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.
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 colo) 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:
$\small{\frac{-6}{7},~\frac{3}{7},~\frac{2}{7}}$
In the next section, we will see a few more solved examples.
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