Chapter 27 - Three Dimensional Geometry

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 his 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.

Previous

Contents

Next 

Copyright©2026 Higher secondary mathematics.blogspot.com

 

26.11 - Miscellaneous Examples

In the previous section, we completed a discussion on vector algebra. We saw many solved examples also. In this section, we will see some miscellaneous examples.

Solved example 26.64
Write down a unit vector in XY-plane, making an angle of 30^o with the positive direction of x-axis
Solution:
1. Fig.26.36 below shows a unit vector $\mathbf\small{\vec{AB}}$ making an angle of 30^o with the positive direction of x-axis.

Fig.26.36

2. It is clear that, projection of $\mathbf\small{AB}$ on the x-axis is $\mathbf\small{\left|\vec{AB} \right| \cos 30~=~(1)\left(\frac{\sqrt{3}}{2} \right)~=~\frac{\sqrt{3}}{2}}$
• So the x-component of $\mathbf\small{\vec{AB}}$ is $\mathbf\small{\left(\frac{\sqrt{3}}{2} \right)\hat{i}}$   

3. Also, it is clear that, projection of $\mathbf\small{AB}$ on the y-axis is $\mathbf\small{\left|\vec{AB} \right| \cos 60~=~(1)\left(\frac{1}{2} \right)~=~\frac{1}{2}}$
• So the y-component of $\mathbf\small{\vec{AB}}$ is $\mathbf\small{\left(\frac{1}{2} \right)\hat{j}}$

4. So the required vector is:
$\mathbf\small{\left(\frac{\sqrt{3}}{2} \right)\hat{i}~+~\left(\frac{1}{2} \right)\hat{j}}$

Solved example 26.65
A girl walks 4 km towards west, then she walks 3 km in a direction 30^o east of north and stops. Determine the girl's displacement from her initial point of departure.
Solution:
1. Fig.26.37 below shows the rough sketch

Fig.26.37

   ♦ $\mathbf\small{\vec{AB}}$ indicates the initial travel
   ♦ $\mathbf\small{\vec{BC}}$ indicates the final travel
• So $\mathbf\small{\vec{AC}}$ indicates the displacement vector. We are asked to find this vector.

2. Based on the fig., we can write:
• Projection of $\small{\vec{AB}}$ on the x-axis is 4 km.
• This vector does not have any projection on the y-axis.
• So we can write: $\small{\vec{AB} = -4\hat{i}}$
• The −ve sign is required because, this vector is pointing towards the −ve side of the x-axis

3. Also, based on the fig., we can write:
• Projection of $\small{\vec{BC}}$ on the x-axis
= DC = $\small{\left|\vec{BC} \right| \sin 30 = 3 \sin 30 = \frac{3}{2}}$
• Projection of $\small{\vec{BC}}$ on the y-axis
= BD = $\small{\left|\vec{BC} \right| \cos 30 = 3 \cos 30 = \frac{3 \sqrt{3}}{2}}$
So we can write: $\small{\vec{BC} = \frac{3}{2} \hat{i} + \frac{3 \sqrt{3}}{2} \hat{j}}$

4. By triangle law of vector addition, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{AC}}    & {~=~}    &{\vec{AB}+\vec{BC}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-4\hat{i} ~+~\frac{3}{2}\hat{i} + \frac{3 \sqrt{3}}{2} \hat{j} }
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-\frac{5}{2}\hat{i} + \frac{3 \sqrt{3}}{2} \hat{j} }
\\ \end{array}}$

Solved example 26.66
Write all the unit vectors in the XY-plane.
Solution:
1. Fig.26.38 below shows a random vector $\mathbf\small{\vec{AB}}$ making an angle of $\small{\theta}$ with the positive direction of x-axis.

Fig.26.38

2. It is clear that, projection of $\mathbf\small{AB}$ on the x-axis is $\mathbf\small{\left|\vec{AB} \right| \cos \theta~=~(1)\left(\cos \theta \right)~=~\cos \theta}$
• So the x-component of $\mathbf\small{\vec{AB}}$ is $\mathbf\small{\left(\cos \theta \right)\hat{i}}$   

3. Also, it is clear that, projection of $\mathbf\small{AB}$ on the y-axis is $\mathbf\small{\left|\vec{AB} \right| \sin \theta~=~(1)\left(\sin \theta \right)~=~\sin \theta}$
• So the y-component of $\mathbf\small{\vec{AB}}$ is $\mathbf\small{\left(\sin \theta \right)\hat{j}}$

4. So we can write:
$\mathbf\small{\vec{AB} = \left(\cos \theta \right)\hat{i} + \left(\sin \theta \right)\hat{j}}$

5. The above result is the general form. For any particular unit vector, all we need to do is to input the value of $\small{\theta}$ of that particular unit vector.
• We know that, $\small{\theta}$ can vary from zero to $\small{2 \pi}$.

6. So we can write:
• All unit vectors in the XY plane, are represented by the vector:
$\mathbf\small{\left(\cos \theta \right)\hat{i} + \left(\sin \theta \right)\hat{j}}$, where $\small{\theta}$ falls in the interval $\small{\left[0, 2 \pi \right]}$

Solved example 26.67
If $\small{\hat{i}+\hat{j}+\hat{k},~2\hat{i}+2\hat{j},~3\hat{i}+5\hat{j}-3\hat{k}}$ and $\small{\hat{i}-6\hat{j}-\hat{k}}$ are the position vectors of points A, B, C and D respectively, then find the angle between $\small{\vec{AB}~\text{and}~\vec{CD}}$. Deduce that $\small{\vec{AB}~\text{and}~\vec{CD}}$ are collinear.
Solution:
1.Based on the given position vectors, we get:
$\small{\vec{AB} = \hat{i}+4\hat{j}-\hat{k}}$
$\small{\vec{CD} = -2\hat{i}-8\hat{j}+2\hat{k}}$
2. Let $\small{\theta}$ be the angle between $\small{\vec{AB}~\text{and}~\vec{CD}}$.
We have: $\small{\theta=\cos^{-1}\left(\frac{\vec{AB}.\vec{CD}}{\left|\vec{AB}\right|\,\left|\vec{CD}\right|} \right)}$
3. So first we need to find $\small{\vec{AB}.\vec{CD}}$
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{AB}.\vec{CD}}    & {~=~}    &{a_1 b_1 + a_2 b_2 + a_3 b_3}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{(1)(-2)+(4)(-8)+(-1)(2)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{-2-32-2}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{-36}
\\ \end{array}}$
4. Next we need to find the magnitudes:
$\small{\left|\vec{AB}\right|=\sqrt{1^2 + 4^2 + (-1)^2}=\sqrt{18}}$
$\small{\left|\vec{CD}\right|=\sqrt{(-2)^2 + (-8)^2 + 2^2}=\sqrt{72}}$
5. Substituting the above values in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\theta}    & {~=~}    &{\cos^{-1}\left(\frac{\vec{AB}.\vec{CD}}{\left|\vec{AB}\right|\,\left|\vec{CD}\right|} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{-36}{(\sqrt{18})(\sqrt{72})} \right) = \cos^{-1}\left(\frac{-36}{(\sqrt{18})(\sqrt{(2)(36)})} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(\frac{(-1)\sqrt{(36)(36)}}{(\sqrt{18})(\sqrt{(2)(36)})} \right) = \cos^{-1}\left(\frac{(-1)\sqrt{(36)}}{(\sqrt{18})(\sqrt{(2)})} \right)}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\cos^{-1}\left(-1 \right)~=~\pi}
\\ \end{array}}$
6. The angle between $\small{\vec{AB}~\text{and}~\vec{CD}}$ is $\small{\pi}$. So the two vectors are collinear.

Alternate method to prove collinearity:

1. Take the ratios of the scalar components of $\small{\vec{AB}~\text{and}~\vec{CD}}$:
$\small{\frac{1}{-2},~\frac{4}{-8},~\text{and}~\frac{1}{-2}}$

2. We see that, all ratios are the same, which is $\small{-\frac{1}{2}}$
• So we can write: $\small{\vec{AB}=-\frac{1}{2} \vec{CD}}$
Which implies that, the two vectors are collinear.

Solved example 26.68
Let $\small{\vec{a},~\vec{b}~\text{and}~\vec{c}}$ be three vectors such that $\small{\left|\vec{a} \right|=3,~\left|\vec{b} \right|=4,~\left|\vec{c} \right|=5}$ and each one of them being perpendicular to the sum of the other two, find  $\small{\left|\vec{a}+\vec{b}+\vec{c} \right|}$.
Solution:
1.Given that:
Each vector is perpendicular to the sum of the other two.
• So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}.\left(\vec{b}+\vec{c} \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{{}}    &{\vec{b}.\left(\vec{a}+\vec{c} \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{{}}    &{\vec{c}.\left(\vec{a}+\vec{b} \right)}    & {~=~}    &{0}
\\ \end{array}}$
2. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left[\left|\vec{a}+\vec{b}+\vec{c} \right| \right]^2}    & {~=~}    &{\left(\vec{a}+\vec{b}+\vec{c} \right).\left(\vec{a}+\vec{b}+\vec{c} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{~~~~~~~\vec{a}.\vec{a}+\vec{a}.\left(\vec{b}+\vec{c} \right)}
\\ {    }    &{}    &{}    & {}    &{~+~\vec{b}.\vec{b}+\vec{b}.\left(\vec{a}+\vec{c} \right)}
\\ {    }    &{}    &{}    & {}    &{~+~\vec{c}.\vec{c}+\vec{c}.\left(\vec{a}+\vec{b} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{~~~~~~~\vec{a}.\vec{a}+0}
\\ {    }    &{}    &{}    & {}    &{~+~\vec{b}.\vec{b}+0}
\\ {    }    &{}    &{}    & {}    &{~+~\vec{c}.\vec{c}+0}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|^2+~\left|\vec{b} \right|^2~+~\left|\vec{c} \right|^2}
\\ {~\color{magenta}    5    }    &{}    &{}    & {~=~}    &{9 + 16 + 25 = 50}
\\ {~\color{magenta}    6    }    &{\Rightarrow}    &{\left|\vec{a}+\vec{b}+\vec{c} \right|}    & {~=~}    &{\sqrt{50}~=~5\sqrt{2}}
\\ \end{array}}$

Solved example 26.69
Three vectors $\small{\vec{a},~\vec{b},~\vec{c}}$ satisfy the condition $\small{\vec{a}+\vec{b}+\vec{c}=\vec{0}}$. Evaluate the quantity $\small{\mu=\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}}$, if $\small{\left|\vec{a} \right|=1,~\left|\vec{b} \right|=4,~\left|\vec{c} \right|=2}$
Solution:
1. Given that:

$\small{\vec{a}+\vec{b}+\vec{c}=\vec{0}}$
Multiplying both sides by $\small{\vec{a}}$, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}.\left[\vec{a}+\vec{b}+\vec{c} \right]}    & {~=~}    &{\vec{a}.\vec{0}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{a}.\vec{c}}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{1+\vec{a}.\vec{b}+\vec{a}.\vec{c}}    & {~=~}    &{0}
\\ \end{array}}$

2. Multiplying both sides by $\small{\vec{b}}$, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{b}.\left[\vec{a}+\vec{b}+\vec{c} \right]}    & {~=~}    &{\vec{b}.\vec{0}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{b}.\vec{a}+\vec{b}.\vec{b}+\vec{b}.\vec{c}}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\vec{b}.\vec{a}+16+\vec{b}.\vec{c}}    & {~=~}    &{0}
\\ \end{array}}$

3. Multiplying both sides by $\small{\vec{c}}$, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c}.\left[\vec{a}+\vec{b}+\vec{c} \right]}    & {~=~}    &{\vec{c}.\vec{0}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{c}.\vec{a}+\vec{c}.\vec{b}+\vec{c}.\vec{c}}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\vec{c}.\vec{a}+\vec{c}.\vec{b}+4}    & {~=~}    &{0}
\\ \end{array}}$

4. We have three results:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{1+\vec{a}.\vec{b}+\vec{a}.\vec{c}}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{}    &{\vec{b}.\vec{a}+16+\vec{b}.\vec{c}}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{}    &{\vec{c}.\vec{a}+\vec{c}.\vec{b}+4}    & {~=~}    &{0}
\\ \end{array}}$

5. Adding the L.H.S together, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{21 + 2\vec{a}.\vec{b} + 2\vec{b}.\vec{c} + 2\vec{c}.\vec{a}}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{21 + 2\left(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{ 2\left(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right)}    & {~=~}    &{-21}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{ \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} }    & {~=~}    &{\frac{-21}{2}}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{\mu}    & {~=~}    &{\frac{-21}{2}}
\\ \end{array}}$

Solved example 26.70
Let $\small{\vec{a}=\hat{i}+4\hat{j}+2\hat{k},~\vec{b}=3\hat{i}-2\hat{j}+7\hat{k}~\text{and}~\vec{c}=2\hat{i}-\hat{j}+4\hat{k}}$. Find a vector $\small{\vec{d}}$, which is perpendicular to both $\small{\vec{a}~\text{and}~\vec{b},~\text{and}~\vec{c}.\vec{d}=15}$
Solution:
1. We want a vector perpendicular to both $\small{\vec{a}~\text{and}~\vec{b}}$
• We know that $\small{\vec{a}\times\vec{b}}$ is a vector which is perpendicular to both $\small{\vec{a}~\text{and}~\vec{b}}$
• So first, we will find this cross product.

2. The cross product is:
$\small{\vec{a}\times\vec{b}=32\hat{i}-\hat{j}-14\hat{k}}$
• The reader may write all the steps related to this cross product

3. So $\small{32\hat{i}-\hat{j}-14\hat{k}}$ is perpendicular to both $\small{\vec{a}~\text{and}~\vec{b}}$
• Then $\small{\lambda \left(32\hat{i}-\hat{j}-14\hat{k} \right)}$ is also perpendicular to both $\small{\vec{a}~\text{and}~\vec{b}}$, where $\small{\lambda}$ is a scalar.
• That means, $\small{32 \lambda \hat{i}-\lambda \hat{j}-14 \lambda \hat{k}}$ is perpendicular to both $\small{\vec{a}~\text{and}~\vec{b}}$

4. We will assume that, the vector written above is the required vector $\small{\vec{d}}$.
• That is., we assume:
$\small{\vec{d} = 32 \lambda \hat{i}-\lambda \hat{j}-14 \lambda \hat{k}}$

5. But $\small{\vec{d}}$ must satisfy the condition:
$\small{\vec{c}.\vec{d}=15}$
• So we can write:
$\small{\left(2\hat{i}-\hat{j}+4\hat{k} \right).\left(32 \lambda \hat{i}-\lambda \hat{j}-14 \lambda \hat{k} \right)=15}$
• Therefore we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{(2)(32 \lambda)+(-1)(-1)(\lambda)+(4)(-1)(14 \lambda)}    & {~=~}    &{15}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{64\lambda + \lambda - 56\lambda}    & {~=~}    &{15}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{9 \lambda}    & {~=~}    &{15}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{ \lambda}    & {~=~}    &{\frac{15}{9}}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{ \lambda}    & {~=~}    &{\frac{5}{3}}
\\ \end{array}}$

6. So based on step (4), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{d}}    & {~=~}    &{32 \lambda \hat{i}-\lambda \hat{j}-14 \lambda \hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{32 \left(\frac{5}{3} \right) \hat{i}-\left(\frac{5}{3} \right) \hat{j}-14 \left(\frac{5}{3} \right) \hat{k}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{1}{3}\left( 160  \hat{i}- 5\hat{j}-70  \hat{k} \right)}
\\ \end{array}}$

Solved example 26.71
If with reference to the right handed system of mutually perpendicular unit vectors $\small{\hat{i},~\hat{j}~\text{and}~\hat{k},~\vec{\alpha}=3\hat{i}-\hat{j},~\vec{\beta}=2\hat{i}+\hat{j}-3\hat{k}}$, then express $\small{\vec{\beta}}$ in the form $\small{\vec{\beta}=\vec{\beta_1}+\vec{\beta_2}}$, where $\small{\vec{\beta_1}}$ is parallel to $\small{\vec{\alpha}}$ and $\small{\vec{\beta_2}}$ is perpendicular to $\small{\vec{\alpha}}$
Solution:
1. We want $\small{\vec{\beta_1}}$ to be parallel to $\small{\vec{\alpha}}$.
• So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{\beta_1}}    & {~=~}    &{\lambda\,\vec{\alpha}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left(3\lambda \right)\hat{i}-\left(\lambda \right)\hat{j}}
\\ \end{array}}$

2. Given that: $\small{\vec{\beta}=\vec{\beta_1}+\vec{\beta_2}}$
• So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{\beta}}    & {~=~}    &{\vec{\beta_1}+\vec{\beta_2}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{\beta_2}}    & {~=~}    &{\vec{\beta}~-~\vec{\beta_1}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\vec{\beta_2}}    & {~=~}    &{2\hat{i}+\hat{j}-3\hat{k}~-~\left[\left(3\lambda \right)\hat{i}-\left(\lambda \right)\hat{j} \right]}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{\vec{\beta_2}}    & {~=~}    &{\left(2 - 3\lambda \right)\hat{i}+\left(1+\lambda \right)\hat{j}-3\hat{k}}
\\ \end{array}}$

3. $\small{\vec{\beta_2}}$ should be perpendicular to $\small{\vec{\alpha}}$. So their dot product will be zero.
• We can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{\alpha}.\vec{\beta_2}}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left[\left(2 - 3\lambda \right)\hat{i}+\left(1+\lambda \right)\hat{j}-3\hat{k} \right].\left[3\hat{i}-\hat{j} \right]}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left(2 - 3\lambda \right)(3)~+~\left(1+\lambda \right)(-1)}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{6 - 9\lambda - 1 - \lambda}    & {~=~}    &{0}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{5 - 10\lambda}    & {~=~}    &{0}
\\ {~\color{magenta}    6    }    &{\Rightarrow}    &{\lambda}    & {~=~}    &{\frac{1}{2}}
\\ \end{array}}$

4. So from step (1), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{\beta_1}}    & {~=~}    &{\left(3\lambda \right)\hat{i}-\left(\lambda \right)\hat{j}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left(\frac{3}{2} \right)\hat{i}-\left(\frac{1}{2} \right)\hat{j}}
\\ \end{array}}$

5. Also, from step (2), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{\beta_2}}    & {~=~}    &{\left(2 - 3\lambda \right)\hat{i}+\left(1+\lambda \right)\hat{j}-3\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left(2 - 3\left(\frac{1}{2} \right) \right)\hat{i}+\left(1+\left(\frac{1}{2} \right) \right)\hat{j}-3\hat{k}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left(\frac{1}{2}  \right)\hat{i}+\left(\frac{3}{2}  \right)\hat{j}-3\hat{k}}
\\ \end{array}}$

---

The link below gives a few more miscellaneous examples:

Miscellaneous Exercise

---

In the next chapter, we will see Three dimensional geometry.

Previous

Contents

Next 

Copyright©2026 Higher secondary mathematics.blogspot.com

26.10 - Cross Product of Two Vectors

In the previous section, we completed a discussion on dot product of two vectors. We saw projection of a vector also. In this section, we will see cross product of two vectors.

Vector (or cross) product of two vectors

We have already seen the basic details about vector product in chapter 7 of our physics classes. Let us recall:
1. We have seen right handed screws and left handed screws in section 7.10.
2. We have seen the direction of the cross product in section 7.11.
3. We have seen the magnitude of the cross product in section 7.12.
4. So we can write:
$\mathbf\small{\vec{a}\times \vec{b}~=~\left|\vec{a} \right|\left|\vec{b} \right|\sin\theta \,\hat{n}}$

---

Let us write the important properties of cross product:
1. $\mathbf\small{\vec{a}\times\vec{b}}$ is a vector

2. If $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ are two nonzero vectors, then $\mathbf\small{\vec{a}\times\vec{b}~=~\vec{0}}$ if and only if $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ are parallel (or collinear) to each other. We can write:
$\mathbf\small{\vec{a}\times\vec{b}~=~\vec{0}~\Leftrightarrow~\vec{a}{\parallel} \vec{b}}$
Let us see two specific cases:
Case 1:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times \vec{a}}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{a} \right|\sin(0) \,\hat{n}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{a} \right|(0) \,\hat{n}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\vec{0}}
\\ \end{array}}$

Case 2:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times \left(-\vec{a} \right)}    & {~=~}    &{\left|\vec{a} \right|\left|-\vec{a} \right|\sin(\pi) \,\hat{n}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|\left|-\vec{a} \right|(0) \,\hat{n}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\vec{0}}
\\ \end{array}}$

3. Let $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ be two nonzero vectors. Let the angle $\mathbf\small{\theta}$ between them be $\mathbf\small{\frac{\pi}{2}}$. Then we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times \vec{b}}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{b} \right|\sin(\frac{\pi}{2}) \,\hat{n}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{b} \right|(1) \,\hat{n}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{b} \right| \,\hat{n}}
\\ \end{array}}$

4. Based on (2) and (3), we get some interesting results:
• Based on (2), we get:
$\mathbf\small{\hat{i}\times\hat{i}~=~\hat{j}\times\hat{j}~=~\hat{k}\times\hat{k}~=~\vec{0}}$
• Based on (3), we get:
    ♦ $\mathbf\small{\hat{i}\times\hat{j}~=~\hat{k}}$
    ♦ $\mathbf\small{\hat{j}\times\hat{k}~=~\hat{i}}$
    ♦ $\mathbf\small{\hat{k}\times\hat{i}~=~\hat{j}}$
Also see fig.7.63 and fig.7.64(a) of section 7.12 of physics notes

5. Let $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ be two nonzero vectors. Let the angle $\mathbf\small{\theta}$ between them be $\mathbf\small{\theta}$. Then we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times \vec{b} \right|}    & {~=~}    &{\left|\vec{a} \right|\left|\vec{b} \right|\sin\theta}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\sin \theta}    & {~=~}    &{\frac{\left|\vec{a}\times \vec{b} \right|}{\left|\vec{a} \right|\left|\vec{b} \right|}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{ \theta}    & {~=~}    &{\sin^{-1}\left(\frac{\left|\vec{a}\times \vec{b} \right|}{\left|\vec{a} \right|\left|\vec{b} \right|} \right)}
\\ \end{array}}$

6. It is always true that, vector product is not commutative. This is because, $\mathbf\small{\vec{a}\times \vec{b}~=~-\left(\vec{b}\times \vec{a} \right)}$
• Assume that, both $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ lie on the plane of the paper.
• Also assume that, $\mathbf\small{\vec{a}\times\vec{b}}$ is directed towards the upper side of the plane of the paper.
• Then $\mathbf\small{\vec{b}\times\vec{a}}$ will be directed towards the bottom side of the plane of the paper.
• This is because:
    ♦ For $\mathbf\small{\vec{a}\times\vec{b}}$, we rotate the right handed screw from $\mathbf\small{\vec{a}~\text{to}~\vec{b}}$
    ♦ For $\mathbf\small{\vec{b}\times\vec{a}}$, we rotate the right handed screw from $\mathbf\small{\vec{b}~\text{to}~\vec{a}}$

7. Based on (4) and (6), we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{i}\times \hat{j} }    & {~=~}    &{\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\hat{j}\times \hat{i}}    & {~=~}    &{-\left(\hat{i}\times \hat{j} \right)}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\hat{j}\times \hat{i}}    & {~=~}    &{-\hat{k}}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{j}\times \hat{k} }    & {~=~}    &{\hat{i}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\hat{k}\times \hat{j}}    & {~=~}    &{-\left(\hat{j}\times \hat{k} \right)}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\hat{k}\times \hat{j}}    & {~=~}    &{-\hat{i}}
\\ \end{array}}$

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\hat{k}\times \hat{i} }    & {~=~}    &{\hat{j}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\hat{i}\times \hat{k}}    & {~=~}    &{-\left(\hat{k}\times \hat{i} \right)}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\hat{i}\times \hat{k}}    & {~=~}    &{-\hat{j}}
\\ \end{array}}$

8. The fig.26.30 below, shows triangle ABC.

Fig.26.30

• $\mathbf\small{\vec{AC}= \vec{a}~\text{and}~\vec{AB}=\vec{b}}$
• So $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ represent the adjacent sides of the triangle ABC
• We know that, area of the triangle ABC = $\mathbf\small{\frac{1}{2}\left(\left|\vec{AB} \right| \right)\left(CD \right)}$
• But $\mathbf\small{CD = \left|\vec{AC} \right|\sin\theta=\left|\vec{a} \right|\sin\theta}$
• So area of the triangle ABC
=  $\mathbf\small{\frac{1}{2}\left(\left|\vec{AB} \right| \right)\left(\left|\vec{a} \right|\sin\theta \right)}$
=  $\mathbf\small{\frac{1}{2}\left(\left|\vec{b} \right| \right)\left(\left|\vec{a} \right|\sin\theta \right)}$
=  $\mathbf\small{\frac{1}{2}\left(\left|\vec{a}\times\vec{b} \right| \right)}$
• We can write:
Area of the triangle is equal to half of the magnitude of the cross product

9. The fig.26.31 below, shows parallelogram ABCD.

Fig.26.31

• $\mathbf\small{\vec{AD}= \vec{a}~\text{and}~\vec{AB}=\vec{b}}$
• So $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ represent the adjacent sides of the parallelogram ABCD.
• We know that, area of the parallelogram ABCD = $\mathbf\small{\left(\left|\vec{AB} \right| \right)\left(DE \right)}$
• But $\mathbf\small{DE = \left|\vec{AD} \right|\sin\theta=\left|\vec{a} \right|\sin\theta}$
• So area of the parallelogram ABCD
=  $\mathbf\small{\left(\left|\vec{AB} \right| \right)\left( \left|\vec{a} \right| \sin\theta \right)}$
=  $\mathbf\small{\frac{1}{2}\left(\left|\vec{b} \right| \right)\left( \left|\vec{a} \right|\sin\theta \right)}$
=  $\mathbf\small{\left(\left|\vec{a}\times\vec{b} \right| \right)}$
• We can write:
Area of the parallelogram is equal to the magnitude of the cross product

10. Vector product of two vectors is also known as cross product of two vectors.

---

Two important properties of cross product

Property I: Distributivity of cross product over addition
This can be explained as below:
Let $\small{\vec{a},~\vec{b},~\vec{c}}$ be any three vectors. Then we can write:
$\small{\vec{a}\times\left(\vec{b}+\vec{c} \right)~=~\vec{a}\times\vec{b}+\vec{a}\times\vec{c}}$

Property II: Distributivity of cross product over multiplication
This can be explained as below:
Let $\small{\vec{a}~\text{and}~\vec{b}}$ be any two vectors and $\small{\lambda}$ be any scalar. Then we can write:
$\small{\lambda \left(\vec{a}\times\vec{b} \right) ~=~\left(\lambda \vec{a}\right)\times\vec{b} ~=~\vec{a}\times\left(\lambda\vec{b} \right)}$

---

Cross product when vectors are given in component form

• Let the two vectors be:
$\small{\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}}$
$\small{\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}}$
• Then we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \right)\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{a_1\hat{i}\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_2\hat{j}\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_3\hat{k}\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{a_1 b_1\left(\hat{i}\times\hat{i}\right)+a_1 b_2\left(\hat{i}\times\hat{j}\right)+a_1 b_3\left(\hat{i}\times\hat{k}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_2 b_1\left(\hat{j}\times\hat{i}\right)+a_2 b_2\left(\hat{j}\times\hat{j}\right)+a_2 b_3\left(\hat{j}\times\hat{k}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_3 b_1\left(\hat{k}\times\hat{i}\right)+a_3 b_2\left(\hat{k}\times\hat{j}\right)+a_3 b_3\left(\hat{k}\times\hat{k}\right)}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{a_1 b_1\left(\vec{0}\right)+a_1 b_2\left(\hat{k}\right)+a_1 b_3\left(-\hat{j}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_2 b_1\left(-\hat{k}\right)+a_2 b_2\left(\vec{0}\right)+a_2 b_3\left(\hat{i}\right)}
\\ {~\color{magenta}    {}   }    &{}    &{}    & {}    &{+~a_3 b_1\left(\hat{j}\right)+a_3 b_2\left(-\hat{i}\right)+a_3 b_3\left(\vec{0}\right)}
\\ {~\color{magenta}    5   }    &{}    &{}    & {~=~}    &{\left(a_2 b_3 - a_3 b_2 \right)\hat{i} + \left(a_3 b_1 - a_1 b_3 \right)\hat{j} + \left(a_1 b_2 - a_2 b_1 \right)\hat{k}}
\\ {~\color{magenta}    6   }    &{}    &{}    & {~=~}    &{\left(a_2 b_3 - a_3 b_2 \right)\hat{i} - \left(a_1 b_3 - a_3 b_1 \right)\hat{j} + \left(a_1 b_2 - a_2 b_1 \right)\hat{k}}
\\ \end{array}}$

• The above result is equivalent to the determinant as shown below:

$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ a_{1}      &{    a_{2}     }    &{   a_{3}      }    
\\ b_{1}      &{    b_{2}     }    &{ b_{3}        }    
\\ \end{array}\right|$

---

Now we will see some solved examples.

Solved example 26.55
Find $\mathbf\small{\left|\vec{a}\times\vec{b} \right|}$, if $\mathbf\small{\vec{a}=2\hat{i}+\hat{j}+\hat{k}~\text{and}~\vec{b}=3\hat{i}+5\hat{j}-2\hat{k}}$
Solution:
1. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 2      &{    1     }    &{   3      }    
\\ 3      &{    5     }    &{ -2        }    
\\ \end{array}\right|$

2. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(-2 - 15 \right)\hat{i} - \left(-4 - 9 \right)\hat{j} + \left(10 - 3 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-17\hat{i}+13\hat{j}+7\hat{k}}
\\ \end{array}}$

3. Therefore,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(-17)^2 + (13)^2 + (7)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{507}}
\\ \end{array}}$

Solved example 26.56
Find $\mathbf\small{\left|\vec{a}\times\vec{b} \right|}$, if $\mathbf\small{\vec{a}=\hat{i}-7\hat{j}+7\hat{k}~\text{and}~\vec{b}=3\hat{i}-2\hat{j}+2\hat{k}}$
Solution:
1. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 1      &{    -7     }    &{   7      }    
\\ 3      &{    -2     }    &{ 2        }    
\\ \end{array}\right|$

2. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(-2 - 15 \right)\hat{i} - \left(-4 - 9 \right)\hat{j} + \left(10 - 3 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-17\hat{i}+13\hat{j}+7\hat{k}}
\\ \end{array}}$

3. Therefore,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(19)^2 + (19)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{19\sqrt{2}}
\\ \end{array}}$

Solved example 26.57
Find a unit vector perpendicular to each of the vectors $\mathbf\small{\left(\vec{a}+\vec{b} \right)~\text{and}~\left(\vec{a}-\vec{b} \right)}$, where $\mathbf\small{\vec{a}=\hat{i}+\hat{j}+\hat{k}~\text{and}~\vec{b}=\hat{i}+2\hat{j}+3\hat{k}}$
Solution:
1. We have:
• $\mathbf\small{\vec{c}=\left(\vec{a}+\vec{b} \right)=2\hat{i}+3\hat{j}+4\hat{k}}$
• $\mathbf\small{\vec{d}=\left(\vec{a}-\vec{b} \right)=-\hat{j}-2\hat{k}}$

2. We are asked to find a unit vector perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• We know that:
$\mathbf\small{\vec{c}\times\vec{d}}$ will be perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• So we will first write this cross product. We have:
$\vec{c}\times\vec{d}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 2      &{    3     }    &{   4      }    
\\ 0      &{    -1     }    &{ -2        }    
\\ \end{array}\right|$

• So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c}\times\vec{d}}    & {~=~}    &{\left(-6 + 4 \right)\hat{i} - \left(-4 - 0 \right)\hat{j} + \left(-2 - 0 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-2\hat{i}+4\hat{j}-2\hat{k}}
\\ \end{array}}$

3. Let $\mathbf\small{\vec{f}~=~\vec{c}\times\vec{d}}$
• Then $\mathbf\small{\hat{f}}$ is a required vector.
We have: $\mathbf\small{\hat{f}=\frac{\vec{f}}{\left|\vec{f} \right|}}$
= $\mathbf\small{\frac{-2\hat{i}+4\hat{j}-2\hat{k}}{\sqrt{(-2)^2 + (4)^2+(-2)^2}}~=~\frac{-2\hat{i}+4\hat{j}-2\hat{k}}{\sqrt{24}}}$
= $\mathbf\small{\frac{-2\hat{i}+4\hat{j}-2\hat{k}}{2\sqrt{6}}~=~\frac{-\hat{i}+2\hat{j}-\hat{k}}{\sqrt{6}}}$

4. Note:
We have a plane in which $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$ lie. The unit vector $\mathbf\small{\hat{f}}$ is perpendicular to that plane. Consequently, $\mathbf\small{-\hat{f}}$ will also be perpendicular to that plane. We would have obtained $\mathbf\small{-\hat{f}}$, if we had started with $\mathbf\small{\vec{d}\times\vec{c}}$ instead of $\mathbf\small{\vec{c}\times\vec{d}}$ in step (2)

Solved example 26.58
Find a unit vector perpendicular to each of the vectors $\mathbf\small{\left(\vec{a}+\vec{b} \right)~\text{and}~\left(\vec{a}-\vec{b} \right)}$, where $\mathbf\small{\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}~\text{and}~\vec{b}=\hat{i}+2\hat{j}-2\hat{k}}$
Solution:
1. We have:
• $\mathbf\small{\vec{c}=\left(\vec{a}+\vec{b} \right)=4\hat{i}+4\hat{j}}$
• $\mathbf\small{\vec{d}=\left(\vec{a}-\vec{b} \right)=-2\hat{i}+4\hat{k}}$

2. We are asked to find a unit vector perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• We know that:
$\mathbf\small{\vec{c}\times\vec{d}}$ will be perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• So we will first write this cross product. We have:
$\vec{c}\times\vec{d}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 4      &{    4     }    &{   0      }    
\\ -2      &{    0     }    &{ 4        }    
\\ \end{array}\right|$
• So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{c}\times\vec{d}}    & {~=~}    &{\left(16 - 0 \right)\hat{i} - \left(16 - 0 \right)\hat{j} + \left(0+8 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{16\hat{i}-16\hat{j}+8\hat{k}}
\\ \end{array}}$

3. Let $\mathbf\small{\vec{f}~=~\vec{c}\times\vec{d}}$
• Then $\mathbf\small{\hat{f}}$ is a required unit vector.
We have: $\mathbf\small{\hat{f}=\frac{\vec{f}}{\left|\vec{f} \right|}}$
= $\mathbf\small{\frac{16\hat{i}-16\hat{j}+8\hat{k}}{\sqrt{(16)^2 + (-16)^2+(8)^2}}~=~\frac{16\hat{i}-16\hat{j}+8\hat{k}}{24}}$
= $\mathbf\small{\frac{2\hat{i}-2\hat{j}+\hat{k}}{3}}$

4. Note:
We have a plane in which $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$ lie. The unit vector $\mathbf\small{\hat{f}}$ is perpendicular to that plane. Consequently, $\mathbf\small{-\hat{f}}$ will also be perpendicular to that plane. We would have obtained $\mathbf\small{-\hat{f}}$, if we had started with $\mathbf\small{\vec{d}\times\vec{c}}$ instead of $\mathbf\small{\vec{c}\times\vec{d}}$ in step (2)

Solved example 26.59
Find the area of a triangle having the points A(1,1,1), B(1,2,3) and C(2,3,1) as it's vertices
Solution:
1. Fig.26.32 below shows the rough sketch:

Fig.26.32

2. Based on the rough sketch, we can write:
$\mathbf\small{\vec{a}= \vec{AB}=\hat{j}+2\hat{k}}$
$\mathbf\small{\vec{b}= \vec{AC}=\hat{i}+2\hat{j}}$

3. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 0      &{    1     }    &{   2      }    
\\ 1      &{    2     }    &{ 0        }    
\\ \end{array}\right|$

4. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(0 - 4 \right)\hat{i} - \left(0 - 2 \right)\hat{j} + \left(0 - 1 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-4\hat{i}+2\hat{j}-\hat{k}}
\\ \end{array}}$

5. Then,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(-4)^2 + (2)^2 + (-1)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{21}}
\\ \end{array}}$

6. Therefore,
Area of triangle ABC
= $\mathbf\small{\frac{1}{2}\left(\left|\vec{a}\times\vec{b} \right| \right)~=~\frac{\sqrt{21}}{2}}$

Solved example 26.60
Find the area of a triangle with vertices A(1,1,2), B(2,3,5) and C(1,5,5).
Solution:
1. Fig.26.33 below shows the rough sketch:

Fig.26.33

2. Based on the rough sketch, we can write:
$\mathbf\small{\vec{a}= \vec{AB}=\hat{i}+2\hat{j}+3\hat{k}}$
$\mathbf\small{\vec{b}= \vec{AC}=4\hat{j}+3\hat{k}}$

3. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 1      &{    2     }    &{   3      }    
\\ 0      &{    4     }    &{ 3        }    
\\ \end{array}\right|$

4. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(6 - 12 \right)\hat{i} - \left(3 - 0 \right)\hat{j} + \left(4 - 0 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{-6\hat{i}-3\hat{j}+4\hat{k}}
\\ \end{array}}$

5. Then,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(-6)^2 + (-3)^2 + (4)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{61}}
\\ \end{array}}$

6. Therefore,
Area of triangle ABC
= $\mathbf\small{\frac{1}{2}\left(\left|\vec{a}\times\vec{b} \right| \right)~=~\frac{\sqrt{61}}{2}}$

Solved example 26.61
Find the area of the parallelogram whose adjacent sides are given by the vectors $\small{\vec{a}=3\hat{i}+\hat{j}+4\hat{k}~\text{and}~\vec{b}=\hat{i}-\hat{j}+\hat{k}}$.
Solution:
1. Fig.26.34 below shows the rough sketch:

Fig.26.34

2. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 3      &{    1     }    &{   4      }    
\\ 1      &{    -1     }    &{ 1        }    
\\ \end{array}\right|$

3. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(1 + 4 \right)\hat{i} - \left(3 - 4 \right)\hat{j} + \left(-3 - 1 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{5\hat{i}+\hat{j}-4\hat{k}}
\\ \end{array}}$

4. Then,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(5)^2 + (1)^2 + (-4)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{42}}
\\ \end{array}}$

5. Therefore,
Area of parallelogram ABCD
= $\mathbf\small{\left|\vec{a}\times\vec{b} \right| ~=~\sqrt{42}}$

Solved example 26.62
Find the area of the parallelogram whose adjacent sides are determined by the vectors $\small{\vec{a}=\hat{i}-\hat{j}+3\hat{k}~\text{and}~\vec{b}=2\hat{i}-7\hat{j}+\hat{k}}$.
Solution:
1. Fig.26.35 below shows the rough sketch:

Fig.26.35

2. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r}                             \hat{i}     &{    \hat{j}     }    &{    \hat{k}      }    \\ 1      &{    -1     }    &{   3      }    
\\ 2      &{    -7     }    &{ 1        }    
\\ \end{array}\right|$

3. So we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}\times\vec{b}}    & {~=~}    &{\left(-1 + 21 \right)\hat{i} - \left(1 - 6 \right)\hat{j} + \left(-7 + 2 \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{20\hat{i}+5\hat{j}-5\hat{k}}
\\ \end{array}}$

4. Then,
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a}\times\vec{b} \right|}    & {~=~}    &{\sqrt{(20)^2 + (5)^2 + (-5)^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\sqrt{450}}
\\ \end{array}}$

5. Therefore,
Area of parallelogram ABCD
= $\mathbf\small{\left|\vec{a}\times\vec{b} \right| ~=~\sqrt{450}~=~15\sqrt{2}}$

Solved example 26.63
If a unit vector $\small{\vec{a}}$ makes angles $\small{\frac{\pi}{3}~\text{with}~\hat{i},~~\frac{\pi}{4}~\text{with}~\hat{j}}$ and an acute angle $\small{\theta}$ with $\small{\hat{k}}$, then find $\small{\theta}$ and hence, the components of $\small{\vec{a}}$.
Solution:
1. $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{3}~\text{with}~\hat{i}}$.
• $\small{\hat{i}}$ lies along the x-axis. That means, $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{3}}$ with the x-axis.
• So the x-component of $\small{\vec{a}}$
= $\small{\left|\vec{a} \right|\,\cos\left(\frac{\pi}{3} \right)\hat{i}}$ 
= $\small{(1)\left(\frac{1}{2} \right)\hat{i}}$ 
= $\small{\left(\frac{1}{2} \right)\hat{i}}$  

2. $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{4}~\text{with}~\hat{j}}$.
• $\small{\hat{j}}$ lies along the y-axis. That means, $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{4}}$ with the y-axis.
• So the y-component of $\small{\vec{a}}$
= $\small{\left|\vec{a} \right|\,\cos\left(\frac{\pi}{4} \right)\hat{j}}$ 
= $\small{(1)\left(\frac{1}{\sqrt{2}} \right)\hat{j}}$ 
= $\small{\left(\frac{1}{\sqrt{2}} \right)\hat{j}}$  

3. $\small{\vec{a}}$ makes an acute angle $\small{\theta~\text{with}~\hat{k}}$.
• $\small{\hat{k}}$ lies along the z-axis. That means, $\small{\vec{a}}$ makes an acute angle $\small{\theta}$ with the z-axis.
• So the z-component of $\small{\vec{a}}$
= $\small{\left|\vec{a} \right|\,\cos\left(\theta \right)\hat{k}}$ 
= $\small{(1)\cos\left(\theta \right)\hat{k}}$ 
= $\small{\cos\left(\theta \right)\hat{k}}$

4. We wrote the three components. So we can write:
$\small{\vec{a}=\left(\frac{1}{2} \right)\hat{i}+\left(\frac{1}{\sqrt{2}} \right)\hat{j}+\cos\left(\theta \right)\hat{k}}$

5. Given that, $\small{\vec{a}}$ is a unit vector. So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left|\vec{a} \right|}    & {~=~}    &{1}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\sqrt{\left(\frac{1}{2} \right)^2+\left(\frac{1}{\sqrt{2}} \right)^2+\left[\cos\left(\theta \right) \right]^2}}    & {~=~}    &{1}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\left(\frac{1}{2} \right)^2+\left(\frac{1}{\sqrt{2}} \right)^2+\left[\cos\left(\theta \right) \right]^2}    & {~=~}    &{1}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{\cos^2\left(\theta \right)}    & {~=~}    &{1-\frac{1}{4}-\frac{1}{2}~=~\frac{1}{4}}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{\cos\left(\theta \right)}    & {~=~}    &{\pm\frac{1}{2}}
\\ \end{array}}$

6. So we have two equations:
(i) $\small{\cos \theta~=~\frac{1}{2}}$ 
(ii) $\small{\cos \theta~=~-\frac{1}{2}}$

• Solving the first equation, we get: $\small{\theta~=~\frac{\pi}{3}}$ 
• Solving the second equation, we get: $\small{\theta~=~\frac{2\pi}{3}}$  

• Given that, $\small{\theta}$ is an acute angle. So we can write:
$\small{\theta~=~\frac{\pi}{3}}$

7. So based on step (4), we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\vec{a}}    & {~=~}    &{\left(\frac{1}{2} \right)\hat{i}+\left(\frac{1}{\sqrt{2}} \right)\hat{j}+\cos\left(\frac{\pi}{3} \right)\hat{k}}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\vec{a}}    & {~=~}    &{\left(\frac{1}{2} \right)\hat{i}+\left(\frac{1}{\sqrt{2}} \right)\hat{j}+\left(\frac{1}{2} \right)\hat{k}}
\\ \end{array}}$

8. Therefore, the components of $\small{\vec{a}}$ are:
$\small{\frac{1}{2},~\frac{1}{\sqrt{2}}~\text{and}~\frac{1}{2}}$

---

The link below gives a few more solved examples:

Exercise 10.4

---

In the next section, we will see some miscellaneous examples.

Previous

Contents

Next 

Copyright©2026 Higher secondary mathematics.blogspot.com