27.12 - Intercept Form of The Equation of a Plane

In the previous section, we saw the plane passing through three non collinear points. In this section, we will see intercept form.

Some basic details can be written in 2 steps:
1. Consider our familiar 2D plane with the x and y axes.
We can write three facts:
(i) A line parallel to the x-axis will never intersect the x-axis.
• But that line will definitely intersect the y-axis.
(ii) A line parallel to the y-axis will never intersect the y-axis.
• But that line will definitely intersect the x-axis.
(iii) A line parallel to neither x-axis nor y-axis will definitely intersect both the axes.

2. In a similar way, for the 3D space, we can write four facts:
(i) A plane parallel to the XOY plane will never intersect the x or y-axes
• But that plane will definitely intersect the z-axis.
(ii)  A plane parallel to the XOZ plane will never intersect the x or z-axes
• But that plane will definitely intersect the y-axis.
(iii) A plane parallel to the YOZ plane will never intersect the y and z-axes
• But that plane will definitely intersect the x-axis.
(iv) A plane which is not parallel to any one of XOY, XOZ or YOZ planes, will definitely intersect all the three axes.

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In this section, we consider the plane which is written in 2(iv) above. It’s Cartesian  equation can be derived in 4 steps:
1. Let the equation of the plane be:
$\small{Ax + By + Cz + D = 0,~(D \ne 0)}$

2. Let the plane make intercepts a,b,c on the x, y and z-axes respectively. This is shown in the fig.27.20 below:

Fig.27.20

• Based on the intercepts, we can write:
    ♦ The plane meets the x-axis at P(a,0,0)
    ♦ The plane meets the y-axis at Q(0,b,0)
    ♦ The plane meets the z-axis at R(0,0,c)

3. Any point on the plane will satisfy the equation of the plane.
• Substituting the coordinates of P in the equation of the plane, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{Ax + By + Cz + D }    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{A(a) + B(0) + C(0) + D }    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{A(a)}    & {~=~}    &{-D}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{A}    & {~=~}    &{\frac{-D}{a}}
\\ \end{array}}$

• Substituting the coordinates of Q in the equation of the plane, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{Ax + By + Cz + D }    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{A(0) + B(b) + C(0) + D }    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{B(b)}    & {~=~}    &{-D}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{B}    & {~=~}    &{\frac{-D}{b}}
\\ \end{array}}$

• Substituting the coordinates of R in the equation of the plane, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{Ax + By + Cz + D }    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{A(0) + B(0) + C(c) + D }    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{C(c)}    & {~=~}    &{-D}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{C}    & {~=~}    &{\frac{-D}{c}}
\\ \end{array}}$

4. Substituting the above values of A, B and C in the original equation, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{Ax + By + Cz + D }    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(\frac{-D}{a} \right)x+ \left(\frac{-D}{b} \right)y+ \left(\frac{-D}{c} \right)z+ D }    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}-1}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}}    & {~=~}    &{1}
\\ \end{array}}$
• This is the equation of the plane in intercept form.

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Now we will see some solved examples

Solved example 27.43
Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and z-axes respectively.
Solution:
1. The general intercept form is:
$\small{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1}$
• Where a, b and c are the intercepts on the x, y and z-axes respectively.

2. Substituting the given intercepts, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{x}{2}+\frac{y}{3}+\frac{z}{4}}    & {~=~}    &{1}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{x}{2}(12)+\frac{y}{3}(12)+\frac{z}{4}(12)}    & {~=~}    &{12}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{6x+4y+3z}    & {~=~}    &{12}
\\ \end{array}}$

◼ Remarks:
• 3 (magenta color): Here we multiply throughout by the L.C.M of 2, 3 and 4, which is 12.

Solved example 27.44
Find the intercepts cut off by the plane 2x + y −z = 5.
Solution:
1. The given equation can be divided throughout by 5 and rearranged as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{2x}{5}+\frac{y}{5}+\frac{(-1)z}{5}}    & {~=~}    &{\frac{5}{5}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{x}{5/2}+\frac{y}{5}+\frac{z}{(-5)}}    & {~=~}    &{1}
\\ \end{array}}$

2. Now the equation is in the general intercept form.
So the intercepts are:
5/2, 5, −5

3. Fig.27.21 below shows the plane and the intercepts

Fig.27.21

Solved example 27.45
Find the equation of the plane with intercept 3 on the y axis and parallel to ZOX plane.
Solution:
1. The general intercept form is:
$\small{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1}$
• Where a, b and c are the intercepts on the x, y and z-axes respectively.

2. In our present case, the plane is parallel to the ZOX plane. For such a plane, there are no intercepts on the z and x axes. So we can ignore the first and last terms on the L.H.S.

3. Therefore, the required equation is:
$\small{\frac{y}{3}=1}$, which is same as y=3

Solved example 27.46
Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then
$\small{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{p^2}}$.
Solution:
1. Based on the general intercept form, the equation of the given plane is:
$\small{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1}$

2. This can be rearranged as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{(abc)x}{a}+\frac{(abc)y}{b}+\frac{(abc)z}{c}}    & {~=~}    &{abc}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{(bc)x+(ac)y+(ab)z}    & {~=~}    &{abc}
\\ \end{array}}$

3. The above result can be written in vector form:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{(bc)x+(ac)y+(ab)z}    & {~=~}    &{abc}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\left(x\hat{i}+y\hat{j}+z\hat{k} \right).\left[(bc)\hat{i}+(ac)\hat{j}+(ab)\hat{k} \right]}    & {~=~}    &{abc}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\left(\hat{r} \right).\left[(bc)\hat{i}+(ac)\hat{j}+(ab)\hat{k} \right]}    & {~=~}    &{abc}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\left(\hat{r} \right).\frac{\left[(bc)\hat{i}+(ac)\hat{j}+(ab)\hat{k} \right]}{\sqrt{b^2 c^2 + a^2 c^2 + a^2 b^2}}}    & {~=~}    &{\frac{abc}{\sqrt{b^2 c^2 + a^2 c^2 + a^2 b^2}}}
\\ \end{array}}$

◼ Remarks:
4 (magenta color): Here, in the L.H.S, we do a division by the magnitude. This is to obtain the unit vector $\small{\hat{n}}$. But then, the R.H.S must also be divided by the same magnintude. 

4. The above result in (3), is in the form: $\small{\vec{r}\hat{n}=d}$
• Where:
    ♦ $\small{\hat{n}}$ is the unit vector normal to the plane
    ♦ $\small{d}$ is distance of the plane from the origin

5. So the R.H.S of the result in (3), is the distance of the plane from the origin. But this distance is given to us. It is p. So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{abc}{\sqrt{b^2 c^2 + a^2 c^2 + a^2 b^2}}}    & {~=~}    &{p}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{a^2 b^2 c^2}{b^2 c^2 + a^2 c^2 + a^2 b^2}}    & {~=~}    &{p^2}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{b^2 c^2 + a^2 c^2 + a^2 b^2}{a^2 b^2 c^2}}    & {~=~}    &{\frac{1}{p^2}}
\\ {~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}    & {~=~}    &{\frac{1}{p^2}}
\\ \end{array}}$

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In the next section, we will see plane passing through the intersection of two planes.

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

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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.

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

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