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