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.
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.
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
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:
$\left|\begin{array}{r} x-x_1 &{
y-y_1 } &{ z-z_1 } \\ x_2-x_1 &{
y_2-y_1 } &{ z_2-z_1 } \\ x_3-x_1 &{
y_3-y_1 } &{ z_3-z_1 } \\ \end{array}\right|~=~0$
$\Rightarrow\left|\begin{array}{r} x-1 &{ y-1 } &{ z-0 }
\\ 1-1 &{ 2-1 } &{ 1-0 }
\\ -2-1 &{ 2-1 } &{ -1-0 } \\ \end{array}\right|~=~0$
$\Rightarrow\left|\begin{array}{r} x-1 &{ y-1 } &{ z-0 }
\\ 0 &{ 1 } &{ 1 }
\\ -3 &{ 1 } &{ -1 } \\ \end{array}\right|~=~0$
$\small{\Rightarrow(x-1)(-1-1)-(y-1)(0+3)+z(0+3)=0}$
$\small{\Rightarrow(x-1)(-2)-(y-1)(3)+z(3)=0}$
$\small{\Rightarrow -2x+2-3y+3+3z=0}$
$\small{\Rightarrow 2x+3y-3z=5}$
In the next section, we will see intercept form.
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