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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22.5 - Solved Examples on Tangents And Normals

In the previous section, we saw Tangents and Normals. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 22.18
Find the equation of the tangent to the curve
$\rm{y\,=\,\frac{x-7}{(x-2)(x-3)}}$ at the point where it cuts the x-axis.
Solution:
1. The derivative can be used to find the slope of tangent at any point. So we will first find the derivative.

2. We want the point at which the curve cuts the x-axis.
At that point, the y-coordinate will be zero. So in the equation of the curve, we substitute y by zero. We get:
$\rm{0\,=\,\frac{x-7}{(x-2)(x-3)}}$
⇒ x − 7 = 0
⇒ x = 7
• So the required point is: (7,0)

3. Next we want the slope at (7,0). We have:

4. So the equation of the tangent can be written as:
$\rm{y - y_0 ~=~m(x-x_0)}$
⇒ $\rm{y - 0 ~=~\frac{1}{20}(x-7)}$
⇒ 20y = x − 7
⇒ 20y − x + 7 = 0

Fig.22.18

• The graph is shown in fig.22.18 below:
    ♦ The curve is drawn in red color.
    ♦ The tangent is drawn in green color.

• The tangent is drawn at (7,0)
• The slope triangle has a height of 0.2 units and base of 4 units. So the slope of tangent is 0.2/4 = 1/20

Solved example 22.19
Find the equation of the tangent and normal to the curve
$\rm{x^{2/3} \,+\, y^{2/3}\,=\,2}$ at (1,1).
Solution:
1. The derivative can be used to find the slope of tangent at any point. So we will first find the derivative.

 

2. So we can write the slope of the tangent at (1,1):

$\rm{\left. \frac{dy}{dx} \right|_{(1,1)}~=~(-1) \left(\frac{1}{1} \right)^{1/3}~=~-1}$

3. Now we can write the equation of the tangent at (1,1).
y − y₀ = m(x − x₀)
⇒ y − 1 = (−1)(x − 1)
⇒ y − 1 = −x + 1
⇒ y + x − 2 = 0

4. Slope of the normal is equal to the negative reciprocal of that of the tangent. So slope of the normal is 1.
• Now we can write the equation of the tangent at (1,1).
y − y₀ = m(x − x₀)
⇒ y − 1 = (1)(x − 1)
⇒ y − 1 = x − 1
⇒ y − x = 0

• The graph is shown in fig.22.19 below:
    ♦ The curve is drawn in red color.
    ♦ The tangent is drawn in green color.

Fig.22.19

• The tangent is drawn at (1,1)
• The slope triangle has a height of −2 units and base of 2 units. So the slope of tangent is −2/2 = −1

Solved example 22.20
Find the equation of the tangent to the curve given by
$\rm{x \,=\, a \sin^3 t,~~y\,=\,b \cos^3 t}$ at  a point where t = π/2.
Solution:
1. The derivative can be used to find the slope of tangent at any point. So we will first find the derivative.

2. Next we want the slope at t = π/2. We have:
$\rm{\frac{dy}{dx}\,=\,\frac{-a \sin t}{b \cos t}}$
$\rm{~=\,\frac{-a \sin (\pi/2)}{b \cos (\pi/2)}}$
$\rm{~=\,\frac{-b \cos (\pi/2)}{a \sin (\pi/2)}}$
$\rm{~=\,\frac{-b (0)}{a (1)}}~=~0$

3. Next we want the (x,y) coordinates at t = π/2
$\rm{x \,=\, a \sin^3 (\pi/2)\,=\,a (1)^3 \,=\,a}$
$\rm{y\,=\,b \cos^3 (\pi/2)\,=\,b(0)^3 = 0}$

4. The slope of the tangent is zero. That means, the tangent is horizontal. So we can write the equation of the tangent just by using the y-coordinate obtained in (4).
• We get: y = 0

5. Fig.22.20 below shows the graph of the given function.
• It is assumed that, a = 3 and b = 4
• So the point (a,0) is (3,0)

Fig.22.20

• We see that:
If we draw the tangent at (3,0), it will be same as the x-axis.
• So the equation of the tangent is: y = 0

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The link below gives a few more solved examples:

Exercise 22.3

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In the next section, we will see Approximations.

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