In the previous section, we completed a discussion on dot product of two vectors. In this section, we will see projection of a vector.
Projection of a vector on a line
This can be explained in steps:
1.
In fig.26.29(i) below, $\small{\vec{AB}}$ makes an angle
$\small{\theta}$ with a directed line l. The angle $\small{\theta}$ is
measured in the anticlockwise direction.
 |
| Fig.26.29 |
2.
Drop a perpendicular (green dashed line) from B on to l. Let C be the
foot of the perpendicular. Then length AC will be equal to
$\small{\left|\vec{AB} \right|\cos\theta}$.
3. We can think of a vector $\small{\vec{AC}}$.
♦ Initial point of this vector is A
♦ Terminal point of this vector is C.
•
Magnitude of this vector will be $\small{\left|\vec{AB} \right|\cos\theta}$.
◼ $\small{\vec{AC}}$ is called:
Projection vector of $\small{\vec{AB}}$ on the directed line l.
♦ It is denoted as $\small{\vec{p}}$
♦ It is a vector
◼ $\small{\left|\vec{p} \right|}$ is called:
Projection of $\small{\vec{AB}}$ on the directed line l.
♦ It is a distance
♦ It is not a vector
◼ So we can write:
♦ $\small{\vec{p}}$ is the Projection vector of $\small{\vec{AB}}$ on the directed line l.
♦ $\small{\left|\vec{p} \right|}$ is the Projection of $\small{\vec{AB}}$ on the directed line l.
4. Since $\small{\vec{p}}$ is aligned with l, the direction of $\small{\vec{p}}$ has only two possibilities:
♦ same as the direction of l
♦ opposite to the direction of l
5. In fig.26.29(i) above, $\small{0 < \theta <\frac{\pi}{2}}$. In such a situation, $\small{\vec{p}}$ will have the same direction as l
This can be proved as shown below:
•
Let $\small{\hat{p}}$ be the unit vector which has the same direction
as l. Multiplying this unit vector by $\small{\left|\vec{p}\right|}$
will give $\small{\vec{p}}$.
•
So we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1
} &{{}} &{\vec{p} } & {~=~}
&{\left[\left|\vec{p} \right| \right]\hat{p}}
\\
{~\color{magenta} 2 } &{} &{} & {~=~}
&{\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}
\\ \end{array}}$
♦ In the interval $\small{\left(0,\frac{\pi}{2} \right) }$,
$\small{\cos \theta}$ is +ve. So $\small{\left[\left|\vec{AB}
\right|\cos \theta \right]}$ is +ve.
♦ $\small{\hat{p}}$ has the same direction as l
•
Therefore, $\small{\vec{p}}$ has the same direction as l
6. In fig.26.29(ii) above, $\small{\frac{\pi}{2} < \theta < \pi}$. In such a situation, $\small{\vec{p}}$ will have the opposite direction of l.
This can be proved as shown below:
•
As before, we have:
$\small{\vec{p}~=~\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}$
♦ In the interval $\small{\left(\frac{\pi}{2},\pi \right) }$,
$\small{\cos \theta}$ is -`ve. So $\small{\left[\left|\vec{AB}
\right|\cos \theta \right]}$ is -`ve.
♦ $\small{\hat{p}}$ has the same direction as l
•
Therefore, $\small{\vec{p}}$ has the opposite direction of l
•
Recall that, we saw the trigonometric ratios of obtuse angles in section 3.5
7. In fig.26.29(iii) below, $\small{\pi < \theta < \frac{3\pi}{2}}$. In such a situation, $\small{\vec{p}}$ will have the opposite direction of l
 |
| Fig.26.29 |
This can be proved as shown below:
•
As before, we have:
$\small{\vec{p}~=~\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}$
♦ In the interval $\small{\left(\pi,\frac{3\pi}{2}\right) }$,
$\small{\cos \theta}$ is -`ve. So $\small{\left[\left|\vec{AB}
\right|\cos \theta \right]}$ is -`ve.
♦ $\small{\hat{p}}$ has the same direction as l
•
Therefore, $\small{\vec{p}}$ has the opposite direction of l
8. In fig.26.29(iv) above, $\small{\frac{3\pi}{2} < \theta < 2 \pi}$. In such a situation, $\small{\vec{p}}$ will have the same direction as l
This can be proved as shown below:
•
As before, we have:
$\small{\vec{p}~=~\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}$
♦ In the interval $\small{\left(\frac{3\pi}{2},2\pi \right) }$,
$\small{\cos \theta}$ is +ve. So $\small{\left[\left|\vec{AB}
\right|\cos \theta \right]}$ is +ve.
♦ $\small{\hat{p}}$ has the same direction as l
•
Therefore, $\small{\vec{p}}$ has the same direction as l
9.
Consider the scalar product $\small{\vec{AB}.\hat{p}}$, where
$\small{\hat{p}}$ is the unit vector which has the same direction as l'.
We can write:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{\vec{AB}.\hat{p}}
& {~=~} &{\left|\vec{AB} \right|\,\left|\hat{p} \right|\cos
\theta}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\left|\vec{AB} \right|(1)\cos \theta}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left|\vec{AB} \right|\cos \theta}
\\ {~\color{magenta} 4 } &{} &{} & {~=~} &{\text{Projection of}~\vec{AB}~\text{on}~l}
\\ \end{array}}$
•
Thus we get a second method to find the projection.
♦ First method is to find $\small{\left|\vec{AB} \right|\cos\theta}$
♦ Second method is to find the dot product $\small{\vec{AB}.\hat{p}}$
•
Remember that, $\small{\vec{AB}.\hat{p}}$ is the dot product of two
vectors. So it is a real number. It is the "projection". Not the
projection vector".
10. Suppose that, we want the projection of a
vector $\small{\vec{a}~\text{on another vector}~\vec{b}}$. We can make
use of the result in (9) above. The steps are as follows:
(i) Assume a directed line l in the same direction as $\small{\vec{b}}$
(ii) Next, we want a unit vector in the same direction as l. Such a unit vector is readily available. It is $\small{\hat{b}}$
(iii) So based on (9), the required projection is: $\small{\vec{a}.\hat{b}}$
(iv) But $\small{\hat{b}~=~\frac{\vec{b}}{\left|\vec{b} \right|}}$
(v)
So the required projection =
$\small{\vec{a}.\left(\frac{\vec{b}}{\left|\vec{b} \right|}
\right)~=~\frac{1}{\left|\vec{b} \right|}\left(\vec{a}.\vec{b} \right)}$
•
Remember that, $\small{\frac{1}{\left|\vec{b}
\right|}\left(\vec{a}.\vec{b} \right)}$ is the scalar product of two
vectors. So it is a real number. It is the "projection". Not the
projection vector".
11. In fig.26.29 above, we considered $\small{\theta}$ to be in the intervals:
$\small{\left(0,\frac{\pi}{2}
\right), \left(\frac{\pi}{2},\pi \right), \left(\pi,\frac{3\pi}{2}
\right), \left(\frac{3\pi}{2},2\pi \right)}$
We did not consider the
cases when $\small{\theta}$ is equal to
$\small{0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi}$. We will see those
cases now.
(i) $\small{\theta=0}$
Consider fig.26.29(i). We want $\small{\vec{p}~\text{and}~\left|\vec{p} \right|}$ when $\small{\theta=0}$
Based on (5), we have:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{\vec{p} } &
{~=~} &{\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}
\\
{~\color{magenta} 2 } &{} &{} & {~=~}
&{\left[\left|\vec{AB} \right|\cos (0) \right]\hat{p}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left[\left|\vec{AB} \right|(1) \right]\hat{p}}
\\ {~\color{magenta} 4 } &{} &{} & {~=~} &{\left(\left|\vec{AB} \right| \right)\hat{p}}
\\ \end{array}}$
•
$\small{\hat{p}}$
is a unit vector. It gives us the direction only. In our present case,
it has the same direction as l. So we can write:
When
$\small{\theta=0}$, the projection vector has the same magnitude as
$\small{\vec{AB}}$. And it has the same direction as l.
(ii) $\small{\theta=\frac{\pi}{2}}$
Consider fig.26.29(i or ii). We want $\small{\vec{p}~\text{and}~\left|\vec{p} \right|}$ when $\small{\theta=\frac{\pi}{2}}$
Based on (5), we have:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{\vec{p} } &
{~=~} &{\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}
\\
{~\color{magenta} 2 } &{} &{} & {~=~}
&{\left[\left|\vec{AB} \right|\cos \left(\frac{\pi}{2}\right)
\right]\hat{p}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left[\left|\vec{AB} \right|(0) \right]\hat{p}}
\\ {~\color{magenta} 4 } &{} &{} & {~=~} &{\left(0 \right)\hat{p}}
\\ \end{array}}$
•
$\small{\hat{p}}$
is a unit vector. It gives us the direction only. In our present case,
it has the same direction as l. So we can write:
When $\small{\theta=\frac{\pi}{2}}$, the projection vector is a zero vector.
(iii) $\small{\theta=\pi}$
Consider fig.26.29(ii or iii ). We want $\small{\vec{p}~\text{and}~\left|\vec{p} \right|}$ when $\small{\theta=\pi}$
Based on (5), we have:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{\vec{p} } &
{~=~} &{\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}
\\
{~\color{magenta} 2 } &{} &{} & {~=~}
&{\left[\left|\vec{AB} \right|\cos (\pi) \right]\hat{p}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left[\left|\vec{AB} \right|(-1) \right]\hat{p}}
\\ {~\color{magenta} 4 } &{} &{} & {~=~} &{(-1)\left(\left|\vec{AB} \right| \right)\hat{p}}
\\ \end{array}}$
•
$\small{\hat{p}}$
is a unit vector. It gives us the direction only. In our present case,
it has the same direction as l. So we can write:
When
$\small{\theta=\pi}$, the projection vector has the same magnitude as
$\small{\vec{AB}}$. And it has the opposite direction of l.
(iv) $\small{\theta=\frac{3\pi}{2}}$
Consider fig.26.29(iii or iv). We want $\small{\vec{p}~\text{and}~\left|\vec{p} \right|}$ when $\small{\theta=\frac{3\pi}{2}}$
Based on (5), we have:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{\vec{p} } &
{~=~} &{\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}
\\
{~\color{magenta} 2 } &{} &{} & {~=~}
&{\left[\left|\vec{AB} \right|\cos \left(\frac{3\pi}{2}\right)
\right]\hat{p}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left[\left|\vec{AB} \right|(0) \right]\hat{p}}
\\ {~\color{magenta} 4 } &{} &{} & {~=~} &{\left(0 \right)\hat{p}}
\\ \end{array}}$
•
$\small{\hat{p}}$
is a unit vector. It gives us the direction only. In our present case,
it has the same direction as l. So we can write:
When $\small{\theta=\frac{3\pi}{2}}$, the projection vector is a zero vector.
(v) $\small{\theta=2\pi}$
Consider fig.26.29(i). We want $\small{\vec{p}~\text{and}~\left|\vec{p} \right|}$ when $\small{\theta=2\pi}$
Based on (5), we have:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{\vec{p} } &
{~=~} &{\left[\left|\vec{AB} \right|\cos \theta \right]\hat{p}}
\\
{~\color{magenta} 2 } &{} &{} & {~=~}
&{\left[\left|\vec{AB} \right|\cos (2\pi) \right]\hat{p}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left[\left|\vec{AB} \right|(1) \right]\hat{p}}
\\ {~\color{magenta} 4 } &{} &{} & {~=~} &{\left(\left|\vec{AB} \right| \right)\hat{p}}
\\ \end{array}}$
•
$\small{\hat{p}}$
is a unit vector. It gives us the direction only. In our present case,
it has the same direction as l. So we can write:
When
$\small{\theta=2\pi}$, the projection vector has the same magnitude as
$\small{\vec{AB}}$. And it has the same direction as l.
Now we will see the projection when a vector is given in component form. It can be written in 4 steps:
1. Consider a vector $\small{\vec{a}~=~a_1\hat{i}+a_2\hat{j}+a_3\hat{k}}$.
We know that, this vector has three components:
♦ $\small{a_1\hat{i}}$, which lies along the x-axis
♦ $\small{a_2\hat{j}}$, which lies along the y-axis
♦ $\small{a_3\hat{k}}$, which lies along the z-axis
2. Consider the component $\small{a_1\hat{i}}$, which lies along the x-axis.
We know that, magnitude of this vector is $\small{a_1}$.
Since this vector lies on the x-axis, we can write:
♦ Projection vector of $\small{\vec{a}}$ on the x-axis is $\small{a_1\hat{i}}$
♦ Projection of $\small{\vec{a}}$ on the x-axis is $\small{a_1}$
3. Similarly, we can write:
♦ Projection vector of $\small{\vec{a}}$ on the y-axis is $\small{a_2\hat{i}}$
♦ Projection of $\small{\vec{a}}$ on the y-axis is $\small{a_2}$
4. Similarly, we can write:
♦ Projection vector of $\small{\vec{a}}$ on the z-axis is $\small{a_3\hat{i}}$
♦ Projection of $\small{\vec{a}}$ on the z-axis is $\small{a_3}$
Now we will see the relation between projection and direction cosines. It can be written in 7 steps:
1. Consider a vector $\small{\vec{a}~=~a_1\hat{i}+a_2\hat{j}+a_3\hat{k}}$. We know the direction cosines:
♦ $\small{l~=~\frac{a_1}{\left|\vec{a} \right|}}$
♦ $\small{m~=~\frac{a_2}{\left|\vec{a} \right|}}$
♦ $\small{n~=~\frac{a_3}{\left|\vec{a} \right|}}$
See solved example 26.19 in section 26.5
2. Consider the first direction cosine: $\small{l~=~\frac{a_1}{\left|\vec{a} \right|}}$
•
Here $\small{a_1}$ is the projection of $\small{\vec{a}}$ on the x-axis.
3. We have seen a second method to find projection. See step (9) above.
•
By that method, the "projection of $\small{\vec{a}}$ on the x-axis" is:
$\small{\vec{a}.\hat{i}}$
4. So the result in (2) becomes: $\small{l~=~\frac{\vec{a}.\hat{i}}{\left|\vec{a} \right|}}$
5. Similarly we can obtain the other two direction cosines also. Let us write all the three together:
•
$\small{l~=~\frac{a_1}{\left|\vec{a} \right|}~=~\frac{\vec{a}.\hat{i}}{\left|\vec{a} \right|}}$
•
$\small{m~=~\frac{a_2}{\left|\vec{a} \right|}~=~\frac{\vec{a}.\hat{j}}{\left|\vec{a} \right|}}$
•
$\small{r~=~\frac{a_3}{\left|\vec{a} \right|}~=~\frac{\vec{a}.\hat{k}}{\left|\vec{a} \right|}}$
6. Suppose that, $\small{\vec{a}}$ is a unit vector.
•
Then $\small{\left|\vec{a}\right|=1}$
•
In such a situation, step (1) above becomes:
♦ $\small{l~=~\frac{a_1}{\left|\vec{a} \right|}~=~\frac{a_1}{1}~=~a_1}$
♦ $\small{m~=~\frac{a_2}{\left|\vec{a} \right|}~=~\frac{a_1}{1}~=~a_2}$
♦ $\small{n~=~\frac{a_3}{\left|\vec{a} \right|}~=~\frac{a_1}{1}~=~a_3}$
7. So if $\small{\vec{a}}$ is a unit vector, then it's component form can be written as:
$\small{\vec{a}~=~a_1\hat{i}+a_2\hat{j}+a_3\hat{k}~=~l\hat{i}+m\hat{j}+n\hat{k}}$
Now we will see some solved examples
Solved example 26.51
Show that each of the given three vectors is a unit vector:
$\small{\frac{1}{7}\left(2\hat{i}+3\hat{j}+6\hat{k} \right)}$
$\small{\frac{1}{7}\left(3\hat{i}-6\hat{j}+2\hat{k} \right)}$
$\small{\frac{1}{7}\left(6\hat{i}-3\hat{k} \right)}$
Also show that they are mutually perpendicular to each other.
Solution:
Part (i): Showing that the vectors are unit vectors
•
We have seen that:
For a unit vector, the scalar components are same as the direction cosines.
1. Let $\small{\vec{a}=\frac{1}{7}\left(2\hat{i}+3\hat{j}+6\hat{k} \right)}$
•
The scalar components are:
♦ $\small{a_1 = \frac{2}{7}}$
♦ $\small{a_2 = \frac{3}{7}}$
♦ $\small{a_3 = \frac{6}{7}}$
•
$\small{\left|\vec{a}\right|=\sqrt{\frac{2^2 + 3^2 + 6^2}{7^2}}=\sqrt{\frac{49}{49}}=1}$
•
Therefore, the direction cosines are:
♦ $\small{l = \frac{a_1}{\left|\vec{a}\right|}=\frac{2/7}{1}=\frac{2}{7}}$
♦ $\small{m = \frac{a_2}{\left|\vec{a}\right|}=\frac{3/7}{1}=\frac{3}{7}}$
♦ $\small{n = \frac{a_3}{\left|\vec{a}\right|}=\frac{6/7}{1}=\frac{6}{7}}$
•
The scalar components are same as the direction cosines. So $\small{\vec{a}}$ is a unit vector.
• The above elaborate steps are written to show the relation between direction cosines and scalar components of a unit vector. To show that $\small{\vec{a}}$ is a unit vector, all we need to do is to show that $\small{\left|\vec{a}\right|=1}$
2. Let $\small{\vec{b}=\frac{1}{7}\left(3\hat{i}-6\hat{j}+2\hat{k} \right)}$
•
The scalar components are:
♦ $\small{b_1 = \frac{3}{7}}$
♦ $\small{b_2 = \frac{-6}{7}}$
♦ $\small{b_3 = \frac{2}{7}}$
•
$\small{\left|\vec{b}\right|=\sqrt{\frac{3^2 + (-6)^2 + 2^2}{7^2}}=\sqrt{\frac{49}{49}}=1}$
•
Therefore, the direction cosines are:
♦ $\small{l = \frac{b_1}{\left|\vec{b}\right|}=\frac{3/7}{1}=\frac{3}{7}}$
♦ $\small{m = \frac{b_2}{\left|\vec{b}\right|}=\frac{-6/7}{1}=\frac{-6}{7}}$
♦ $\small{n = \frac{b_3}{\left|\vec{b}\right|}=\frac{2/7}{1}=\frac{2}{7}}$
•
The scalar components are same as the direction cosines. So $\small{\vec{b}}$ is a unit vector.
3. Let $\small{\vec{c}=\frac{1}{7}\left(6\hat{i}+2\hat{j}-3\hat{k} \right)}$
•
The scalar components are:
♦ $\small{c_1 = \frac{6}{7}}$
♦ $\small{c_2 = \frac{2}{7}}$
♦ $\small{c_3 = \frac{-3}{7}}$
•
$\small{\left|\vec{c}\right|=\sqrt{\frac{6^2 + 2^2 + (-3)^2}{7^2}}=\sqrt{\frac{49}{49}}=1}$
•
Therefore, the direction cosines are:
♦ $\small{l = \frac{c_1}{\left|\vec{c}\right|}=\frac{6/7}{1}=\frac{6}{7}}$
♦ $\small{m = \frac{c_2}{\left|\vec{c}\right|}=\frac{2/7}{1}=\frac{2}{7}}$
♦ $\small{n = \frac{c_3}{\left|\vec{c}\right|}=\frac{-3/7}{1}=\frac{-3}{7}}$
•
The scalar components are same as the direction cosines. So $\small{\vec{c}}$ is a unit vector.
Part (ii): Showing that the given three vectors are mutually perpendicular to each other
•
We have seen that:
If two vectors are mutually perpendicular to each other, their dot product will be zero.
•
For three vectors, three combinations are possible. So we have to calculate three dot products.
1. First we calculate $\small{\vec{a}.\vec{b}}$
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\vec{a}.\vec{b}} & {=} &{\left(\frac{2}{7} \right)\left(\frac{3}{7} \right)+\left(\frac{3}{7} \right)\left(\frac{-6}{7} \right)+\left(\frac{6}{7} \right)\left(\frac{2}{7} \right)}
\\ {~\color{magenta} 2 } &{}&{} & {=} &{\frac{6-18+12}{49}=\frac{0}{49}}
\\ {~\color{magenta} 3 } &{}&{} & {=} &{0}
\\ \end{array}}$
•
We see that: $\small{\vec{a}.\vec{b}=0}$
So $\small{\vec{a}~\text{and}~\vec{b}}$ are mutually perpendicular to each other.
2. Next we calculate $\small{\vec{b}.\vec{c}}$
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\vec{b}.\vec{c}} & {=} &{\left(\frac{3}{7} \right)\left(\frac{6}{7} \right)+\left(\frac{-6}{7} \right)\left(\frac{2}{7} \right)+\left(\frac{2}{7} \right)\left(\frac{-3}{7} \right)}
\\ {~\color{magenta} 2 } &{}&{} & {=} &{\frac{18-12-6}{49}=\frac{0}{49}}
\\ {~\color{magenta} 3 } &{}&{} & {=} &{0}
\\ \end{array}}$
• We see that: $\small{\vec{b}.\vec{c}=0}$
So $\small{\vec{b}~\text{and}~\vec{c}}$ are mutually perpendicular to each other.
3. Finally we calculate $\small{\vec{c}.\vec{a}}$
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\vec{c}.\vec{a}} & {=} &{\left(\frac{6}{7} \right)\left(\frac{2}{7} \right)+\left(\frac{2}{7} \right)\left(\frac{3}{7} \right)+\left(\frac{-3}{7} \right)\left(\frac{6}{7} \right)}
\\ {~\color{magenta} 2 } &{}&{} & {=} &{\frac{12+6-18}{49}=\frac{0}{49}}
\\ {~\color{magenta} 3 } &{}&{} & {=} &{0}
\\ \end{array}}$
• We see that: $\small{\vec{c}.\vec{a}=0}$
So $\small{\vec{c}~\text{and}~\vec{a}}$ are mutually perpendicular to each other.
Solved example 26.52
Find the projection of
the vector $\small{\vec{a}=2\hat{i}+3\hat{j}+2\hat{k}}$ on the vector
$\small{\vec{b}=\hat{i}+2\hat{j}+\hat{k}}$
Solution:
1. We have:
Projection of $\small{\vec{a}~\text{on}~\vec{b}=\frac{1}{\left|\vec{b} \right|}\left(\vec{a}.\vec{b} \right)}$
2. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\frac{1}{\left|\vec{b} \right|}\left(\vec{a}.\vec{b} \right)} & {=} &{\left(\frac{1}{\sqrt{1^2 + 2^2+1^2}} \right)\left[\left(2 \right)\left(1 \right)+\left(3 \right)\left(2 \right)+\left(2 \right)\left(1 \right) \right]}
\\ {~\color{magenta} 2 } &{}&{} & {=} &{\frac{2+6+2}{\sqrt{6}}=\frac{10}{\sqrt{6}}}
\\ {~\color{magenta} 3 } &{}&{} & {=} &{\frac{5\sqrt{6}}{3}}
\\ \end{array}}$
Solved example 26.53
Find the projection of
the vector $\small{\vec{a}=\hat{i}-\hat{j}}$ on the vector
$\small{\vec{b}=\hat{i}+\hat{j}}$
Solution:
1. We have:
Projection of $\small{\vec{a}~\text{on}~\vec{b}=\frac{1}{\left|\vec{b} \right|}\left(\vec{a}.\vec{b} \right)}$
2. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\frac{1}{\left|\vec{b} \right|}\left(\vec{a}.\vec{b} \right)} & {=} &{\left(\frac{1}{\sqrt{1^2 +1^2}} \right)\left[\left(1 \right)\left(1 \right)+\left(-1 \right)\left(1 \right) \right]}
\\ {~\color{magenta} 2 } &{}&{} & {=} &{\frac{0}{\sqrt{2}}}
\\ {~\color{magenta} 3 } &{}&{} & {=} &{0}
\\ \end{array}}$
Solved example 26.54
Find the projection of the vector $\small{\vec{a}=\hat{i}+3\hat{j}+7\hat{k}}$ on the vector $\small{\vec{b}=7\hat{i}-\hat{j}+8\hat{k}}$
Solution:
1. We have:
Projection of $\small{\vec{a}~\text{on}~\vec{b}=\frac{1}{\left|\vec{b} \right|}\left(\vec{a}.\vec{b} \right)}$
2. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\frac{1}{\left|\vec{b} \right|}\left(\vec{a}.\vec{b} \right)} & {=} &{\left(\frac{1}{\sqrt{7^2 + (-1)^2+8^2}} \right)\left[\left(1 \right)\left(7 \right)+\left(3 \right)\left(-1 \right)+\left(7 \right)\left(8 \right) \right]}
\\ {~\color{magenta} 2 } &{}&{} & {=} &{\frac{7-3+56}{\sqrt{114}}}
\\ {~\color{magenta} 3 } &{}&{} & {=} &{\frac{60}{\sqrt{114}}}
\\ \end{array}}$
The link below gives a few more solved examples:
Exercise 10.3
In the next section, we will see cross product.
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