In the previous section, we saw scalar multiplication and unit vector. In this section, we will see components of a vector.
First we will see the unit vectors along the three axes. It can be explained in 4 steps:
1. A rectangular coordinate system with origin O is shown in fig.26.18 below:
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| Fig.26.18 |
2. Let us mark three points A(1,0,0), B(0,1,0) and C(0,0,1). • Based on the coordinates, we can easily see that:
♦ A will be at a distance of 1 unit from O, and on the +ve side of the x-axis (OX axis)
♦ B will be at a distance of 1 unit from O, and on the +ve side of the y-axis (OY axis)
♦ C will be at a distance of 1 unit from O, and on the +ve side of the z-axis (OZ axis)
3. Now we can think of three vectors: $\small{\vec{OA},\vec{OB},\vec{OC}}$
• For $\small{\vec{OA}}$, magnitude is 1. So it is a unit vector. Its direction is towards the +ve side of x-axis (OX direction)
• For $\small{\vec{OB}}$, magnitude is 1. So it is a unit vector. Its direction is towards the +ve side of y-axis (OY direction)
• For $\small{\vec{OC}}$, magnitude is 1. So it is a unit vector. Its direction is towards the +ve side of z-axis (OZ direction)
4. These three vectors are very important in vector algebra. They are given special names:
• Unit vector $\small{\vec{OA}}$ along the OX axis is denoted as $\small{\hat{i}}$
• Unit vector $\small{\vec{OB}}$ along the OY axis is denoted as $\small{\hat{j}}$
• Unit vector $\small{\vec{OC}}$ along the OZ axis is denoted as $\small{\hat{k}}$
Now we will see the components of a position vector. It can be written in 9 steps:
1. In fig.26.19 below, P(x,y,z) is a point in space.
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| Fig.26.19 |
• From P, a perpendicular is dropped onto the XOY plane. P1 is the foot of this perpendicular.
• From P1, a perpendicular is dropped on to the OX axis. Q is the foot of this perpendicular.
• From P1, a perpendicular is dropped on to the OY axis. S is the foot of this perpendicular.
• From P, a perpendicular is dropped on to the OZ axis. R is the foot of this perpendicular.
2. Consider $\small{\vec{OP}}$, the position vector of P
This position vector is the resultant of $\small{\vec{OQ}, \vec{QP_1}~\text{and}~\vec{P_1 P}}$, takern in order.
That is., $\small{\vec{OP}~=~\vec{OQ} + \vec{QP_1} + \vec{P_1 P}}$
• Let us find suitable substitutes for each of the three vectors on the R.H.S of this equation.
3. $\small{\vec{OQ}}$ has a magnitude of x because, Q is the foot of the perpendicular from P1.
• This vector lies along the OX axis. The unit vector corresponding to this axis is $\small{\hat{i}}$
• Both $\small{\hat{i}~\text{and}~\vec{OQ}}$ has the origin at O. So instead of $\small{\vec{OQ}}$, we can write: $\small{x\,\hat{i}}$
4. So the result in (2) becomes:
$\small{\vec{OP}~=~x\,\hat{i} + \vec{QP_1} + \vec{P_1 P}}$
5. $\small{\vec{QP_1}}$ has a magnitude of y because, P1 is the foot of the perpendicular from P.
• $\small{\vec{QP_1} = \vec{OS}}$ because, S is the foot of the perpendicular from P1.
• $\small{\vec{OS}}$ lies along the OY axis. The unit vector corresponding to this axis is $\small{\hat{j}}$
• Both $\small{\hat{j}~\text{and}~\vec{OS}}$ has the origin at O. So instead of $\small{\vec{QP_1}}$, we can write: $\small{y\,\hat{j}}$
6. So the result in (4) becomes:
$\small{\vec{OP}~=~x\,\hat{i} + y\,\hat{j} + \vec{P_1 P}}$
7. $\small{\vec{P_1 P}}$ has a magnitude of z because, P1 is the foot of the perpendicular from P.
• $\small{\vec{P_1 P} = \vec{OR}}$ because, R is the foot of the perpendicular from P.
• $\small{\vec{OR}}$ lies along the OZ axis. The unit vector corresponding to this axis is $\small{\hat{k}}$
• Both
$\small{\hat{k}~\text{and}~\vec{OR}}$ has the origin at O. So instead
of $\small{\vec{P_1 P}}$, we can write: $\small{z\,\hat{k}}$
8. So the result in (6) becomes the final result:
$\small{\vec{OP}~=~x\,\hat{i} + y\,\hat{j} + z\,\hat{k}}$
9. The result in (8) is called the component form of $\small{\vec{OP}}$
• x, y and z are called the scalar components of $\small{\vec{OP}}$
• They are also known as the rectangular components of $\small{\vec{OP}}$
• $\small{x\hat{i}, y\hat{j}~\text{and}~z\hat{k}}$ are called the vector components of $\small{\vec{OP}}$
If we are given a vector in the component form, we will be able to find the magnitude of that vector. The method can be explained in 3 steps:
1. In fig.26.19 above, points O, Q and P1 are vertices of a right triangle. Side OP1 is the hypotenuse. So by applying Pythagoras theorem, we get:
$\small{\left|\vec{OP_1} \right|^2 ~=~ \left|\vec{OQ} \right|^2 + \left|\vec{QP_1} \right|^2 ~=~ x^2 + y^2}$
2. Similarly, points O, P1 and P are vertices of a right triangle. Side OP is the hypotenuse. So by applying Pythagoras theorem, we get:
$\small{\left|\vec{OP} \right|^2 ~=~ \left|\vec{OP_1} \right|^2 + \left|\vec{P_1 P} \right|^2}$
3. Based on the result in (1), the result in (2) becomes:
$\small{\left|\vec{OP} \right|^2 ~=~ x^2 + y^2 + \left|\vec{P_1 P} \right|^2}$
$\small{\Rightarrow \left|\vec{OP} \right|^2 ~=~ x^2 + y^2 + z^2}$
$\small{\Rightarrow \left|\vec{OP} \right| ~=~\left|x\,\hat{i} + y\,\hat{j} + z\,\hat{k} \right|~=~ \sqrt{x^2 + y^2 + z^2}}$
Now we will see seven useful formulas that can be applied when two vectors $\small{\vec{a}~\text{and}~\vec{b}}$ are given in component form:
• $\small{\vec{a}~=~a_1\hat{i} + a_2\hat{j} + a_3\hat{k}}$
• $\small{\vec{b}~=~b_1\hat{i} + b_2\hat{j} + b_3\hat{k}}$
1. $\small{\vec{a}+\vec{b}~=~\left( a_1 + b_1 \right)\hat{i} + \left( a_2 + b_2 \right)\hat{j} + \left( a_3 + b_3 \right)\hat{k}}$
2. $\small{\vec{a}-\vec{b}~=~\left( a_1 - b_1 \right)\hat{i} + \left( a_2 - b_2 \right)\hat{j} + \left( a_3 - b_3 \right)\hat{k}}$
3. $\small{\vec{a}~\text{and}~\vec{b}}$ are equal if and only if:
$\small{a_1 = b_1,~~a_2 = b_2,~~\text{and}~~a_3 = b_3}$
4. $\small{\lambda\vec{a}~=~\left(\lambda a_1 \right)\hat{i} + \left(\lambda a_2 \right)\hat{j} + \left(\lambda a_3 \right)\hat{k}}$
Here $\small{\lambda}$ is a scalar
5. $\small{k\vec{a} + m\vec{a}~=~(k+m)\vec{a}}$
Here $\small{k~\text{and}~m}$ are scalars
6. $\small{k\left(m\vec{a} \right)~=~(km)\vec{a}}$
7. $\small{k\left(\vec{a} + \vec{b} \right)~=~k\vec{a} + k\vec{b}}$
Now we can write the condition for two vectors to be collinear. It can be written in 4 steps
1. If we multiply $\small{\vec{a}}$ by a scalar $\small{\lambda}$, we get a new vector $\small{\lambda\vec{a}}$. In the new vector, only the magnitude has changed. Direction remains the same. So $\small{\vec{a}~\text{and}~\lambda\vec{a}}$ are collinear.
2. We can write the reverse also:
If $\small{\vec{a}~\text{and}~\vec{b}}$ are collinear, then there exists a scalar $\small{\lambda}$ such that, $\small{\vec{b} = \lambda\vec{a}}$
3. We can write this in component form:
If two vectors
$\small{\vec{a}~=~a_1\hat{i} + a_2\hat{j} + a_3\hat{k}}$
$\small{\vec{b}~=~b_1\hat{i} + b_2\hat{j} + b_3\hat{k}}$
are collinear then:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{b}} & {~=~} &{\lambda\vec{a}} \\
{~\color{magenta} 2 } &{\Rightarrow} &{b_1\hat{i} + b_2\hat{j} + b_3\hat{k}} & {~=~} &{\lambda\left(a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \right)} \\
{~\color{magenta} 3 } &{\Rightarrow} &{b_1\hat{i} + b_2\hat{j} + b_3\hat{k}} & {~=~} &{\left(\lambda a_1\right)\hat{i} + \left(\lambda a_2\right)\hat{j} + \left(\lambda a_3\right)\hat{k} } \\
{~\color{magenta} 4 } &{\Rightarrow} &{} & {} &{b_1 = \lambda a_1,~b_2 = \lambda a_2,~b_3= \lambda a_3} \\
{~\color{magenta} 5 } &{\Rightarrow} &{} & {} &{\frac{b_1}{a_1} = \frac{b_2}{a_2} = \frac{b_3}{a_3} =\lambda} \\
\end{array}}$
4. That is.,
If $\small{\vec{a}~\text{and}~\vec{b}}$ are collinear, then the all three ratios of the scalar components will be the same.
Next we will see the relation between unit vectors and direction cosines. It can be written in 6 steps:
1. We have seen that, $\small{\hat{a}}$ is the unit vector in the direction of $\small{\vec{a}}$, and can be obtained using the formula: $\small{\hat{a} = \frac{\vec{a}}{\left|\vec{a} \right|}}$
2. Let us denote the position vector $\small{\vec{OP}}$ as $\small{\vec{r}}$.
Then we can write the unit vector in the direction of the position vector:
$\small{\hat{r} = \frac{\vec{r}}{\left|\vec{r} \right|}}$
3. If the coordinates of P are (x,y,z), then:
$\small{\vec{OP} ~=~ \vec{r} ~=~ x\,\hat{i} + y\,\hat{j} + z\,\hat{k}}$
4. Substituting this in (2), we get:
$\small{\hat{r} = \frac{x\,\hat{i} + y\,\hat{j} + z\,\hat{k}}{\left|\vec{r} \right|}~=~\left(\frac{x}{\left|\vec{r} \right|} \right)\hat{i}+\left(\frac{y}{\left|\vec{r} \right|} \right)\hat{j}+\left(\frac{ z}{\left|\vec{r} \right|} \right)\hat{k}}$
5. When we learnt about direction cosines we got the following results:
$\small{x = \cos \alpha \left|\vec{r} \right|,~y = \cos \beta \left|\vec{r} \right|,~x = \cos \gamma \left|\vec{r} \right|}$
See 26.5 of the first section of this chapter
6. Substituting the result from (5) into the result in (4), we get:
$\small{\hat{r}~=~\left(\frac{\cos \alpha \left|\vec{r} \right|}{\left|\vec{r} \right|} \right)\hat{i}+\left(\frac{\cos \beta \left|\vec{r} \right|}{\left|\vec{r} \right|} \right)\hat{j}+\left(\frac{\cos \gamma \left|\vec{r} \right|}{\left|\vec{r} \right|} \right)\hat{k}}$
$\small{\Rightarrow\hat{r}~=~\left(\cos \alpha \right)\hat{i}+\left(\cos \beta \right)\hat{j}+\left(\cos \gamma \right)\hat{k}}$
• So, if we are given the direction cosines of a vector, then we can directly write the unit vector in the direction of that given vector.
Now we will see a solved example
Solved example 26.6
Find the values of x, y and z so that the vectors $\small{\vec{a} = x\hat{i} + 2\hat{j} + z\hat{k}}$ and $\small{\vec{b} = 2\hat{i} + y\hat{j} + \hat{k}}$ are equal
Solution:
1. If two vectors are equal, their corresponding components will be equal.
2. Let us equate the corresponding components:
• Equating the vector components along the x-axis, we get: $\small{x\hat{i} = 2\hat{i}}$. Therefore, x = 2
• Equating the vector components along the y-axis, we get: $\small{2\hat{j} = y\hat{i}}$. Therefore, y = 2
• Equating the vector components along the z-axis, we get: $\small{z\hat{k} = \hat{k}}$. Therefore, z = 1
In the next section, we will see a few more solved examples.
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