In the previous section,
we completed a discussion on vector algebra. In this chapter,
we will see Three dimensional geometry.
In class 11, we saw the basic details about three dimensional geometry. See chapter 12. In the present chapter, we will see some advanced topics related to three dimension. In the previous chapter, we saw vectors in three dimensional space. We can use those vectors for our present discussion. This will make the discussion more interesting.
Direction cosines of a line
This can be explained in 5 steps:
1. Consider a directed line L passing through the origin. It makes the following angles with the axes:
♦ $\small{\alpha}$ with the +ve direction of the x-axis
♦ $\small{\beta}$ with the +ve direction of the y-axis
♦ $\small{\gamma}$ with the +ve direction of the z-axis
• Then $\small{\cos \alpha,~\cos \beta~\text{and}~\cos \gamma}$ are the direction cosines of the line L
2. Suppose that, we reverse the direction of L. We will call the new line as L'.
• L' will make the following angles with the axes:
♦ $\small{\left( \pi - \alpha \right)}$ with the +ve direction of the x-axis
♦ $\small{\left( \pi - \beta \right)}$ with the +ve direction of the y-axis
♦ $\small{\left( \pi - \gamma \right)}$ with the +ve direction of the z-axis
• This can be easily proved if L lies in a two dimensional plane. In fig.27.1(a) below, $\small{\alpha~\text{and}~\beta}$ and are shown in white and green colors.
 |
| Fig.27.1 |
In fig.27.1(b), $\small{\alpha~\text{and}~\beta}$ are shown twice, by considering opposite angles. Based on those opposite angles, we can write:
• L' makes the following angles with the axes:
♦ $\small{\left( \pi - \alpha \right)}$ with the +ve direction of the x-axis
♦ $\small{\left( \pi - \beta \right)}$with the +ve direction of the y-axis
• So in the three dimensional space, $\small{\cos \left( \pi - \alpha \right),~\cos \left( \pi - \beta \right)~\text{and}~\cos \left( \pi - \gamma \right) }$ are the direction cosines of L'.
• We know that:
♦ $\small{\cos\left( \pi - \alpha \right)~=~-\cos \alpha}$
♦ $\small{\left(\cos \pi - \beta \right)~=~-\cos \beta}$
♦ $\small{\left(\cos \pi - \gamma \right)~=~-\cos \gamma}$
• That means, when a directed line is reversed, the direction cosines are also reversed.
3. If we are given a simple line in space, it can be extended in two opposite directions. So a simple line will have two sets of direction cosines.
• While doing problems, we must know which set to use. So we must specify the direction of the line.
• In other words, while doing problems in three dimensional geometry, we deal with directed lines only.
4. For a directed line, there will be a unique set of direction cosines. Those direction cosines are denoted by $\small{l,~m~\text{and}~n}$
5. Suppose that, the given line does not pass through the origin. Then we draw a line through the origin and parallel to the given line. The direction cosines of this new line will be same as those of the given line.
Relation between direction cosines of a line
This can be explained in 9 steps:
1. In fig.27.2 below, a line L is shown. This line does not pass through the origin. So we draw a line L1 passing through the origin and parallel to L
 |
| Fig.27.2 |
2. Mark any convenient point P(x,y,z) on L1
• Imagine a vector $\small{\vec{OP}}$ with initial point O and terminal point P. Since the coordinates of P are (x,y,z), we can write:
$\small{\vec{OP}~=~x\hat{i}+y\hat{j}+z\hat{k}}$
3. Based on the above component form, we can write:
♦ Projection of $\small{\vec{OP}}$ on the x-axis = x
♦ Projection of $\small{\vec{OP}}$ on the y-axis = y
♦ Projection of $\small{\vec{OP}}$ on the z-axis = z
4. The line L1 makes an angle $\small{\alpha}$ with the +ve side of the x-axis. So $\small{\vec{OP}}$ also makes the same angle $\small{\alpha}$ with the +ve side of the x-axis. Then we can write: $\small{\left|\vec{OP} \right| \cos \alpha = x}$. Recall a similar fig., based on which, we discussed about vectors. (See fig.26.4 of the first section of the previous chapter 26)
• From this, we get:
$\small{\cos \alpha = \frac{x}{\left|\vec{OP} \right|} \Rightarrow l = \frac{x}{\left|\vec{OP} \right|}}$
5. The line L1 makes an angle $\small{\beta}$ with the +ve side of the y-axis. So $\small{\vec{OP}}$ also makes the same angle $\small{\beta}$ with the +ve side of the y-axis. Then we can write: $\small{\left|\vec{OP} \right| \cos \beta = y}$.
• From this, we get:
$\small{\cos \beta= \frac{y}{\left|\vec{OP} \right|} \Rightarrow m = \frac{y}{\left|\vec{OP} \right|}}$
(See fig.26.5 of the first section of the previous chapter 26)
6. The line L1 makes an angle $\small{\gamma}$ with the +ve side of the z-axis. So $\small{\vec{OP}}$ also makes the same angle $\small{\gamma}$ with the +ve side of the z-axis. Then we can write: $\small{\left|\vec{OP} \right| \cos \gamma = z}$.
• From this, we get:
$\small{\cos \gamma = \frac{z}{\left|\vec{OP} \right|} \Rightarrow n = \frac{z}{\left|\vec{OP} \right|}}$
(See fig.26.6 of the first section of the previous chapter 26)
7. In the above steps (4), (5) and (6), we made use of our knowledge about vectors. Thereby, we obtained the direction cosines of L1.
• We can write:
To obtain the direction cosines of L1, we require three items:
♦ Any convenient point P on L1
♦ The coordinates of P
♦ The length OP
8. Our next task is to find $\small{\left|\vec{OP} \right|}$
• This is the magnitude of the vector $\small{\vec{OP}}$.
• So it is simply, the distance between O and P.
• We have the coordinates of both O and P. So the distance between them is $\small{\sqrt{x^2 + y^2 + z^2}}$
That is., $\small{\left|\vec{OP} \right| = \sqrt{x^2 + y^2 + z^2}}$
9. Now we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{l^2 + m^2 + n^2} & {~=~} &{\frac{x^2}{\left|\vec{OP} \right|^2}~+~\frac{y^2}{\left|\vec{OP} \right|^2}~+~\frac{z^2}{\left|\vec{OP} \right|^2}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{x^2 ~+~y^2~+~z^2}{\left|\vec{OP} \right|^2}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\frac{x^2 ~+~y^2~+~z^2}{x^2 ~+~y^2~+~z^2}}
\\ {~\color{magenta} 4 } &{} &{} & {~=~} &{1}
\\ \end{array}}$
Direction Ratios
This can be explained in 6 steps:
1. Consider a line with direction cosines l, m and n
2. Suppose that, there exist three numbers a, b and c which satisfy the following condition:
$\small{\frac{l}{a}~=~\frac{m}{b}~=~\frac{n}{c}}$
3. If the above condition is satisfied, then we say that:
a, b and c are direction ratios of the line.
4. In step (2), there are three fractions. Since they are equal, we can write:
$\small{\frac{l}{a}~=~\frac{m}{b}~=~\frac{n}{c}~=~k}$, where k is a real number.
• So we get:
$\small{l=ak,~m=bk~\text{and}~n=ck}$
5. Now we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{l^2 + m^2 + n^2} & {~=~} &{a^2 k^2~+~b^2 k^2~+~c^2 k^2}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{1} & {~=~} &{k^2(a^2~+~b^2~+~c^2)}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{k^2} & {~=~} &{\frac{1}{a^2 ~+~b^2~+~c^2}}
\\ {~\color{magenta} 4 } &{\Rightarrow} &{k} & {~=~} &{\pm\frac{1}{\sqrt{a^2 ~+~b^2~+~c^2}}}
\\ \end{array}}$
◼ Remarks:
• 2 (magenta color): Here we use the fact that, $\small{l^2 + m^2 + n^2\,=\,1}$
6. So based on step (4), we get the direction cosines:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{l} & {~=~} &{\pm\frac{a}{\sqrt{a^2 ~+~b^2~+~c^2}}}
\\ {~\color{magenta} 2 } &{{}} &{m} & {~=~} &{\pm\frac{b}{\sqrt{a^2 ~+~b^2~+~c^2}}}
\\ {~\color{magenta} 3 } &{{}} &{n} & {~=~} &{\pm\frac{c}{\sqrt{a^2 ~+~b^2~+~c^2}}}
\\ \end{array}}$
• We must select the sign carefully. It depends on the direction of the line. This will become more clear when we do some solved examples
Solved example 27.1
If a line makes angles 90°, 60° and 30° with the positive direction of x, y and z axis respectively, find its direction cosines.
Solution:
1. The line makes angle 90° with the positive direction of the x axis. So $\small{\alpha = 90^{\circ}}$
• Then the direction cosine $\small{l = \cos 90 = 0}$
2. The line makes angle 60° with the positive direction of the y axis. So $\small{\beta = 60^{\circ}}$
• Then the direction cosine $\small{m = \cos 60 = \frac{1}{2}}$
3. The line makes angle 30° with the positive direction of the z axis. So $\small{\gamma = 30^{\circ}}$
• Then the direction cosine $\small{n = \cos 30 = \frac{\sqrt{3}}{2}}$
Solved example 27.2
If a line has direction ratios 2, −1, −2, determine its direction cosines.
Solution:
1. We have:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{l} & {~=~} &{\pm\frac{a}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{2}{\sqrt{2^2 ~+~(-1)^2~+~(-2)^2}}~=~\pm\frac{2}{3}}
\\ {~\color{magenta} 2 } &{{}} &{m} & {~=~} &{\pm\frac{b}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{-1}{3}}
\\ {~\color{magenta} 3 } &{{}} &{n} & {~=~} &{\pm\frac{c}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{-2}{3}}
\\ \end{array}}$
2. So the two sets of direction cosines are:
$\small{\frac{2}{3},~\frac{-1}{3},~\frac{-2}{3}~\text{and}~\frac{-2}{3},~\frac{1}{3},~\frac{2}{3}}$
3. Depending on the direction of the directed line, we must choose the appropriate set.
• Let there be two points A and B on the line.
• If a person traveling from A to B has the direction cosines $\small{\frac{2}{3},~\frac{-1}{3},~\frac{-2}{3}}$, then a person traveling from B to A will have the direction cosines $\small{\frac{-2}{3},~\frac{1}{3},~\frac{2}{3}}$.
Solved example 27.3
If a line has direction ratios −18, 12, −4, then what are its direction cosines.
Solution:
1. We have:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{l} & {~=~} &{\pm\frac{a}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{-18}{\sqrt{(-18)^2 ~+~12^2~+~(-4)^2}}~=~\pm\frac{-18}{22}~=~\pm\frac{-9}{11}}
\\ {~\color{magenta} 2 } &{{}} &{m} & {~=~} &{\pm\frac{b}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{12}{22}~=~\pm\frac{6}{11}}
\\ {~\color{magenta} 3 } &{{}} &{n} & {~=~} &{\pm\frac{c}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{-4}{22}~=~\pm\frac{-2}{11}}
\\ \end{array}}$
2. So the two sets of direction cosines are:
$\small{\frac{-9}{11},~\frac{6}{11},~\frac{-2}{11}~\text{and}~\frac{9}{11},~\frac{-6}{11},~\frac{2}{11}}$
3. Depending on the direction of the directed line, we must choose the appropriate set.
• Let there be two points A and B on the line.
• If a person traveling from A to B has the direction cosines $\small{\frac{-9}{11},~\frac{6}{11},~\frac{-2}{11}}$, then a person traveling from B to A will have the direction cosines $\small{\frac{9}{11},~\frac{-6}{11},~\frac{2}{11}}$.
Direction cosines of a line passing through two points
This can be explained in 9 steps:
1. In fig.27.3 below, $\small{P(x_1,x_2,x_3)~\text{and}~Q(x_1,x_2,x_3)}$ are two points in space.
 |
| Fig.27.3 |
• There can be one and only one line through both P and Q. That line is shown in magenta color. We want the direction cosines of this magenta line.
2. Imagine that, a vector $\small{\vec{PQ}}$ is present between P and Q. Based on our knowledge in vector algebra, we can write the component form of $\small{\vec{PQ}}$:
$\small{\vec{PQ}~=~\left(x_2 - x_1 \right)\hat{i}+\left(y_2 - y_1 \right)\hat{j}+\left(z_2 - z_1 \right)\hat{k}}$
3. Let $\small{\vec{PQ}}$ make angles $\small{\alpha,~\beta~\text{and}~\gamma}$ with the positive side of the x, y and z-axis respectively. Then the magenta line will also be making the same angles with the three axes.
4. $\small{\left(x_2 - x_1 \right)}$ is the projection of $\small{\vec{PQ}}$ on the x-axis. So we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x_2 - x_1} & {~=~} &{\left|\vec{PQ} \right|\cos \alpha}
\\ {~\color{magenta} 2 } &{{\Rightarrow}} &{\cos \alpha} & {~=~} &{\frac{x_2 - x_1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$
5. $\small{\left(y_2 - y_1 \right)}$ is the projection of $\small{\vec{PQ}}$ on the y-axis. So we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y_2 - y_1} & {~=~} &{\left|\vec{PQ} \right|\cos \beta}
\\ {~\color{magenta} 2 } &{{\Rightarrow}} &{\cos \beta} & {~=~} &{\frac{y_2 - y_1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$
6. $\small{\left(z_2 - z_1 \right)}$ is the projection of $\small{\vec{PQ}}$ on the z-axis. So we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{z_2 - z_1} & {~=~} &{\left|\vec{PQ} \right|\cos \gamma}
\\ {~\color{magenta} 2 } &{{\Rightarrow}} &{\cos \gamma} & {~=~} &{\frac{z_2 - z_1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$
7. So we obtained the three direction cosines of $\small{\vec{PQ}}$. Note that, $\small{\left|\vec{PQ} \right|}$ can be easily determined because, we have the coordinates of both P and Q. We get:
$\small{\left|\vec{PQ} \right| = \sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}$
8. The magenta line has the same direction cosines as that of $\small{\vec{PQ}}$. So we can compile the result.
• The direction cosines of a line through two points $\small{P(x_1,x_2,x_3)~\text{and}~Q(x_1,x_2,x_3)}$ are:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos \alpha} & {~=~} &{\frac{x_2 - x_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ {~\color{magenta} 2 } &{} &{\cos \beta} & {~=~} &{\frac{y_2 - y_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ {~\color{magenta} 3 } &{} &{\cos \gamma} & {~=~} &{\frac{z_2 - z_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ \end{array}}$
9. The direction cosines written in (8) above, are the ones for a person traveling from P to Q. If he travels from Q to P, he must use the reversed direction cosines:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos \alpha} & {~=~} &{\frac{x_1 - x_2}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ {~\color{magenta} 2 } &{} &{\cos \beta} & {~=~} &{\frac{y_1 - y_2}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ {~\color{magenta} 3 } &{} &{\cos \gamma} & {~=~} &{\frac{z_1 - z_2}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ \end{array}}$
Solved example 27.4
Find the direction cosines of the line passing through the two points (−2,4,−5) and (1,2,3).
Solution:
1. Let the two points be: P(−2,4,−5) and Q(1,2,3).
2. So for a person traveling in the direction P to Q, the direction cosines are:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos \alpha} & {~=~} &{\frac{x_2 - x_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}~=~\frac{1-(-2)}{\sqrt{\left(1-(-2) \right)^2 + \left(2 - 4 \right)^2 + \left(3 - (-5) \right)^2}}~=~\frac{3}{\sqrt{\left(3 \right)^2 + \left(-2 \right)^2 + \left(8 \right)^2}}~=~\frac{3}{\sqrt{77}}}
\\ {~\color{magenta} 2 } &{} &{\cos \beta} & {~=~} &{\frac{y_2 - y_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}~=~\frac{2-4}{\sqrt{77}}~=~\frac{-2}{\sqrt{77}}}
\\ {~\color{magenta} 3 } &{} &{\cos \gamma} & {~=~} &{\frac{z_2 - z_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}~=~\frac{3-(-5)}{\sqrt{77}}~=~\frac{8}{\sqrt{77}}}
\\ \end{array}}$
3. Also, for a person traveling in the direction Q to P, the direction cosines are:
$\small{\frac{-3}{\sqrt{77}},~\frac{2}{\sqrt{77}},~\frac{-8}{\sqrt{77}}}$
Solved example 27.5
Find the direction cosines of x, y and z-axis.
Solution:
Part (a): Direction cosines of x-axis
1. The x-axis makes angle 0° with the positive direction of the x axis. So $\small{\alpha = 0^{\circ}}$
• Then the direction cosine $\small{l = \cos 0 = 1}$
2. The x-axis makes angle 90° with the positive direction of the y axis. So $\small{\beta = 90^{\circ}}$
• Then the direction cosine $\small{m = \cos 90 = 0}$
3. The x-axis makes angle 90° with the positive direction of the z axis. So $\small{\gamma = 90^{\circ}}$
• Then the direction cosine $\small{n = \cos 90 = 0}$
4. So the direction cosines of the x-axis are: 1,0,0
Part (b): Direction cosines of y-axis
1. The y-axis makes angle 90° with the positive direction of the x axis. So $\small{\alpha = 90^{\circ}}$
• Then the direction cosine $\small{l = \cos 90 = 0}$
2. The y-axis makes angle 0° with the positive direction of the y axis. So $\small{\beta = 0^{\circ}}$
• Then the direction cosine $\small{m = \cos 0 = 1}$
3. The y-axis makes angle 90° with the positive direction of the z axis. So $\small{\gamma = 90^{\circ}}$
• Then the direction cosine $\small{n = \cos 90 = 0}$
4. So the direction cosines of the y-axis are: 0,1,0
Part (b): Direction cosines of z-axis
1. The z-axis makes angle 90° with the positive direction of the x axis. So $\small{\alpha = 90^{\circ}}$
• Then the direction cosine $\small{l = \cos 90 = 0}$
2. The z-axis makes angle 90° with the positive direction of the y axis. So $\small{\beta = 90^{\circ}}$
• Then the direction cosine $\small{m = \cos 90 = 0}$
3. The y-axis makes angle 0° with the positive direction of the z axis. So $\small{\gamma = 0^{\circ}}$
• Then the direction cosine $\small{n = \cos 0 = 1}$
4. So the direction cosines of the z-axis are: 0,0,1
In the next section, we will see more details about direction ratios.
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