Solution:

Magnitude of $$\overrightarrow{a}$$ = $$\sqrt{x^2 + y^2}$$

or, $$\overrightarrow{a}$$ = $$\sqrt{3^2 + 4^2}$$

or, $$\overrightarrow{a}$$ = $$\sqrt{9 + 16}$$

or, $$\overrightarrow{a}$$ = $$\sqrt{25}$$

$$\therefore$$ $$\overrightarrow{a}$$ = 5 units.

Solution:

Magnitude of $$\overrightarrow{b}$$ = $$\sqrt{x^2 + y^2}$$

or, $$\overrightarrow{b}$$ = $$\sqrt{(-3)^2 + 4^2}$$

or, $$\overrightarrow{b}$$ = $$\sqrt{9 + 16}$$

or, $$\overrightarrow{b}$$ = $$\sqrt{25}$$

$$\therefore$$ $$\overrightarrow{b}$$ = 5 units

Solution:

Magnitude of $$\overrightarrow{g}$$ = $$\sqrt{x^2 + y^2}$$

or, $$\overrightarrow{g}$$ = $$\sqrt{0^2 + 4^2}$$

or, $$\overrightarrow{g}$$ = $$\sqrt{0 + 16}$$

or, $$\overrightarrow{g}$$ = $$\sqrt{16}$$

$$\therefore$$ $$\overrightarrow{g}$$ = 4units.

Solution:

Magnitude of $$\overrightarrow{AB}$$ = $$\sqrt{x^2 + y^2}$$

or, $$\overrightarrow{AB}$$ = $$\sqrt{5^2 + 3^2}$$

or, $$\overrightarrow{AB}$$ = $$\sqrt{25 + 9}$$

$$\therefore$$$$\overrightarrow{AB}$$ = $$\sqrt{34}$$ units

Solution:

Magnitude of $$\overrightarrow{CD}$$ = $$\sqrt{x^2 + y^2}$$

or, $$\overrightarrow{CD}$$ = $$\sqrt{2^2 + (-3)^2}$$

or, $$\overrightarrow{CD}$$ = $$\sqrt{4 + 9}$$

$$\therefore$$ $$\overrightarrow{CD}$$ = $$\sqrt{13}$$ units

Solution:

Magnitude of $$\overrightarrow{MN}$$ = $$\sqrt{x^2 + y^2}$$

or, $$\overrightarrow{MN}$$ = $$\sqrt{(-4)^2 + (5)^2}$$

or, $$\overrightarrow{MN}$$ = $$\sqrt{16 + 25}$$

$$\therefore$$ $$\overrightarrow{MN}$$ = $$\sqrt{41}$$ units

Solution:

Magnitude of $$\overrightarrow{XY}$$ = $$\sqrt{x^2 + y^2}$$

or, $$\overrightarrow{XY}$$ = $$\sqrt{(-4)^2 + (-6)^2}$$

or, $$\overrightarrow{XY}$$ = $$\sqrt{16 + 36}$$

$$\therefore$$ $$\overrightarrow{XY}$$ = $$\sqrt{52}$$ units

Solution:

Here,
or, $$\begin{vmatrix} a \end{vmatrix}$$ = $$\sqrt{x^2 + y^2}$$
or, $$\begin{vmatrix} a \end{vmatrix}$$ = $$\sqrt{0^2 + 5^2}$$
or, $$\begin{vmatrix} a \end{vmatrix}$$ = $$\sqrt{0 + 25}$$
or, $$\begin{vmatrix} a \end{vmatrix}$$ = 5 units
Now,
Unit Vector of $$\begin{vmatrix} a \end{vmatrix}$$ = $$\frac{\overrightarrow{a}}{\begin{vmatrix} a \end{vmatrix}}$$
= $$\frac{1}{5}$$$$\begin{pmatrix} 0 \\ 5 \end{pmatrix}$$
= $$\begin{pmatrix} 0 \\ 1 \end{pmatrix}$$

Solution:

Here,
M(2, 5) = (x1, y1)
N(4, 8) = (x2, y2)
Now,
or, $$\overrightarrow{MN}$$ = $$\begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$$
or,$$\overrightarrow{MN}$$ = $$\begin{pmatrix} 4- 2\\ 8- 5\end{pmatrix}$$
$$\therefore$$ $$\overrightarrow{MN}$$ = $$\begin{pmatrix} 2\\ 3\end{pmatrix}$$

Solution:

Here,
$$\overrightarrow{a}$$ = $$\begin{pmatrix} 3 \\ 4 \end{pmatrix}$$ = $$\begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$$
$$\overrightarrow{b}$$ = $$\begin{pmatrix} -6 \\ 8 \end{pmatrix}$$ = $$\begin{pmatrix} x_2 \\ y_2 \end{pmatrix}$$
Now,
or, $$\overrightarrow{a}$$ + $$\overrightarrow{b}$$ = $$\begin{pmatrix} x_1 + x_2 \\ y_1 + y_2 \end{pmatrix}$$
or, $$\overrightarrow{a}$$ + $$\overrightarrow{b}$$ = $$\begin{pmatrix} 3 - 6 \\ 4 + 8 \end{pmatrix}$$
or, $$\overrightarrow{a}$$ + $$\overrightarrow{b}$$ = $$\begin{pmatrix} -3 \\ 12 \end{pmatrix}$$

$$\therefore$$ The sum of $$\overrightarrow{a}$$ + $$\overrightarrow{b}$$ is$$\begin{pmatrix} -3 \\ 12 \end{pmatrix}$$.

Solution:

Given,
For $$\overrightarrow{AB}$$
A(3, 4) = (x1, y1)
B(m, n) = (x2, y2)
Now,
or, $$\overrightarrow{AB}$$ = $$\begin{pmatrix} x_2 - x_1 \\ y_2 - y_2 \end{pmatrix}$$
or, $$\overrightarrow{AB}$$ = $$\begin{pmatrix} m- 3\\ n- 4\end{pmatrix}$$

For $$\overrightarrow{CD}$$
or, $$\overrightarrow{CD}$$ = $$\begin{pmatrix} x_2 - x_1 \\ y_2 - y_2 \end{pmatrix}$$
or, $$\overrightarrow{CD}$$ = $$\begin{pmatrix} 4- 2\\ 6- 3\end{pmatrix}$$
or, $$\overrightarrow{CD}$$ = $$\begin{pmatrix} 2\\ 3\end{pmatrix}$$

So,
or, $$\overrightarrow{AB}$$ = $$\overrightarrow{CD}$$
or, m - 3 = 2 or, n - 4 = 3
or, m = 2 + 3 or, n = 3 + 4
$$\therefore$$ m = 5 $$\therefore$$ n = 7

Solution:

$$\overrightarrow{a}$$ = $$\begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$$
$$\overrightarrow{b}$$ = $$\begin{pmatrix} x_2 \\ y_2 \end{pmatrix}$$
We know that,
or, $$\overrightarrow{a}$$ + $$\overrightarrow{b}$$ = $$\begin{pmatrix} x_1 + x_2 \\ y_1 + y_2 \end{pmatrix}$$
or, $$\overrightarrow{a}$$ + $$\overrightarrow{b}$$ = $$\begin{pmatrix} 2+ 4\\ 3 + 5\end{pmatrix}$$
or, $$\overrightarrow{a}$$ + $$\overrightarrow{b}$$ = $$\begin{pmatrix} 6\\ 8\end{pmatrix}$$
Now,
or, |$$\overrightarrow{a}$$ + $$\overrightarrow{b}$$| = $$\sqrt{x^2 + y^2}$$
or, |$$\overrightarrow{a}$$ + $$\overrightarrow{b}$$| = $$\sqrt{6^2 + 8^2}$$
or, |$$\overrightarrow{a}$$ + $$\overrightarrow{b}$$| = $$\sqrt{36 + 64}$$
or, |$$\overrightarrow{a}$$ + $$\overrightarrow{b}$$| = $$\sqrt{100}$$
or, |$$\overrightarrow{a}$$ + $$\overrightarrow{b}$$| = 10 units
Again,
or, unit vector = $$\frac{\overrightarrow{a} + \overrightarrow{b}}{|\overrightarrow{a} + \overrightarrow{b}|}$$
or, $$\widehat{v}$$ = $$\frac{\begin{pmatrix} 6\\ 8\end{pmatrix}}{10}$$
or,$$\widehat{v}$$ = $$\begin{pmatrix} \frac{6}{10} \\ \frac{8}{10} \end{pmatrix}$$
$$\therefore$$ $$\widehat{v}$$ = $$\begin{pmatrix} \frac{3}{5} \\ \frac{4}{5} \end{pmatrix}$$

Solution:

Here,
$$\overrightarrow{u}$$ = $$\begin{pmatrix} 8 \\ 10 \end{pmatrix}$$
$$\overrightarrow{v}$$ = $$\begin{pmatrix} 5 \\ 6 \end{pmatrix}$$
Now,
or, $$\overrightarrow{u}$$ - $$\overrightarrow{v}$$ = $$\begin{pmatrix} 8 \\ 10 \end{pmatrix}$$ -$$\begin{pmatrix} 5 \\ 6 \end{pmatrix}$$
or, $$\overrightarrow{u}$$ - $$\overrightarrow{v}$$ = $$\begin{pmatrix} 8 - 5 \\ 10 - 6 \end{pmatrix}$$
or, $$\overrightarrow{u}$$ - $$\overrightarrow{v}$$ = $$\begin{pmatrix} 3 \\ 4 \end{pmatrix}$$
Then,
Magnitude of $$\overrightarrow{u}$$ - $$\overrightarrow{v}$$ = $$\sqrt{x^2 + y^2}$$
= $$\sqrt{3^2 + 4^2}$$
= $$\sqrt{9 + 16}$$
= $$\sqrt{25}$$
= 5 units.

Solution:

$$\overrightarrow{p}$$ = $$\begin{pmatrix} 4 \\ 5 \end{pmatrix}$$
$$\overrightarrow{q}$$ = $$\begin{pmatrix} 4 \\ 1 \end{pmatrix}$$
Now,
or, $$\overrightarrow{p}$$ - $$\overrightarrow{q}$$ = $$\begin{pmatrix} 4 \\ 5 \end{pmatrix}$$ -$$\begin{pmatrix} 4 \\ 1 \end{pmatrix}$$
or, $$\overrightarrow{p}$$ - $$\overrightarrow{q}$$ = $$\begin{pmatrix} 4 - 4 \\ 5 - 4 \end{pmatrix}$$
or, $$\overrightarrow{p}$$ - $$\overrightarrow{q}$$ = $$\begin{pmatrix} 0 \\ 4 \end{pmatrix}$$
Again,
Magnitude of $$\overrightarrow{p}$$ - $$\overrightarrow{q}$$ = $$\sqrt{x^2 + y^2}$$
= $$\sqrt{0^2 + 4^2}$$
= $$\sqrt{4^2}$$
= 4 units
Then,
Unit vector of $$\overrightarrow{p}$$ - $$\overrightarrow{q}$$ = $$\frac{\overrightarrow{p} + \overrightarrow{q}}{|\overrightarrow{p} + \overrightarrow{q}|}$$
or, $$\widehat{v}$$ = $$\begin{pmatrix}\frac{6}{8} \\ \frac{1}{4} \end{pmatrix}$$
or, $$\widehat{v}$$ = ($$\frac{0}{4}$$ $$\times$$ $$\frac{4}{1}$$)
$$\therefore$$ $$\widehat{v}$$ = ($$\frac{0}{1}$$)

Solution:

Given,
P(a, b), Q(4, 5), R(2, 1), S(0, 1) and $$\overrightarrow{PQ}$$ = $$\overrightarrow{RS}$$

For $$\overrightarrow{PQ}$$
P(a, b) = (x1, y1)
Q(4, 5) = (x2, y2)
or, $$\overrightarrow{PQ}$$ = $$\begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$$
or, $$\overrightarrow{PQ}$$ = $$\begin{pmatrix} 4 - a \\ 5 - b\end{pmatrix}$$

For$$\overrightarrow{RS}$$
R(2, 1) = (x1, y1)
S(0, 1) = (x2, y2)
or, $$\overrightarrow{RS}$$ = $$\begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$$
or, $$\overrightarrow{RS}$$ = $$\begin{pmatrix} 0 - 2 \\ 1- 1\end{pmatrix}$$
or, $$\overrightarrow{RS}$$ = $$\begin{pmatrix} - 2\\ 0 \end{pmatrix}$$

Now,
or, $$\overrightarrow{PQ}$$ = $$\overrightarrow{RS}$$
or, $$\begin{pmatrix} 4 - a\\ 5 - b\end{pmatrix}$$ = $$\begin{pmatrix} - 2 \\ 0 \end{pmatrix}$$
or, 4 - a = -2 or, 5 - b = 0
or, 4 + 2 = a $$\therefore$$ b = 5
$$\therefore$$ a = 6