In the previous section, we saw addition of vectors. In this section, we will see multiplication.
Multiplication of a vector by a scalar
This can be explained in 3 steps:
1. Suppose that, we are given a vector $\small{\vec{a}}$.
• We can multiply $\small{\vec{a}}$ by a scalar $\small{\lambda}$
• The product obtained is denoted as $\small{\lambda\vec{a}}$
2. $\small{\lambda\vec{a}}$ is also a vector. It has the following four properties:
(i) $\small{\vec{a}~\text{and}~\lambda\vec{a}}$ are collinear
(ii) $\small{\vec{a}~\text{and}~\lambda\vec{a}}$ have the same direction if $\small{\lambda}$ is +ve
(iii) $\small{\vec{a}~\text{and}~\lambda\vec{a}}$ have opposite directions if $\small{\lambda}$ is −ve
(iv)
Magnitude of $\small{\lambda\vec{a}}$ is $\small{\left|\lambda
\right|}$ times the magnitude of $\small{\vec{a}}$. That is.,
$\small{\left|\lambda\vec{a} \right|~=~\left|\lambda \right|\,\left|\vec{a} \right|}$
3. Fig.26.17 below gives a geometrical explanation of the multiplication process.
 |
| Fig.26.17 |
•
In fig(i), we have the original $\small{\vec{a}}$. The single
perpendicular white line indicates that, the vector has a magnitude of
two units.
• In fig(ii), the vector has a magnitude of one unit. That
is., half the magnitude of the original $\small{\vec{a}}$. Direction of
this vector is same as that of $\small{\vec{a}}$
• In fig(iii), the
vector has a magnitude of four units. That is., two times the magnitude
of the original $\small{\vec{a}}$. Direction of this vector is same as
that of $\small{\vec{a}}$
• In fig(iv), the vector has a magnitude of
one unit. That is., half the magnitude of the original
$\small{\vec{a}}$. Direction of this vector is opposite of
$\small{\vec{a}}$. This is because, $\small{\vec{a}}$ is multiplied by a
−ve scalar.
• In fig(v), the vector has a magnitude of three unit.
That is., (3/2) times the magnitude of the original $\small{\vec{a}}$.
Direction of this vector is opposite of $\small{\vec{a}}$. This is
because, $\small{\vec{a}}$ is multiplied by a −ve scalar.
• The vectors in all five figs are parallel to each other
Now we can discuss about additive inverse. It can be explained in 6 steps:
1. Consider the two vectors $\small{\vec{a}~\text{and}~\lambda \vec{a}}$
2. Suppose that, $\small{\lambda = -1}$. Then $\small{\lambda \vec{a} = -\vec{a}}$
3. Let us compare $\small{\vec{a}~\text{and}~- \vec{a}}$:
♦ $\small{-\vec{a}}$ has the same magnitude as $\small{\vec{a}}$
♦ $\small{-\vec{a}}$ has the direction opposite to that of $\small{\vec{a}}$
4. We can add the two vectors. We will get:
$\small{\vec{a} + \left(-\vec{a} \right)~=~\left(-\vec{a} \right) + \vec{a}~=~\vec{0}}$
5. So $\small{-\vec{a}}$ is called the additive inverse of $\small{\vec{a}}$
6. $\small{-\vec{a}}$ is also called the negative of $\small{\vec{a}}$
Next we will discus about unit vectors. It can be explained in 6 steps:
1. Consider the two vectors $\small{\vec{a}~\text{and}~\lambda \vec{a}}$
2.
Suppose that, $\small{\lambda = \frac{1}{\left|\vec{a} \right|}}$. Then
$\small{\lambda \vec{a} = \frac{\vec{a}}{\left|\vec{a} \right|}}$
3. We know that, $\small{\vec{a}}$ can be written as the product of two items:
(i) $\small{\left|\vec{a} \right|}$
(ii) A vector of magnitude one unit and direction same as that of $\small{\vec{a}}$
4. So step (2) can be modified as:
If
$\small{\lambda = \frac{1}{\left|\vec{a} \right|}}$, then
$\small{\lambda \vec{a} = \frac{\vec{a}}{\left|\vec{a}
\right|}~=~\frac{\left|\vec{a} \right|[\text{A vector of magnitude one
unit and direction same as that of}\,\vec{a}]}{\left|\vec{a} \right|}}$
$\small{\Rightarrow\frac{\vec{a}}{\left|\vec{a}
\right|}~=~\text{A vector of magnitude one unit and direction same as
that of}\,\,\vec{a}}$
5. $\small{\text{A vector of magnitude one unit and direction same as that of}\,\,\vec{a}}$ is called:
$\bf{\text{A unit vector in the direction of}\,\,\vec{a}}$
• Such a vector is denoted as: $\bf{\hat{a}}$
6. So we can modify step (4) as:
$\small{\frac{\vec{a}}{\left|\vec{a} \right|}~=~\hat{a}}$
In the next section, we will see components of a vector.
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