7.0 Subtraction of vectors

  Basic Vectors

Consider two vectors $\overrightarrow A $ and $\overrightarrow B $ making an angle of $\theta$ with each other.

Let $\overrightarrow R $ be the resultant vector and it is given as,
$$\overrightarrow R = \overrightarrow A - \overrightarrow B $$
The above equation can also be written as,
$$\overrightarrow R = \overrightarrow A + \left( { - \overrightarrow B } \right)$$

From parallelogram law of vector addition diagonal $OR$ is the resultant vector $\overrightarrow R $.

So, the magnitude of $\overrightarrow R $ is given as,
$$\left| {\overrightarrow R } \right| = \sqrt {{A^2} + {B^2} + 2AB\cos \left( {\pi - \theta } \right)} $$$$\left| {\overrightarrow R } \right| = \sqrt {{A^2} + {B^2} - 2AB\cos \theta } $$

$\theta:$ angle between $\overrightarrow A $ and $\overrightarrow B $
$\left| {\overrightarrow A } \right| = A:$ Magnitude of vector $\overrightarrow A $
$\left| {\overrightarrow B } \right| = B:$ Magnitude of vector $\overrightarrow B $
$\left| {\overrightarrow R } \right| = B:$ Magnitude of resultant vector $\overrightarrow R $

Note:

In addition of vectors,
$$\overrightarrow R = \overrightarrow A + \overrightarrow B $$$$\left| {\overrightarrow R } \right| = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } $$
When $\theta = 0^\circ $ (Maximum value of $\left| {\overrightarrow R } \right|$)
$$\left| {\overrightarrow R } \right| = \sqrt {{A^2} + {B^2} + 2AB} $$$$\left| {\overrightarrow R } \right| = \sqrt {{{\left( {A + B} \right)}^2}} $$$$\left| {\overrightarrow R } \right| = A + B$$

When $\theta = 90^\circ \,{\text{or}}\,\frac{\pi }{2}$,
$$\left| {\overrightarrow R } \right| = \sqrt {{A^2} + {B^2}} $$

When $\theta = 180^\circ \,{\text{or}}\,\pi $ (Minimum value of $\left| {\overrightarrow R } \right|$)
$$\left| {\overrightarrow R } \right| = \sqrt {{A^2} + {B^2} - 2AB} $$$$\left| {\overrightarrow R } \right| = \sqrt {{{\left( {A - B} \right)}^2}} $$$$\left| {\overrightarrow R } \right| = A - B$$

Question: Consider vector $\overrightarrow A $ and $\overrightarrow B $ which makes an angle $\theta$ with each other. In parallelogram law of vector addition, one of the diagonal gives $\overrightarrow A + \overrightarrow B $, then the other diagonal will be equal to?

Solution: For the parallelogram law of vector addition we can draw,

Given, $$\overrightarrow {OQ} = \overrightarrow A + \overrightarrow B $$
We have to find diagonal $\overrightarrow {PR} $.

Let $\overrightarrow {PR} = \overrightarrow C $.

In $\Delta PQR$ we can write,

From triangle law of addition of addition, we can write,
$$\overrightarrow C + \overrightarrow A = \overrightarrow B $$$$\overrightarrow C = \overrightarrow B - \overrightarrow A $$ or $$\overrightarrow {PR} = \overrightarrow B - \overrightarrow A $$