\hat{k}}{\sqrt{3^2 + 5^2 +(-6)^2}} \right)}    & {~=~}    &{7}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{3\hat{i}+5\hat{j}-6\hat{k}}{\sqrt{70}} \right)}    & {~=~}    &{7}
\\ \end{array}}$

Solved example 27.36
Find the Cartesian equation of the following planes:
$\small{\text{(a)}~~\left(\vec{r} \right).\left(\hat{i}+\hat{j}-\hat{k}\right)=2~~~~\text{(b)}~~\left(\vec{r} \right).\left(2\hat{i}+3\hat{j}-4\hat{k}\right)=1}$
$\small{\text{(c)}~~\left(\vec{r} \right).\left[(s-2t)\hat{i}+(3-t)\hat{j}+(2s+t)\hat{k}\right]=15}$
Solution:
In each case, the scalar multiplication will give the Cartesian equation.
Part (a):
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r} \right).\left(\hat{i}+\hat{j}-\hat{k}\right)}    & {~=~}    &{2}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(\hat{i}+\hat{j}-\hat{k}\right)}    & {~=~}    &{2}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{x+y-z}    & {~=~}    &{2}
\\ \end{array}}$

Part (b):
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r} \right).\left(2\hat{i}+3\hat{j}-4\hat{k}\right)}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(2\hat{i}+3\hat{j}-4\hat{k}\right)}    & {~=~}    &{1}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{2x+3y-4z}    & {~=~}    &{1}
\\ \end{array}}$

Part (c):
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r} \right).\left[(s-2t)\hat{i}+(3-t)\hat{j}+(2s+t)\hat{k}\right]}    & {~=~}    &{15}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left[(s-2t)\hat{i}+(3-t)\hat{j}+(2s+t)\hat{k}\right]}    & {~=~}    &{15}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{(s-2t)x+(3-t)y+(2s+t)z}    & {~=~}    &{15}
\\ \end{array}}$

---

Now we will see the method to find the coordinates of the foot of the perpendicular.
• Consider the fig.27.13 that we saw in the previous section. For convenience, it is shown again below:

Fig.27.13

• We know that, ON is the perpendicular drawn from the origin. We want a method to find the coordinates of N. It can be written in 5 steps:
1. Write the equation of the given plane in the vector form:
$\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$
2. From the above equation, we get the unit vector:
$\small{\frac{\vec{n}}{\left|\vec{n} \right|}}$
• We get the distance $\small{d}$from the origin also.
3. Write down the coefficients of the above unit vector. These coefficients are the direction cosines of $\small{\vec{ON}}$
4. The magnitude of $\small{\vec{ON}}$ is $\small{d}$.Thus we get:
• x coordinate of N = $\small{d ~\times ~\text{first direction cosine}}$
• y coordinate of N = $\small{d ~\times ~\text{second direction cosine}}$
• z coordinate of N = $\small{d ~\times ~\text{third direction cosine}}$
5. Note that, $\small{\vec{ON}}$ is simply, the position vector of N. In the above steps, we are actually calculating the components of $\small{\vec{ON}}$.
• We know that:
    ♦ the coefficients of the components
    ♦ of the position vector of a point
    ♦ are the coordinates of that point 

---

Now we will see some solved examples:

Solved example 27.37
In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
$\small{\text{(a)}~~2x+3y+4z-12=0~~~~\text{(b)}~~3y+4z-6=0}$
$\small{\text{(c)}~~x+y+z=1~~~~\text{(d)}~~5y+8=0}$
$\small{\text{(e)}~~2x-3y+4z-6=0}$
Solution:
Part (a):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. So we convert the given Cartesian form to vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2x+3y+4z-12}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{2x+3y+4z}    & {~=~}    &{12}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(2\hat{i}+3\hat{j}+4\hat{k}\right)}    & {~=~}    &{12}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{2\hat{i}+3\hat{j}+4\hat{k}}{\sqrt{2^2 + 3^2 + 4^2}}}    & {~=~}    &{\frac{12}{\sqrt{2^2 + 3^2 + 4^2}}}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{2\hat{i}+3\hat{j}+4\hat{k}}{\sqrt{29}}}    & {~=~}    &{\frac{12}{\sqrt{29}}}
\\ \end{array}}$

3. Let N be the foot of the perpendicular drawn from the origin.
• Then the direction cosines of $\small{\vec{ON}}$ are:
$\small{l=\frac{2}{\sqrt{29}},~m=\frac{3}{\sqrt{29}},~n=\frac{4}{\sqrt{29}}}$
• Also, $\small{\left|\vec{ON} \right| = \frac{12}{\sqrt{29}}}$

4. Now we can write the coordinates:
• x-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)l = \left(\frac{12}{\sqrt{29}} \right)\left(\frac{2}{\sqrt{29}} \right) = \frac{24}{29}}$   
• y-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)m = \left(\frac{12}{\sqrt{29}} \right)\left(\frac{3}{\sqrt{29}} \right) = \frac{36}{29}}$   
• z-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)n = \left(\frac{12}{\sqrt{29}} \right)\left(\frac{4}{\sqrt{29}} \right) = \frac{48}{29}}$

Part (b):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. So we convert the given Cartesian form to vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{3y+4z-6}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{3y+4z}    & {~=~}    &{6}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(0\hat{i}+3\hat{j}+4\hat{k}\right)}    & {~=~}    &{6}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{0\hat{i}+3\hat{j}+4\hat{k}}{\sqrt{3^2 + 4^2}}}    & {~=~}    &{\frac{6}{\sqrt{3^2 + 4^2}}}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{0\hat{i}+3\hat{j}+4\hat{k}}{5}}    & {~=~}    &{\frac{6}{5}}
\\ \end{array}}$

3. Let N be the foot of the perpendicular drawn from the origin.
• Then the direction cosines of $\small{\vec{ON}}$ are:
$\small{l=0,~m=\frac{3}{5},~n=\frac{4}{5}}$
• Also, $\small{\left|\vec{ON} \right| = \frac{6}{5}}$

4. Now we can write the coordinates:
• x-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)l = \left(\frac{6}{5} \right)\left(0 \right) = 0}$   
• y-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)m = \left(\frac{6}{5} \right)\left(\frac{3}{5} \right) = \frac{18}{25}}$   
• z-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)n = \left(\frac{6}{5} \right)\left(\frac{4}{5} \right) = \frac{24}{25}}$

Part (c):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. So we convert the given Cartesian form to vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x+y+z}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(\hat{i}+\hat{j}+\hat{k}\right)}    & {~=~}    &{1}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{1^2 + 1^2 + 1^2}}}    & {~=~}    &{\frac{1}{\sqrt{1^2 + 1^2 + 1^2}}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}}    & {~=~}    &{\frac{1}{\sqrt{3}}}
\\ \end{array}}$

3. Let N be the foot of the perpendicular drawn from the origin.
• Then the direction cosines of $\small{\vec{ON}}$ are:
$\small{l=,\frac{1}{\sqrt{3}}~m=\frac{1}{\sqrt{3}},~n=\frac{1}{\sqrt{3}}}$
• Also, $\small{\left|\vec{ON} \right| = \frac{1}{\sqrt{3}}}$

4. Now we can write the coordinates:
• x-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)l = \left(\frac{1}{\sqrt{3}} \right)\left(\frac{1}{\sqrt{3}} \right) = \frac{1}{3}}$   
• y-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)m = \left(\frac{1}{\sqrt{3}} \right)\left(\frac{1}{\sqrt{3}} \right) = \frac{1}{3}}$   
• z-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)n = \left(\frac{1}{\sqrt{3}} \right)\left(\frac{1}{\sqrt{3}} \right) = \frac{1}{3}}$

Part (d):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. So we convert the given Cartesian form to vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{5y+8}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{5y}    & {~=~}    &{-8}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(0\hat{i}+5\hat{j}+0\hat{k}\right)}    & {~=~}    &{-8}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(0\hat{i}-5\hat{j}+0\hat{k}\right)}    & {~=~}    &{8}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{0\hat{i}-5\hat{j}+0\hat{k}}{\sqrt{0^2 + 5^2 + 0^2}}}    & {~=~}    &{\frac{8}{\sqrt{0^2 + 5^2 + 0^2}}}
\\ {~\color{magenta}    6    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{0\hat{i}-5\hat{j}+0\hat{k}}{5}}    & {~=~}    &{\frac{8}{5}}
\\ {~\color{magenta}    6    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(0\hat{i}-\hat{j}+0\hat{k} \right)}    & {~=~}    &{\frac{8}{5}}
\\ \end{array}}$

3. Let N be the foot of the perpendicular drawn from the origin.
• Then the direction cosines of $\small{\vec{ON}}$ are:
$\small{l=0,~m=-1,~n=0}$
• Also, $\small{\left|\vec{ON} \right| = \frac{8}{5}}$

4. Now we can write the coordinates:
• x-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)l = \left(\frac{8}{5} \right)\left(0 \right) = 0}$   
• y-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)m = \left(\frac{8}{5} \right)\left(-1 \right) = \frac{-8}{5}}$   
• z-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)n = \left(\frac{8}{5} \right)\left(0 \right) = 0}$

Part (e):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. So we convert the given Cartesian form to vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2x-3y+4z-6}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{2x-3y+4z}    & {~=~}    &{6}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(2\hat{i}-3\hat{j}+4\hat{k}\right)}    & {~=~}    &{6}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{2^2 + 3^2 + 4^2}}}    & {~=~}    &{\frac{6}{\sqrt{2^2 + 3^2 + 4^2}}}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{29}}}    & {~=~}    &{\frac{6}{\sqrt{29}}}
\\ \end{array}}$

3. Let N be the foot of the perpendicular drawn from the origin.
• Then the direction cosines of $\small{\vec{ON}}$ are:
$\small{l=\frac{2}{\sqrt{29}},~m=\frac{-3}{\sqrt{29}},~n=\frac{4}{\sqrt{29}}}$
• Also, $\small{\left|\vec{ON} \right| = \frac{6}{\sqrt{29}}}$

4. Now we can write the coordinates:
• x-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)l = \left(\frac{6}{\sqrt{29}} \right)\left(\frac{2}{\sqrt{29}} \right) = \frac{12}{29}}$   
• y-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)m = \left(\frac{6}{\sqrt{29}} \right)\left(\frac{-3}{\sqrt{29}} \right) = \frac{-18}{29}}$   
• z-coordinate of N = $\small{\left(\left|\vec{ON} \right| \right)n = \left(\frac{6}{\sqrt{29}} \right)\left(\frac{4}{\sqrt{29}} \right) = \frac{24}{29}}$

Solved example 27.38
If O be the origin and the coordinates of P be (1,2,−3), then find the equation of the plane passing through P and perpendicular to OP.
Solution:
Part (a):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• So we want $\small{\hat{n}~\text{and}~d}$

2. The plane should be perpendicular to OP. That means, $\small{\vec{OP}}$ is normal to the plane
• $\small{\vec{OP}}$ is the position vector of P, which is: $\small{\hat{i}+2\hat{j}-3\hat{k}}$
• So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{n}}    & {~=~}    &{\frac{\vec{OP}}{\left|\vec{OP} \right|}}
\\ {~\color{magenta}    2    }    &{{}}    &{}    & {~=~}    &{\frac{\hat{i}+2\hat{j}-3\hat{k}}{\sqrt{1^2 + 2^2 + (-3)^2}}}
\\ {~\color{magenta}    3    }    &{{}}    &{}    & {~=~}    &{\frac{\hat{i}+2\hat{j}-3\hat{k}}{\sqrt{14}}}
\\ \end{array}}$

3. d= distance OP = $\small{\left|\vec{OP} \right|~=~\sqrt{14}}$

4. Substituting in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\hat{n}}    & {~=~}    &{d}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(\frac{\hat{i}+2\hat{j}-3\hat{k}}{\sqrt{14}} \right)}    & {~=~}    &{\sqrt{14}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(\hat{i}+2\hat{j}-3\hat{k} \right)}    & {~=~}    &{14}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{x+2y-3z}    & {~=~}    &{14}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{x+2y-3z-14}    & {~=~}    &{0}
\\ \end{array}}$

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In the next section, we will see plane perpendicular to a given vector and passing through a given point.

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27.8 - Equation of A Plane in Normal Form

In the previous section, we completed a discussion on the shortest distance between two skew lines. In this section, we will see plane.

In 3D space, there are infinite planes in infinite orientations. But we can obtain a unique plane, if any one of the following three conditions are satisfied:
(i) Drop a perpendicular from the origin O, onto a plane. If the length of that perpendicular is fixed, then we get a unique plane.
(ii) A plane is passing through a given point $\small{\left(x_1,y_1,z_1 \right)}$. Also, the plane is perpendicular to a given vector or line. Then we get a unique plane.
(iii) The plane passes through three given non collinear points.

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Now we will try to derive the vector equation of a plane. It can be done in 4 steps:
1. In fig.27.13 below, a plane passes through 3 points A, B and C.

Fig.27.13

• A perpendicular is dropped from O, onto the plane. The foot of the perpendicular is N. Let the length ON be $\small{d\,(d\ne0)}$.

• Suppose that, $\small{\hat{n}}$ is the unit vector perpendicular to the plane.
Then $\small{\vec{ON}=d\,\hat{n}}$

2. Mark any convenient point P on the plane.
• ON is perpendicular to the plane. So all lines on the plane will be perpendicular to ON. Obviously, NP will be perpendicular to ON.
• We can write: $\small{\vec{NP}~\text{and}~\vec{ON}}$ are perpendicular to each other.
• So we get: $\small{\vec{NP}.\vec{ON}=0}$

3. Let $\small{\vec{r}}$ be the position vector of $\small{P}$
Applying triangle law of vector addition, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{ON}+\vec{NP}}    & {~=~}    &{\vec{OP}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\vec{NP}}    & {~=~}    &{\vec{OP}-\vec{ON}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\vec{NP}}    & {~=~}    &{\vec{r}-d\,\hat{n}}
\\ \end{array}}$

4. Substituting in (2), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{NP}.\vec{ON}}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r}-d\,\hat{n} \right).d\,\hat{n}}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r}-d\,\hat{n} \right).\hat{n}}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\vec{r}.\hat{n}-d\left(\hat{n}.\hat{n} \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    5    }    &{{\Rightarrow}}    &{\vec{r}.\hat{n}-d}    & {~=~}    &{0}
\\ {~\color{magenta}    6    }    &{{\Rightarrow}}    &{\vec{r}.\hat{n}}    & {~=~}    &{d}
\\ \end{array}}$
• This is the vector form of the equation of the plane.
◼ Remarks:
• 3 (magenta color): We are able to obtain this step from 2 (magenta color) because, $\small{d}$ is the distance from origin. We wrote that, it is not zero.
• 5 (magenta color): Here we apply the fact that, $\small{\left(\hat{n}.\hat{n} \right)}$ is 1.

---

Now we will derive the Cartesian form. It can be done in 3 steps:
1. In fig.27.13 above, P is an arbitrary point. So we can write the component form of $\small{\vec{r}}$:
$\small{\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}}$

2. If $\small{l,~m~\text{and}~n}$ are the direction cosines of $\small{\hat{n}}$, then the component form of $\small{\hat{n}}$ can be written as:
$\small{\hat{n}=l\hat{i}+m\hat{j}+n\hat{k}}$

3. Substituting the above two results in the vector form, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\hat{n}}    & {~=~}    &{d}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(l\hat{i}+m\hat{j}+n\hat{k} \right)}    & {~=~}    &{d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{lx + my + nz}    & {~=~}    &{d}
\\ \end{array}}$
• This is the Cartesian form.

---

Suppose that, instead of direction cosines $\small{l,~m~\text{and}~n}$ of $\small{\hat{n}}$, we are given the direction ratios $\small{a,~b~\text{and}~c}$ of the vector perpendicular to the plane. Then we can derive the Cartesian form of the plane in 4 steps:
1. We have the basic vector form:
$\small{\vec{r}.\hat{n}=d}$

2. In the present case,
    ♦ We do not have $\small{\hat{n}}$, which is the unit vector perpendicular to the plane
    ♦ But we do have $\small{\vec{n}}$, which is the vector perpendicular to the plane
• We have $\small{\vec{n}}$ because, we are given the direction ratios $\small{a,~b~\text{and}~c}$ of the vector perpendicular to the plane.
• We can write: $\small{\vec{n}=a\hat{i}+b\hat{j}+c\hat{k}}$

3. But we can obtain $\small{\hat{n}}$ easily:
$\small{\hat{n}=\frac{\vec{n}}{\left|\vec{n} \right|}}$

4. So substituting in the basic vector form, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\hat{n}}    & {~=~}    &{d}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(\frac{a\hat{i}+b\hat{j}+c\hat{k}}{\sqrt{a^2 + b^2 + c^2}} \right)}    & {~=~}    &{d}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{ax + by + cz}    & {~=~}    &{d\,\sqrt{a^2 + b^2 + c^2}}
\\ \end{array}}$
• This is the Cartesian form when the direction ratios are given.

---

Let us see some solved examples

Solved example 27.30
Find the distance of the plane $\small{2x-3y+4z-6=0}$ from the origin
Solution:
1. The given equation can be rearranged as:
$\small{2x-3y+4z=6}$

2. Writing this in vector form, we get:
$\small{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(2\hat{i}-3\hat{j}+4\hat{k} \right)=6}$

3. In the L.H.S, we have the dot product of two vectors.
• The first vector is $\small{\vec{r}}$
• The second vector is not $\small{\hat{n}}$ because, its magnitude is not one.
• The second vector is $\small{\vec{n}}$, and its magnitude is $\small{\sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{29}}$

4. So the vector equation in (2) is comparable to:
$\small{\vec{r}.\vec{n}= 6}$
• But the standard form of a plane is: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$
• That is., to obtain 'd' on the R.H.S, $\small{\vec{n}}$ must be divided by $\small{\left|\vec{n} \right|}$
• But then, the right side also must be divided by $\small{\left|\vec{n} \right|}$

5. Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\vec{n}}    & {~=~}    &{6}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{6}{\left|\vec{n} \right|}~=~d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{6}{\sqrt{29}}~=~d}
\\ \end{array}}$

Solved example 27.31
Find the distance of the plane $\small{3x-4y+12z-3=0}$ from the origin
Solution:
1. The given equation can be rearranged as:
$\small{3x-4y+12z=3}$

2. Writing this in vector form, we get:
$\small{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(3\hat{i}-4\hat{j}+12\hat{k} \right)=3}$

3. In the L.H.S, we have the dot product of two vectors.
• The first vector is $\small{\vec{r}}$
• The second vector is not $\small{\hat{n}}$ because, its magnitude is not one.
• The second vector is $\small{\vec{n}}$, and its magnitude is $\small{\sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{169}=13}$

4. So the vector equation in (2) is comparable to:
$\small{\vec{r}.\vec{n}= 3}$
• But the standard form of a plane is: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$
• That is., to obtain 'd' on the R.H.S, $\small{\vec{n}}$ must be divided by $\small{\left|\vec{n} \right|}$
• But then, the right side also must be divided by $\small{\left|\vec{n} \right|}$

5. Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\vec{n}}    & {~=~}    &{3}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{3}{\left|\vec{n} \right|}~=~d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{3}{13}~=~d}
\\ \end{array}}$

Solved example 27.32
Find the vector equation of the plane which is at a distance of $\small{\frac{6}{\sqrt{29}}}$ from the origin and its normal vector from the origin is $\small{2\hat{i}-3\hat{j}+4\hat{k}}$. Also find its Cartesian form.
Solution:
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$

2. In our present problem, we are given the normal vector: $\small{2\hat{i}-3\hat{j}+4\hat{k}}$
• This given normal vector is not $\small{\hat{n}}$ because, its magnitude is not one.
• This given normal vector is $\small{\vec{n}}$, with magnitude $\small{\sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{29}}$

3. We can obtain $\small{\hat{n}}$ as:
$\small{\hat{n}=\frac{\vec{n}}{\left|\vec{n} \right|}=\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{29}}}$

4. Substituting in (1), we get the vector equation of the plane as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\hat{n}}    & {~=~}    &{d}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{29}} \right)}    & {~=~}    &{\frac{6}{\sqrt{29}}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{2}{\sqrt{29}}\,\hat{i}~-~\frac{3}{\sqrt{29}}\,\hat{j}~+~\frac{4}{\sqrt{29}}\,\hat{k} \right)}    & {~=~}    &{\frac{6}{\sqrt{29}}}
\\ \end{array}}$

5. To write the Cartesian form, we must find the actual scalar multiplication:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r} \right).\left(\frac{2}{\sqrt{29}}\,\hat{i}~-~\frac{3}{\sqrt{29}}\,\hat{j}~+~\frac{4}{\sqrt{29}}\,\hat{k} \right)}    & {~=~}    &{\frac{6}{\sqrt{29}}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(\frac{2}{\sqrt{29}}\,\hat{i}~-~\frac{3}{\sqrt{29}}\,\hat{j}~+~\frac{4}{\sqrt{29}}\,\hat{k} \right)}    & {~=~}    &{\frac{6}{\sqrt{29}}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{2x}{\sqrt{29}}-\frac{3y}{\sqrt{29}}+\frac{4z}{\sqrt{29}}}    & {~=~}    &{\frac{6}{\sqrt{29}}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{2x - 3y + 4z - 6}    & {~=~}    &{0}
\\ \end{array}}$

Solved example 27.33
Find the direction cosines of the unit vector perpendicular to the plane $\small{\left(\vec{r} \right).\left(6\hat{i}-3\hat{j}-2\hat{k} \right)+1 = 0}$ passing through the origin.
Solution:
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. The given vector equation can be rearranged as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(\vec{r} \right).\left(6\hat{i}-3\hat{j}-2\hat{k} \right)+1}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(6\hat{i}-3\hat{j}-2\hat{k} \right)}    & {~=~}    &{-1}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{(-1)\left(\vec{r} \right).\left(6\hat{i}-3\hat{j}-2\hat{k} \right)}    & {~=~}    &{(-1)(-1)}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(-6\hat{i}+3\hat{j}+2\hat{k} \right)}    & {~=~}    &{1}
\\ \end{array}}$

3.In the L.H.S, we have the dot product of two vectors.
• The first vector is $\small{\vec{r}}$
• The second vector is not $\small{\hat{n}}$ because, its magnitude is not one.
• The second vector is $\small{\vec{n}}$, and its magnitude is $\small{\sqrt{(-6)^2 + (3)^2 + (2)^2} = \sqrt{49} = 7}$

4. So the vector equation in (2) is comparable to:
$\small{\vec{r}.\vec{n}= 1}$
• But the general form of a plane is: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$
• That is., to obtain 'd' on the R.H.S, $\small{\vec{n}}$ must be divided by $\small{\left|\vec{n} \right|}$
• But then, the right side also must be divided by $\small{\left|\vec{n} \right|}$

5. Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\vec{n}}    & {~=~}    &{-1}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{-1}{\left|\vec{n} \right|}~=~d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{-6\hat{i}+3\hat{j}+2\hat{k}}{7} \right)}    & {~=~}    &{\frac{-1}{7}~=~d}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{-6\hat{i}}{7}+\frac{3\hat{j}}{7}+\frac{2\hat{k}}{7} \right)}    & {~=~}    &{\frac{-1}{7}~=~d}
\\ \end{array}}$

6. From the above result, it is clear that:
$\small{\hat{n}=\frac{-6\hat{i}}{7}+\frac{3\hat{j}}{7}+\frac{2\hat{k}}{7}}$

7. For any unit vector, the coefficients are the direction cosines.
• So in our present case, the required direction cosines are:
$\small{\frac{-6}{7},~\frac{3}{7},~\frac{2}{7}}$

Solved example 27.34
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin
$\small{\text{(a)}~~z = 2~~\text{(b)}~~x+y+z=1}$
$\small{\text{(c)}~~2x+3y-z =5~~\text{(d)}~~5y+8=0}$
Solution:
Part (a):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. The given Cartesian equation can be converted into vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{z}    & {~=~}    &{2}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(0\hat{i}+0\hat{j}+\hat{k} \right)}    & {~=~}    &{2}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(0\hat{i}+0\hat{j}+\hat{k} \right)}    & {~=~}    &{2}
\\ \end{array}}$

3. In the L.H.S, we have the dot product of two vectors.
• The first vector is $\small{\vec{r}}$
• The second vector is $\small{\hat{n}}$ because, its magnitude is one.

4. So the vector equation in (2) is comparable to the general form:
$\small{\vec{r}.\hat{n}=d}$

5. Thus we get:
$\small{\hat{n}=0\hat{i}+0\hat{j}+\hat{k}}$

6. For any unit vector, the coefficients are the direction cosines. So in our present case, the required direction cosines are:
$\small{0,~0,~1}$

7. The vector equation obtained in (2) is comparable to the general form. That means, the quantity in the R.H.S is d.
• Therefore, the distance of the plane from the origin is 2 units.

Part (b):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. The given Cartesian equation can be converted into vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x+y+z}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(\hat{i}+\hat{j}+\hat{k} \right)}    & {~=~}    &{1}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\hat{i}+\hat{j}+\hat{k} \right)}    & {~=~}    &{1}
\\ \end{array}}$

3. In the L.H.S, we have the dot product of two vectors.
• The first vector is $\small{\vec{r}}$
• The second vector is not $\small{\hat{n}}$ because, its magnitude is not one.
• The second vector is $\small{\vec{n}}$, and its magnitude is $\small{\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}}$

4. So the vector equation in (2) is comparable to:
$\small{\vec{r}.\vec{n}= 1}$
• But the standard form of a plane is: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$
• That is., to obtain 'd' on the R.H.S, $\small{\vec{n}}$ must be divided by $\small{\left|\vec{n} \right|}$
• But then, the right side also must be divided by $\small{\left|\vec{n} \right|}$

5. Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\vec{n}}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{1}{\left|\vec{n} \right|}~=~d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} \right)}    & {~=~}    &{\frac{1}{\sqrt{3}}~=~d}
\\ \end{array}}$
• Thus we get:
$\small{\hat{n}=\frac{\hat{i}}{\sqrt{3}}+\frac{\hat{j}}{\sqrt{3}}+\frac{\hat{k}}{\sqrt{3}}}$

6. For any unit vector, the coefficients are the direction cosines. So in our present case, the required direction cosines are:
$\small{\frac{1}{\sqrt{3}},~\frac{1}{\sqrt{3}},~\frac{1}{\sqrt{3}}}$

7. From (5) we get d. We can write:
Distance of the plane from the origin is $\small{\frac{1}{\sqrt{3}}}$ units.

Part (c):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. The given Cartesian equation can be converted into vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2x+3y-z}    & {~=~}    &{5}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(2\hat{i}+3\hat{j}-\hat{k} \right)}    & {~=~}    &{5}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(2\hat{i}+3\hat{j}-\hat{k} \right)}    & {~=~}    &{5}
\\ \end{array}}$

3. In the L.H.S, we have the dot product of two vectors.
• The first vector is $\small{\vec{r}}$
• The second vector is not $\small{\hat{n}}$ because, its magnitude is not one.
• The second vector is $\small{\vec{n}}$, and its magnitude is $\small{\sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{14}}$

4. So the vector equation in (2) is comparable to:
$\small{\vec{r}.\vec{n}= 5}$
• But the standard form of a plane is: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$
• That is., to obtain 'd' on the R.H.S, $\small{\vec{n}}$ must be divided by $\small{\left|\vec{n} \right|}$
• But then, the right side also must be divided by $\small{\left|\vec{n} \right|}$

5. Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\vec{n}}    & {~=~}    &{5}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{5}{\left|\vec{n} \right|}~=~d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{2\hat{i}+3\hat{j}-\hat{k}}{\sqrt{14}} \right)}    & {~=~}    &{\frac{5}{\sqrt{14}}~=~d}
\\ \end{array}}$
• Thus we get:
$\small{\hat{n}=\frac{2\hat{i}}{\sqrt{14}}+\frac{3\hat{j}}{\sqrt{14}}-\frac{\hat{k}}{\sqrt{14}}}$

6. For any unit vector, the coefficients are the direction cosines. So in our present case, the required direction cosines are:
$\small{\frac{2}{\sqrt{14}},~\frac{3}{\sqrt{14}},~\frac{-1}{\sqrt{14}}}$

7. From (5) we get d. We can write:
Distance of the plane from the origin is $\small{\frac{5}{\sqrt{14}}}$ units.

Part (d):
1. The general vector form is: $\small{\vec{r}.\hat{n}=d}$
• This is same as: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$

2. The given Cartesian equation can be converted into vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{5y}    & {~=~}    &{-8}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left(0\hat{i}+5\hat{j}+0\hat{k} \right)}    & {~=~}    &{-8}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(5\hat{j} \right)}    & {~=~}    &{-8}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(-5\hat{j} \right)}    & {~=~}    &{8}
\\ \end{array}}$

3. In the L.H.S, we have the dot product of two vectors.
• The first vector is $\small{\vec{r}}$
• The second vector is not $\small{\hat{n}}$ because, its magnitude is not one.
• The second vector is $\small{\vec{n}}$, and its magnitude is 5

4. So the vector equation in (2) is comparable to:
$\small{\vec{r}.\vec{n}= 8}$
• But the standard form of a plane is: $\small{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)=d}$
• That is., to obtain 'd' on the R.H.S, $\small{\vec{n}}$ must be divided by $\small{\left|\vec{n} \right|}$
• But then, the right side also must be divided by $\small{\left|\vec{n} \right|}$

5. Thus we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{r}.\vec{n}}    & {~=~}    &{8}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{\vec{n}}{\left|\vec{n} \right|} \right)}    & {~=~}    &{\frac{8}{\left|\vec{n} \right|}~=~d}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\vec{r} \right).\left(\frac{-5\hat{j}}{5} \right)}    & {~=~}    &{\frac{8}{5}~=~d}
\\ \end{array}}$
• Thus we get:
$\small{\hat{n}=\frac{-5\hat{j}}{5}=-\hat{j}=0\hat{i}-\hat{j}+0\hat{k}}$

6. For any unit vector, the coefficients are the direction cosines. So in our present case, the required direction cosines are:
$\small{0,~-1,~0}$

7. From (5) we get d. We can write:
Distance of the plane from the origin is $\small{\frac{8}{5}}$ units.

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In the next section, we will see a few more solved examples. We will see coordinates of the foot of the perpendicular from origin also.

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